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Chapter 1 - Mathematics

1. 3y 32 3 4z 0 Now we are set up for the solution The third row shows that z 0 Substitute that back into the second row to get y 1 and then substitute back into the first row to get x 1 1 7 Example For the Physics problem from the start of this chapter Gauss method gives this 40h 15c 100 5 4p t02 40h 15c 100 50h 25c 50 175 4 c 175 So c 4 and back substitution gives that h 1 The Chemistry problem is solved later 1 8 Example The reduction y z 9 rct y z 9 2r 4y 32 1 ne 2y 5z 17 3a 6y 5z 0 tes 3y 8z 27 TF yt z 9 UE ya s T 1 2 2 3 2 shows that z 3 y 1 and x 7 As these examples illustrate Gauss method uses the elementary reduction operations to set up back substitution 1 9 Definition In each row the first variable with a nonzero coefficient is the row s leading variable A system is in echelon form if each leading variable is to the right of the leading variable in the row above it except for the leading variable in the first row 1 10 Example The only operation needed in the examples above is pivoting Here is a linear system that requires the operation of swapping equations After the first pivot r y 0 TY 0 22 2y z 2w 4 2p1 p2 z 2w 4 y w 0 y w 0 2z w 5 2z w 5 6 Chapter One Linear Systems the second equation has no leading y To get one we look lower d
2. 2y z 2w 0 3y z 0 w 0 We next perform back substitution to express each variable in terms of the free variable z Working from the bottom up we get first that w is 0 z next that y is 1 3 z and then substituting those two into the top equation x 2 1 3 z z 2 0 0 gives x 1 3 z So back substitution gives a paramatrization of the solution set by starting at the bottom equation and using the free variables as the parameters to work row by row to the top The proof below follows this pattern Comment That is this proof just does a verification of the bookkeeping in back substitution to show that we haven t overlooked any obscure cases where this procedure fails say by leading to a division by zero So this argument while quite detailed doesn t give us any new insights Nevertheless we have written it out for two reasons The first reason is that we need the result the computational procedure that we employ must be verified to work as promised The second reason is that the row by row nature of back substitution leads to a proof that uses the technique of mathematical induction This is an important and non obvious proof technique that we shall use a number of times in this book Doing an induction argument here gives us a chance to see one in a setting where the proof material is easy to follow and so the technique can be studied Readers who are unfamiliar with induction arguments should be sure t
3. Topit Cramers Rue m Topic Speed of Calculating Determinants Topic Projective Geometry Chapter Five Similarity I Complex Vector Spaces 2 2 2 2 1 Factoring and Complex Numbers A Review 2 Complex Representations ati 20 de wee ye Ge ee 159 159 159 168 176 176 183 195 195 205 212 212 214 222 231 238 238 242 250 250 254 260 269 274 281 287 293 294 294 299 303 312 319 319 326 326 331 334 337 1 Delnition and Exemples oscar a auae i 353 2 Diagonalizability 222229 234399 edo 355 3 Eigenvalues and Eigenvectors 359 DE Nilpotent 1l k ae a da a a a ee wes 367 1 SelfCompositi n occ scoasa meseta eds m RR 367 DORR A ae X mE A Se NN eS 370 IV Jordan Form a ceo soo bo Rm em we eas 381 1 Polynomials of Maps and Matrices sss 381 2 Jordan Canonical Form 2 22 22 9S 388 Topie Method of Powers a sus an nenn 401 Topic Stable Populations 422222 oder 405 Topie Linear Recurrences vu 292m ss en 407 Appendix A 1 Propositions ee RA ee A 1 rr asus rn aa a een Bees wea A 3 Techniques of Proof 2 404 4444620038 aaa A 5 Sets Functions and Relations 0 A 7 Note starred subsections are optional Chapter Oue Linear Systems I Solving Linear Systems Systems of linear equations are common in science and mathematics These two examples from high school science Onan give a sense
4. 0c4 along with c 0 and c9 0 gives an impossibility The following result shows that this effect always holds It shows that what Gauss linear elimination method eliminates is linear relationships among the rows 2 5 Lemma In an echelon form matrix no nonzero row is a linear combination of the other rows Proor Let R be in echelon form Suppose to obtain a contradiction that some nonzero row is a linear combination of the others pi C1P1 ct ci ipi i Ci41 Pita CmPm We will first use induction to show that the coefficients c1 c _ associated with rows above p are all zero The contradiction will come from consideration of p and the rows below it The base step of the induction argument is to show that the first coefficient cj is zero Let the first row s leading entry be in column number be the column number of the leading entry of the first row and consider the equation of entries in that column Pit C1P1 Ci 1Pi 1 0 Ci41Pi41 1 CmPm L The matrix is in echelon form so the entries 2 0 pmo including p are all zero 0 cipi 6i 1 0 Cr O c 0 Because the entry p e is nonzero as it leads its row the coefficient c must be Zero The inductive step is to show that for each row index k between 1 and 2 if the coefficient c and the coefficients co cj are all zero then cxy1 is also zero That argument and the contradiction that f
5. Ay 14 These pictures don t prove the results from the prior section which apply to any number of linear equations and any number of unknowns but nonetheless they do help us to understand those results This section develops the ideas that we need to express our results from the prior section and from some future sections geometrically In particular while the two dimensional case is familiar to extend to systems with more than two unknowns we shall need some higher dimensional geometry II 1 Vectors in Space Higher dimensional geometry sounds exotic It is exotic interesting and eye opening But it isn t distant or unreachable We begin by defining one dimensional space to be the set R To see that definition is reasonable draw a one dimensional space and make the usual correspondence with R pick a point to label 0 and another to label 1 0 1 Now with a scale and a direction finding the point corresponding to say 4 2 17 is easy start at 0 and head in the direction of 1 i e the positive direction but don t stop there go 2 17 times as far Section II Linear Geometry of n Space 33 The basic idea here combining magnitude with direction is the key to ex tending to higher dimensions An object comprised of a magnitude and a direction is a vector we will use the same word as in the pre
6. d z y 1 e Ay z 20 f 2a z u 5 3x 3y 2 272 2y z 0 y w 1 x 2 5 3x z w 0 a y z 10 di y 224w 9 3 Use the computer to solve these systems from the second subsection a 3a 6y 18 b t y 1 c m o X4 4 x 2 y 6 z y l z 2 2x 3 5 Ari T2 523 17 d 2a b c 2 e r 2y z 3 f x z w 4 2a c 3 2x y w 4 22 y w 2 a b 0 y 2z4 uw 1 3e y 2z 7 4 What does the computer give for the solution of the general 2x2 system ac cy p bx dy q 64 Chapter One Linear Systems Topic Input Output Analysis An economy is an immensely complicated network of interdependences Changes in one part can ripple out to affect other parts Economists have struggled to be able to describe and to make predictions about such a complicated object Mathematical models using systems of linear equations have emerged as a key tool One is Input Output Analysis pioneered by W Leontief who won the 1973 Nobel Prize in Economics Consider an economy with many parts two of which are the steel industry and the auto industry As they work to meet the demand for their product from other parts of the economy that is from users external to the steel and auto sectors these two interact tightly For instance should the external demand for autos go up that would lead to an increase in the auto industry s usage of steel Or should the external demand for steel fall then it would lead to a fall in steel s purchase of
7. Notice that some streets are one way only Hint this will not yield a unique solution since traffic can flow through this network in various ways you should get at least one free variable b Suppose that some construction is proposed for Winooski Avenue East be tween Willow and Jay so traffic on that block will be reduced What is the least amount of traffic flow that can be allowed on that block without disrupting the hourly flow into and out of the network
8. Then y 1 4 z To express x in terms of z substitute for y into the first equation to get x 1 1 2 z The solution set is 1 1 2 z 1 4 z z 1 z R We finish this subsection by developing the notation for linear systems and their solution sets that we shall use in the rest of this book 2 6 Definition An mxn matrix is a rectangular array of numbers with m rows and n columns Each number in the matrix is an entry 14 Chapter One Linear Systems Matrices are usually named by upper case roman letters e g A Bach entry is denoted by the corresponding lower case letter e g a is the number in row 2 and column j of the array For instance 1 22 5 x E 4 has two rows and three columns and so is a 2x3 matrix Read that two by three the number of rows is always stated first The entry in the second row and first column is aa 3 Note that the order of the subscripts matters a1 2 Z a2 1 since a a 2 2 The parentheses around the array are a typo graphic device so that when two matrices are side by side we can tell where one ends and the other starts 2 7 Example We can abbreviate this linear system 11 22 4 t2 Ta 0 T 223 4 with this matrix 1 2 0 4 0 1 1 0 10 2 4 The vertical bar just reminds a reader of the difference between the coefficients on the systems s left hand side and the constants on the right When a bar is used to divide a matrix into p
9. U101 UVa UnUn Note that the dot product of two vectors is a real number not a vector and that the dot product of a vector from IR with a vector from R is defined only when n equals m Note also this relationship between dot product and length dotting a vector with itself gives its length squared uiu Unun 2 4 Remark The wording in that definition allows one or both of the two to be a row vector instead of a column vector Some books require that the first vector be a row vector and that the second vector be a column vector We shall not be that strict Still reasoning with letters but guided by the pictures we use the next theorem to argue that the triangle formed by y and v in R lies in the planar subset of R generated by 4 and v 2 5 Theorem Triangle Inequality For any u v R e e lt lu 11 ple with equality if and only if one of the vectors is a nonnegative scalar multiple of the other one This inequality is the source of the familiar saying The shortest distance between two points is in a straight line finish start o Proor We ll use some algebraic properties of dot product that we have not yet checked for instance that t 6 d 4 b and that y v u See Exercise 17 The desired inequality holds if and only if its square holds e e lt el mg i r v Qi v al 2 a le Ie U T T T US d
10. pi gt pj Show that in A mat 1 757 A this condition is not needed b Write down a 2x2 matrix with nonzero entries and show that the 1 p1 1 operation is not reversed by 1 p p1 c Expand the proof of that lemma to make explicit exactly where the i Z j condition on pivoting is used III 2 Row Equivalence We will close this section and this chapter by proving that every matrix is row equivalent to one and only one reduced echelon form matrix The ideas that appear here will reappear and be further developed in the next chapter The underlying theme here is that one way to understand a mathematical situation is by being able to classify the cases that can happen We have met this theme several times already We have classified solution sets of linear systems into the no elements one element and infinitely many elements cases We have also classified linear systems with the same number of equations as unknowns into the nonsingular and singular cases We adopted these classifications because they give us a way to understand the situations that we were investigating Here where we are investigating row equivalence we know that the set of all matrices breaks into the row equivalence classes When we finish the proof here we will have a way to understand each of those classes its matrices can be thought of as derived by row operations from the unique reduced echelon form matrix in that class To understand how row operation
11. s resistance such that the potential is equal to the flow times the resistance The units of measurement are potential is described in volts the rate of flow is in amperes and resistance to the flow is in ohms These units are defined so that volts amperes ohms Components with this property that the voltage amperage response curve is a line through the origin are called resistors Light bulbs such as the ones shown above are not this kind of component because their ohmage changes as they heat up For example if a resistor measures 2 ohms then wiring it to a 12 volt battery results in a flow of 6 amperes Conversely if we have flow of electrical current of 2 amperes through it then there must be a 4 volt potential Topic Analyzing Networks 73 difference between it s ends This is the voltage drop across the resistor One way to think of a electrical circuits like the one above is that the battery provides a voltage rise while the other components are voltage drops The two facts that we need about networks are Kirchhoff s Laws Current Law For any point in a network the flow in equals the flow out Voltage Law Around any circuit the total drop equals the total rise In the above network there is only one voltage rise at the battery but some networks have more than one For a start we can consider the network below It has a battery that provides the potential to flow and three resistors resistors are drawn as zig za
12. the right node and the bottom node gives these lo 11 i2 i is 0 i2 15 la 13 14 io Kirchhoff s Voltage Law applied to the inside loop the io to i to 3 to io loop the outside loop and the upper loop not involving the battery gives these 541 10i3 10 Mo 4i4 10 541 5015 29 0 Those suffice to determine the solution ig 7 3 i 2 3 ia 5 3 ia 2 3 la 5 3 and i5 0 Networks of other kinds not just electrical ones can also be analyzed in this way For instance networks of streets are given in the exercises Exercises Many of the systems for these problems are mostly easily solved on a computer 1 Calculate the amperages in each part of each network a This is a simple network 3 ohm 9 volt 2 ohm 2 ohm b Compare this one with the parallel case discussed above 3 ohm 9 volt 2 ohm 2 ohm 2 ohm c This is a reasonably complicated network 3 ohm 3 ohm 9 volt 3 ohm 2 ohm 4 ohm 2 ohm 2 ohm 76 Chapter One Linear Systems 2 In the first network that we analyzed with the three resistors in series we just added to get that they acted together like a single resistor of 10 ohms We can do a similar thing for parallel circuits In the second circuit analyzed 20 volt 12 ohm 8 ohm the electric current through the battery is 25 6 amperes Thus the parallel portion is equivalent to a
13. 1 2 to 1 1 38 Chapter One Linear Systems b the vector from 2 1 1 to 3 0 4 and the vector from 5 1 4 to 6 0 7 v 1 3 Does 1 0 2 1 lie on the line through 2 1 1 0 and 5 10 1 4 v 1 4 a Describe the plane through 1 1 5 1 2 2 2 0 and 3 1 0 4 b Is the origin in that plane 1 5 Describe the plane that contains this point and line Qr v 1 6 Intersect these planes 1 0 1 0 2 e 1 orem e 3 4 0 mem 1 3 0 0 4 v 1 7 Intersect each pair if possible 1 0 1 0 a Hem n seR 2 1 2 2 2 1 0 0 b TORIES rem e D o fa s w R 1 i 2 1 1 8 When a plane does not pass through the origin performing operations on vec tors whose bodies lie in it is more complicated than when the plane passes through the origin Consider the picture in this subsection of the plane 2 0 5 0 5 Jo 1 Ju 0 z y 2 R 0 0 1 and the three vectors it shows with endpoints 2 0 0 1 5 1 0 and 1 5 0 1 a Redraw the picture including the vector in the plane that is twice as long as the one with endpoint 1 5 1 0 The endpoint of your vector is not 3 2 0 what is it b Redraw the picture including the parallelogram in the plane that shows the sum of the vectors ending at 1 5 0 1 and 1 5 1 0 The endpoint of the sum on the diagonal is not 3 1 1 what is it 1 9 Show that the line segments a1 a2 b1 52 and c1 c2 di d2 have the same lengths
14. 1 and w 0 The four tuple 1 0 5 4 is not a solution since its first coordinate does not equal its second We refer to a variable used to describe a family of solutions as a parameter and we say that the set above is paramatrized with y and w The terms parameter and free variable do not mean the same thing Above y and w are free because in the echelon form system they do not lead any row They are parameters because they are used in the solution set description We could have instead paramatrized with y and z by rewriting the second equation as w 2 3 1 3 z In that case the free variables are still y and w but the parameters are y and z Notice that we could not have paramatrized with x and y so there is sometimes a restriction on the choice of parameters The terms parameter and free are related because as we shall show later in this chapter the solution set of a system can always be paramatrized with the free variables Consequenlty we shall paramatrize all of our descriptions in this way 2 5 Example This is another system with infinitely many solutions a 2y 1 a 2y 1 2x 2 2 mme Ay 2 0 3ct 2y 2 u 4 m 4y z w e 2y par 4y 2 0 w 1 The leading variables are x y and w The variable z is free Notice here that although there are infinitely many solutions the value of one of the variables is fixed w 1 Write w in terms of z with w 1 0z
15. 14 1 5 1411 and divide by m t4 1 5 141 60 end with x m t 1 variables Because we have shown both the base step and the inductive step by the principle of mathematical induction the proposition is true QED We say that the set Lei peert Ch Bx C1 Ck ER is generated by or spanned by the set of vectors t8 A There is a tricky point to this definition If a homogeneous system has a unique solution the zero vector then we say the solution set is generated by the empty set of vectors This fits with the pattern of the other solution sets in the proof above the solution set is derived by taking the c s to be the free variables and if there is a unique solution then there are no free variables This proof incidentally shows as discussed after Example 2 4 that solution sets can always be paramatrized using the free variables The next lemma finishes the proof of Theorem 3 1 by considering the par ticular solution part of the solution set s description 3 8 Lemma For a linear system where pis any particular solution the solution set equals this set p h h satisfies the associated homogeneous system Section I Solving Linear Systems 25 Proor We will show mutual set inclusion that any solution to the system is in the above set and that anything in the set is a solution to the system For set inclusion the first way that if a vector solves the system then it is in the set described above assume tha
16. 2 0 0 not all of the variables are leading variables The Gauss method theorem showed that a triple satisfies the first system if and only if it satisfies the third Thus the solution set z y z 2x z 3 and z y z 1l and 3z y 4 12 Chapter One Linear Systems can also be described as x y z 2x z 3 and y 32 2 1 2 How ever this second description is not much of an improvement It has two equa tions instead of three but it still involves some hard to understand interaction among the variables To get a description that is free of any such interaction we take the vari able that does not lead any equation z and use it to describe the variables that do lead x and y The second equation gives y 1 2 3 2 z and the first equation gives x 3 2 1 2 z Thus the solution set can be de scribed as x y z 3 2 1 2 z 1 2 3 2 z 2 z R For instance 1 2 5 2 2 is a solution because taking z 2 gives a first component of 1 2 and a second component of 5 2 The advantage of this description over the ones above is that the only variable appearing z is unrestricted it can be any real number 2 2 Definition The non leading variables in an echelon form linear system are free variables In the echelon form system derived in the above example x and y are leading variables and z is free 2 3 Example A linear system can end with more t
17. 323 9 F T T 6x2 9 add 1 times row 1 to row 2 29 203 T 313 9 The third step is the only nontrivial one We ve mentally multiplied both sides of the first row by 1 mentally added that to the old second row and written the result in as the new second row Now we can find the value of each variable The bottom equation shows that 13 3 Substituting 3 for x3 in the middle equation shows that r9 1 Substituting those two into the top equation gives that x 3 and so the system has a unique solution the solution set is 3 1 3 Most of this subsection and the next one consists of examples of solving linear systems by Gauss method We will use it throughout this book It is fast and easy But before we get to those examples we will first show that this method is also safe in that it never loses solutions or picks up extraneous solutions 1 4 Theorem Gauss method If a linear system is changed to another by one of these operations 1 an equation is swapped with another 2 an equation has both sides multiplied by a nonzero constant 3 an equation is replaced by the sum of itself and a multiple of another then the two systems have the same set of solutions Each of those three operations has a restriction Multiplying a row by 0 is not allowed because obviously that can change the solution set of the system Similarly adding a multiple of a row to itself is not allowed because ad
18. Chapter One Linear Systems We often draw the arrow as starting at the origin and we then say it is in the canonical position or natural position When the vector bi 01 b2 a2 is in its canonical position then it extends to the endpoint b a1 b2 az We typically just refer to the point 1 2 rather than the endpoint of the canonical position of that vector Thus we will call both of these sets R z1 22 21 12 R de 21 29 ER In the prior section we defined vectors and vector operations with an alge braic motivation Tr x v2 T2 v2 wa v2 T W2 we can now interpret those operations geometrically For instance if Y repre sents a displacement then 37 represents a displacement in the same direction but three times as far and 1v represents a displacement of the same distance as V but in the opposite direction 30 OO x U And where Y and w represent displacements Y wW represents those displace ments combined VaL ni RE eas u The long arrow is the combined displacement in this sense if in one minute a ship s motion gives it the displacement relative to the earth of Y and a passen ger s motion gives a displacement relative to the ship s deck of w then v 1 is the displacement of the passenger relative to the earth Another way to understand the vector sum is with the parallelogram rule Draw the parallelogram formed by the vec
19. The three operations from Theorem 1 4 are the elementary reduction operations or row operations or Gaussian operations They are swapping multiplying by a scalar or rescaling and pivoting When writing out the calculations we will abbreviate row 7 by p For instance we will denote a pivot operation by kp p with the row that is changed written second We will also to save writing often list pivot steps together when they use the same p 1 6 Example A typical use of Gauss method is to solve this system c Y 0 22 y 32 3 e 2y z 3 The first transformation of the system involves using the first row to eliminate the x in the second row and the x in the third To get rid of the second row s 2r we multiply the entire first row by 2 add that to the second row and write the result in as the new second row To get rid of the third row s x we multiply the first row by 1 add that to the third row and write the result in as the new third row FL y 0 A 3y 32 3 pitps 3y 2 3 Section I Solving Linear Systems 5 Note that the two p steps 2p1 pa and p1 p3 are written as one opera tion In this second system the last two equations involve only two unknowns To finish we transform the second system into a third system where the last equation involves only one unknown This transformation uses the second row to eliminate y from the third row EN rct y 0
20. Topic Dimensional Analysis 152 vil Chapter Three Maps Between Spaces I Isomorphisms 2 2 22 2 nn 1 Definition and Examples 2 Dimension Characterizes Isomorphism IL Hetomarphisms oo cs 5 20664 nz 1 Definition 224 2 202606 be ho ba eS 2 Rangespace and Nullspace III Computing Linear Maps 1 Representing Linear Maps with Matrices 2 Any Matrix Represents a Linear Map IV Matris Operations cocos 1 Sums and Scalar Products 2 Matrix Multiplication 3 Mechanics of Matrix Multiplication 4 Voverses 22 09 909 x X Eoo mom ox Y Chousem Bas san x eek AAA 1 Changing Representations of Vectors 2 Changing Map Representations XI Projectiom 2 222 9000x909 m x Mo m 1 Orthogonal Projection Into a Line 2 Gram Schmidt Orthogonalization 3 Projection Into a Subspace Topie Line of Best Fit cs soa ee 22s Topic Geometry of Linear Maps Topic Markov Chains Topic Orthonormal Matrices Chapter Four Determinants L Debnition ses easa nee he eee ee l Exploration o oc cacet o rennen 2 Properties of Determinants 3 The Permutation Expansion 4 Determinants Exist II Geometry of Determinants 1 Determinants as Size Functions IM Other Formulas 2222 m m omm omn 1 Laplace s Expansion
21. a reader might wonder if some pairs of vectors satisfy the inequality in this way while y is a large number with absolute value bigger than the right hand side it is a negative large number The next result says that no such pair of vectors exists 2 6 Corollary Cauchy Schwartz Inequality For any d U ER av lt un lel with equality if and only if one vector is a scalar multiple of the other Pnoor The Triangle Inequality s proof shows that t y lt t 17 so if d vis positive or zero then we are done If w V is negative then this holds dv u v v lt l lel Ille l The equality condition is Exercise 18 QED 42 Chapter One Linear Systems The Cauchy Schwartz inequality assures us that the next definition makes sense because the fraction has absolute value less than or equal to one 2 7 Definition The angle between two nonzero vectors ti U IR is Uv 6 arccos arecos a the angle between the zero vector and any other vector is defined to be a right angle Thus vectors from R are orthogonal if and only if their dot product is zero 2 8 Example These vectors are orthogonal Z 1 f poe The arrows are shown away from canonical position but nevertheless the vectors are orthogonal 2 9 Example The R angle formula given at the start of this subsection is a special case of the definition Between these two the angle is 1 0
22. answer these questions The answer to each is yes The first question is answered in the last subsection of this section In the second section we give a geometric description of solution sets In the final section of this chapter we tackle the last set of questions Consequently by the end of the first chapter we will not only have a solid grounding in the practice of Gauss method we will also have a solid grounding in the theory We will be sure of what can and cannot happen in a reduction Exercises v 2 15 Find the indicated entry of the matrix if it is defined 1 3 1 i i a a2 1 b 01 2 c 02 2 d a3 1 Y 2 16 Give the size of each matrix 1 1 oki o 06 Y 2 17 Do the indicated vector operation if it is defined N 0 0 a aa v 2 18 Solve each system using matrix notation Express the solution using vec tors a 3a 6y 18 b a y 1 c m 23 4 x 2 y 6 z y l z T2 2 z3 5 4r 22 523 1T d 2a b c 2 e z 2y z 3 f x z w 4 2a c 3 2x y w 4 2x y w 2 a b 0 y z w l sr y z 7 v 2 19 Solve each system using matrix notation Give each solution set in vector notation a 22 y z 1 b x 2 1 c z y z 0 4x y 3 y 22 w 3 y w 0 xz 2y 32 w 7 3r 2y 3z w 0 y w 0 d a 2b 3c d e 1 3a b c d e 3 v 2 20 The vector is in the set What value of the parameters produces that vec tor a E 2 k ER i 2 3 b OLOCIOUCGS 1 0 1
23. autos The type of Input Output model we will consider takes in the external demands and then predicts how the two interact to meet those demands We start with a listing of production and consumption statistics These numbers giving dollar values in millions are excerpted from Leontief 1965 describing the 1958 U S economy Today s statistics would be quite different both because of inflation and because of technical changes in the industries used by used by used by steel auto others total une 5395 2664 25448 steel value af 48 9030 30346 auto For instance the dollar value of steel used by the auto industry in this year is 2 664 million Note that industries may consume some of their own output We can fill in the blanks for the external demand This year s value of the steel used by others this year is 17 389 and this year s value of the auto used by others is 21 268 With that we have a complete description of the external demands and of how auto and steel interact this year to meet them Now imagine that the external demand for steel has recently been going up by 200 per year and so we estimate that next year it will be 17 589 Imagine also that for similar reasons we estimate that next year s external demand for autos will be down 25 to 21 243 We wish to predict next year s total outputs That prediction isn t as simple as adding 200 to this year s steel total and subtracting 25 from this year s auto total For
24. be trusted Mathematicians have made a careful study of how to get the most reliable results A basic improvement on the naive code above is to not simply take the entry in the pivot_row pivot_row position for the pivot but rather to look at all of the entries in the pivot_row column below the pivot_row row and take the one that is most likely to give reliable results e g take one that is not too small This strategy is partial pivoting For example to solve the troublesome system above we start by looking at both equations for a best first pivot and taking the 1 in the second equation as more likely to give good results Then the pivot step of 001p2 p gives a first equation of 1 001y 1 which the computer will represent as 1 0 x 10 y 1 0 x 10 leading to the conclusion that y 1 and after back substitution x 1 both of which are close to right The code from above can be adapted to this purpose for pivot_row 1 pivot_row lt n 1 pivot_rowt find the largest pivot in this column in row max max pivot_row for row_below pivot_row 1 pivot_row lt n row_below 4 if abs a row below pivot row gt abs almax row_below max row_below swap rows to move that pivot entry up for col pivot_row col lt n col temp a pivot row col a pivot row col a max col a max col temp proceed as before for row_below pivot_row t1 row_below lt n row_below multiplier a row_below p
25. eight digits is easy to invent 1 0 x 1072 z 1 0 x 10 y 1 0 x 10 1 0 x 10 a 1 0 x 10 y 0 0 x 10 The computer s row reduction step 1000p1 pa produces a second equation 1001y 999 which the computer rounds to two places as 1 0 x 10 y 1 0 x 10 Then the computer decides from the second equation that y 1 and from the first equation that x 0 This y value is fairly good but the x is quite bad Thus another cause of unreliable output is a mixture of floating point arithmetic and a reliance on pivots that are small An experienced programmer may respond that we should go to double pre cision where sixteen significant digits are retained This will indeed solve many problems However there are some difficulties with it as a general approach For one thing double precision takes longer than single precision on a 486 70 Chapter One Linear Systems chip multiplication takes eleven ticks in single precision but fourteen in dou ble precision Programmer s Ref and has twice the memory requirements So attempting to do all calculations in double precision is just not practical And besides the above systems can obviously be tweaked to give the same trouble in the seventeenth digit so double precision won t fix all problems What we need is a strategy to minimize the numerical trouble arising from solving systems on a computer and some guidance as to how far the reported solutions can
26. example is that from this matrix 2 2 a Gauss method could derive any of these echelon form matrices Ca A 63 The first results from 2p1 pa The second comes from following 1 2 p with 4p1 pa The third comes from 2p1 pa followed by 2p2 p after the first pivot the matrix is already in echelon form so the second one is extra work but it is nonetheless a legal row operation The fact that the echelon form outcome of Gauss method is not unique leaves us with some questions Will any two echelon form versions of a system have the same number of free variables Will they in fact have exactly the same variables free In this section we will answer both questions yes We will do more than answer the questions We will give a way to decide if one linear system can be derived from another by row operations The answers to the two questions will follow from this larger result 111 1 Gauss Jordan Reduction Gaussian elimination coupled with back substitution solves linear systems but it s not the only method possible Here is an extension of Gauss method that has some advantages 1 1 Example To solve a y 22 2 y 3z 7 x gt pl we can start by going to echelon form as usual 1 1 2 2 1 1 aiu uH a cas Sige a Fe pe ng 0 1 a Mat 00 als We can keep going to a second stage by making the leading entries into ones vom f 1 2 2 SP ora la 00 1 2 Section III Reduced Echelon Form 47 an
27. form corresponds to some parametrization but why the same parametrization A solution set can be parametrized in many ways and Gauss method or the Gauss Jordan method can be done in many ways so a first guess might be that we could derive many different reduced echelon form versions of the same starting system and many different parametrizations But we never do Experience shows that starting with the same system and proceeding with row operations in many different ways always yields the same reduced echelon form and the same parametrization using the unmodified free variables In the rest of this section we will show that the reduced echelon form version of a matrix is unique It follows that the parametrization of a linear system in terms of its unmodified free variables is unique because two different ones would give two different reduced echelon forms We shall use this result and the ones that lead up to it in the rest of the book but perhaps a restatement in a way that makes it seem more immediately useful may be encouraging Imagine that we solve a linear system parametrize and check in the back of the book for the answer But the parametrization there appears different Have we made a mistake or could these be different looking descriptions of the same set as with the three descriptions above of 5 The prior paragraph notes that we will show here that different looking parametrizations using the unmodified free variables describ
28. induction on the number of row operations used to reduce one matrix to the other Before we proceed here is an outline of the ar gument readers unfamiliar with induction may want to compare this argument with the one used in the General Particular Homogeneous proof First for the base step of the argument we will verify that the proposition is true when reduction can be done in zero row operations Second for the inductive step we will argue that if being able to reduce the first matrix to the second in some number t gt 0 of operations implies that each row of the second is a linear combination of the rows of the first then being able to reduce the first to the second in t 1 operations implies the same thing Together this base step and induction step prove this result because by the base step the proposition is true in the zero operations case and by the inductive step the fact that it is true in the zero operations case implies that it is true in the one operation case and the inductive step applied again gives that it is therefore true in the two operations case etc Proor We proceed by induction on the minimum number of row operations that take a first matrix A to a second one B More information on mathematical induction is in the appendix 54 Chapter One Linear Systems In the base step that zero reduction operations suffice the two matrices are equal and each row of B is obviously a combination of A
29. no non 0 0 equations There aren t any more tricky points after this one V 3 24 Prove that if and Y satisfy a homogeneous system then so do these vec tors a b 35 c k amp mt for k meR What s wrong with These three show that if a homogeneous system has one solution then it has many solutions any multiple of a solution is another solution Section I Solving Linear Systems 31 and any sum of solutions is a solution also so there are no homogeneous systems with exactly one solution 3 25 Prove that if a system with only rational coefficients and constants has a solution then it has at least one all rational solution Must it have infinitely many 32 Chapter One Linear Systems II Linear Geometry of n Space For readers who have seen the elements of vectors before in calculus or physics this section is an optional review However later work will refer to this material so it is not optional if it is not a review In the first section we had to do a bit of work to show that there are only three types of solution sets singleton empty and infinite But in the special case of systems with two equations and two unknowns this is easy to see Draw each two unknowns equation as a line in the plane and then the two lines could have a unique intersection be parallel or be the same line Unique solution No solutions Infinitely many solutions 32 2y 7 3r 2y 7 3a 2y 7 a y l 3r 2y 4 6x
30. of B is a linear combination of the rows of D and vice versa Bi 4 101 84 262 Sim m and 6 tjiP tj aa Fr tjmbm where the s s and t s are scalars The base step of the induction is to verify the lemma for the first rows of the matrices that is to verify that 41 k If either row is a zero row then the entire matrix is a zero matrix since it is in echelon form and hterefore both matrices are zero matrices by Corollary 2 3 and so both and k are oo For the case where neither 3 nor 6 is a zero row consider the 1 instance of the linear relationship above B 1 101 51 202 Simdm 0 e bia oce 81 1 0 eo dis ce 51 2 0 sea OY ees 81m 0 ede df mis First note that lt k is impossible in the columns of D to the left of column k the entries are are all zeroes as d leads the first row and so if lt ki then the equation of entries from column would be bj 1 1 0 51m gt 0 but bi isn t zero since it leads its row and so this is an impossibility Next a symmetric argument shows that k lt also is impossible Thus the 41 k base case holds Section III Reduced Echelon Form 57 The inductive step is to show that if 0 k and 9 ko and 4 kr then also 1 kr for r in the interval 1 m 1 This argument is saved for Exercise 22 QED That lemma answers two of the questions that we have posed i any tw
31. one dimension then a plane must involve two For example the plane through the points 1 0 5 2 1 3 and 2 4 0 5 consists of endpoints of the vectors in 1 1 3 O t 1 s 4 tseR 5 8 4 5 the column vectors associated with the parameters 1 2 1 3 2 1 1 2 1 0 4 4 0 8 3 5 4 5 0 5 5 are two vectors whose whole bodies lie in the plane As with the line note that some points in this plane are described with negative t s or negative s s or both A description of planes that is often encountered in algebra and calculus uses a single equation as the condition that describes the relationship among the first second and third coordinates of points in a plane dr 22 y 2 4 The translation from such a description to the vector description that we favor in this book is to think of the condition as a one equation linear system and paramatrize x 1 2 4 y z 2 0 5 0 5 1 0 elvre 0 0 1 Generalizing from lines and planes we define a k dimensional linear sur face or k flat in R to be p t101 taU2 tkUk ty tg R where U Uy ER For example in R4 2 1 7 ped jeer 3 0 0 5 0 Section II Linear Geometry of n Space 37 is a line 0 1 2 0 1 0 ol ttl o ta t s R 0 1 0 is a plane and 3 0 1 2 1 0 0 0 Y 7 of testi tela r s t R 0 5 1 0 0 is a three dimensional linear surface Agai
32. one thing a rise in steel will cause that industry to have an increased demand for autos which will mitigate to some extent the loss in external demand for autos On the other hand the drop in external demand for autos will cause the auto industry to use less steel and so lessen somewhat the upswing in steel s business In short these two industries form a system and we need to predict the totals at which the system as a whole will settle Topic Input Output Analysis 65 For that prediction let s be next years total production of steel and let a be next year s total output of autos We form these equations next year s production of steel next year s use of steel by steel next year s use of steel by auto next year s use of steel by others next year s production of autos next year s use of autos by steel next year s use of autos by auto next year s use of autos by others On the left side of those equations go the unknowns s and a At the ends of the right sides go our external demand estimates for next year 17 589 and 21 243 For the remaining four terms we look to the table of this year s information about how the industries interact For instance for next year s use of steel by steel we note that this year the steel industry used 5395 units of steel input to produce 25 448 units of steel output So next year when the steel industry will produce s units out we expect that doing so will
33. single resistor of 20 25 6 4 8 ohms a What is the equivalent resistance if we change the 12 ohm resistor to 5 ohms b What is the equivalent resistance if the two are each 8 ohms c Find the formula for the equivalent resistance if the two resistors in parallel are r1 ohms and r ohms 3 For the car dashboard example that opens this Topic solve for these amperages assume that all resistances are 2 ohms a If the driver is stepping on the brakes so the brake lights are on and no other circuit is closed b If the hi beam headlights and the brake lights are on 4 Show that in this Wheatstone Bridge r2 Ta r2 ri equals r4 ra if and only if the current flowing through rg is zero The way that this device is used in practice is that an unknown resistance at r4 is compared to the other three r1 r2 and ra At rg is placed a meter that shows the current The three resistances r1 r2 and r3 are varied typically they each have a calibrated knob until the current in the middle reads 0 and then the above equation gives the value of r4 There are networks other than electrical ones and we can ask how well Kirchoff s laws apply to them The remaining questions consider an extension to networks of streets 5 Consider this traffic circle North Avenue Main Street Pier Boulevard Topic Analyzing Networks 77 This is the traffic volume in units of cars per five minutes North Pier Main into
34. so si Z d Section I Solving Linear Systems 27 Now apply Lemma 3 8 to conclude that a solution set p h h solves the associated homogeneous system is either empty if there is no particular solution p or has one element if there is a p and the homogeneous system has the unique solution 0 or is infinite if there is a p and the homogeneous system has a non 0 solution and thus by the prior paragraph has infinitely many solutions QED This table summarizes the factors affecting the size of a general solution number of solutions of the associated homogeneous system one infinitely many unique infinitely many p Mn yes solution solutions exists 2 no no no solutions solutions The factor on the top of the table is the simpler one When we perform Gauss method on a linear system ignoring the constants on the right side and so paying attention only to the coefficients on the left hand side we either end with every variable leading some row or else we find that some variable does not lead a row that is that some variable is free Of course ignoring the constants on the right is formalized by considering the associated homogeneous system We are simply putting aside for the moment the possibility of a contradictory equation A nice insight into the factor on the top of this table at work comes from con sidering the case of a system having the same number of equations as variables This sy
35. take s 5395 25 448 units of steel input this is simply the assumption that input is proportional to output We are assuming that the ratio of input to output remains constant over time in practice models may try to take account of trends of change in the ratios Next year s use of steel by the auto industry is similar This year the auto industry uses 2664 units of steel input to produce 30346 units of auto output So next year when the auto industry s total output is a we expect it to consume a 2664 30346 units of steel Filling in the other equation in the same way we get this system of linear equation 5395 2664 IA Td Sa 48 9 030 A 25448 30346 Rounding to four decimal places and putting it into the form for Gauss method gives this 0 78805 0 0879a 17 589 0 0019s 0 7024a 21 268 The solution is s 25 708 and a 30350 Looking back recall that above we described why the prediction of next year s totals isn t as simple as adding 200 to last year s steel total and subtract ing 25 from last year s auto total In fact comparing these totals for next year to the ones given at the start for the current year shows that despite the drop in external demand the total production of the auto industry is predicted to rise The increase in internal demand for autos caused by steel s sharp rise in business more than makes up for the loss in external demand for autos One of the advanta
36. 1 3 0 2 v12 12 02 0 3 22 arccos arccos 3 V2V 13 approximately 0 94 radians Notice that these vectors are not orthogonal Al though the yz plane may appear to be perpendicular to the ry plane in fact the two planes are that way only in the weak sense that there are vectors in each orthogonal to all vectors in the other Not every vector in each is orthogonal to all vectors in the other Exercises Y 2 10 Find the length of each vector Section II Linear Geometry of n Space 43 1 4 0 a i b E c a o e 1 0 0 v 2 11 Find the angle between each two if it is defined 00000 Q v 2 12 During maneuvers preceding the Battle of Jutland the British battle cruiser Lion moved as follows in nautical miles 1 2 miles north 6 1 miles 38 degrees east of south 4 0 miles at 89 degrees east of north and 6 5 miles at 31 degrees east of north Find the distance between starting and ending positions Ohanian 2 13 Find k so that these two vectors are perpendicular O G 2 14 Describe the set of vectors in R orthogonal to this one V 2 15 a Find the angle between the diagonal of the unit square in R and one of the axes b Find the angle between the diagonal of the unit cube in R and one of the axes c Find the angle between the diagonal of the unit cube in R and one of the axes d What is the limit as n goes to oo of the angle between the diagonal of th
37. 100 150 25 out of 75 150 50 We can set up equations to model how the traffic flows a Adapt Kirchoff s Current Law to this circumstance Is it a reasonable mod elling assumption b Label the three between road arcs in the circle with a variable Using the adapted Current Law for each of the three in out intersections state an equa tion describing the traffic flow at that node c Solve that system d Interpret your solution e Restate the Voltage Law for this circumstance How reasonable is it 6 This is a network of streets Shelburne St Willow Jay Ln west gt east M Winooski Ave The hourly flow of cars into this network s entrances and out of its exits can be observed east Winooski west Winooski Willow Jay Shelburne into 80 50 65 40 out of 30 5 70 55 75 Note that to reach Jay a car must enter the network via some other road first which is why there is no into Jay entry in the table Note also that over a long period of time the total in must approximately equal the total out which is why both rows add to 235 cars Once inside the network the traffic may flow in differ ent ways perhaps filling Willow and leaving Jay mostly empty or perhaps flowing in some other way Kirchhoff s Laws give the limits on that freedom a Determine the restrictions on the flow inside this network of streets by setting up a variable for each block establishing the equations and solving them
38. 3 f x 2 w 4 2a c 3 2r y w 4 22 y w 2 a b 0 y z w l 3r y z 7 3 16 Solve each system giving the solution set in vector notation Identify the particular solution and the solution of the homogeneous system a 22 y z 1 b x z 1 c z y z 0 4x y 3 y 22 w 3 Y w 0 x 2y 32 w 7 3x 2y 3z w 0 y w 0 d a 2b 3c d e 1 3a b c d e 3 v 3 17 For the system 272 y w 3 y z 2w 2 x 2y z 1 30 Chapter One Linear Systems which of these can be used as the particular solution part of some general solu tion 0 2 1 3 1 4 all eli ol 0 0 1 v 3 18 Lemma 3 8 says that any particular solution may be used for p Find if possible a general solution to this system r y w 4 2x 3y 2 0 ytz w 4 that uses the given vector as its particular solution 0 5 2 0 1 1 lo m1 Ol 4 10 1 3 19 One of these is nonsingular while the other is singular Which is which 1 3 1 3 a a En 5 b v 3 20 Singular or nonsingular a i a b E 3 c i 3 Careful 1 2 1 2 2 1 d E 1 e 0 347 1 1 4 v 3 21 Is the given vector in the set generated by the given set 3 00 1 2 0 1 less 1 1 2 3 22 Prove that any linear system with a nonsingular matrix of coefficients has a solution and that the solution is unique 3 23 To tell the whole truth there is another tricky point to the proof of Lemma 3 7 What happens if there are
39. Linear Algebra Jim Hefferon G Hei A 6 Notation R real numbers N natural numbers 0 1 2 C complex numbers sunt set of such that sequence like a set but order matters V W U vector spaces U w vectors 0 Oy zero vector zero vector of V D bases En 1 En standard basis for R B 6 basis vectors Repg v matrix representing the vector Pr set of n th degree polynomials Mnxm set of n xm matrices S span of the set S M N direct sum of subspaces V W isomorphic spaces h g homomorphisms linear maps H G matrices t s transformations maps from a space to itself T S square matrices Repg p h matrix representing the map h ij matrix entry from row i column j T determinant of the matrix T R h Y h rangespace and nullspace of the map h Reh Molh generalized rangespace and nullspace Lower case Greek alphabet name character name character name character alpha Qa iota L rho p beta B kappa K sigma c gamma v lambda X tau T delta mu H upsilon v epsilon nu V phi Q zeta xi chi X eta n omicron o psi Y theta 0 pi T omega w Cover This is Cramer s Rule for the system x 2y 6 3x y 8 The size of the first box is the determinant shown the absolute value of the size is the area The size of the second box is x times that and equals the size of the final box Hence x is the final determinant divided by
40. Section I Solving Linear Systems 19 0 1 2 c 2 05 1 rines 2 0 1 2 21 Decide if the vector is in the set a E Ju lem b OKA 2 0 1 c 1 3 a rinem 1 7 3 1 2 3 d OKOLICO UEI 1 1 1 2 22 Paramatrize the solution set of this one equation system z t r2r2 In 0 v 2 23 a Apply Gauss method to the left hand side to solve x 2y w a 2x 2 b s y 2w c for x y z and w in terms of the constants a b and c b Use your answer from the prior part to solve this x 2y w 3 2x 2 1 y 2w 2 v 2 24 Why is the comma needed in the notation a for matrix entries v 2 25 Give the 4x4 matrix whose i j th entry is a i 9 b 1 to the i j power 2 26 For any matrix A the transpose of A written A 5 is the matrix whose columns are the rows of A Find the transpose of each of these 1 1 2 3 2 3 5 10 V 2 27 a Describe all functions f x ax bx c such that f 1 2 and f 1 6 b Describe all functions f x ax bx c such that f 1 2 2 28 Show that any set of five points from the plane R lie on a common conic section that is they all satisfy some equation of the form ax by cay da ey f 0 where some of a f are nonzero 2 29 Make up a four equations four unknowns system having a a one parameter solution set b a two parameter solution set c a three parameter solution set 2 80 a Solve the system of equations ac y
41. The array of coefficients can be entered in this way gt A array 1 1 1 0 0 0 0 0 1 0 1 0 1 0 0 0 1 0 1 1 0 0 0 0 1 1 0 1 0 5 0 10 0 0 0 0 0 2 0 4 0 01 0 5 1 0 0 10 0 putting the rows on separate lines is not necessary but is done for clarity The vector of constants is entered similarly gt u array 0 0 0 0 10 10 0 Then the system is solved like magic gt linsolve A u 2 75 2 75 x p ac MEME OR g 3 3 3 3 7 3 Systems with infinitely many solutions are solved in the same way the com puter simply returns a parametrization Exercises Answers for this Topic use Maple as the computer algebra system In particular all of these were tested on Maple V running under MS DOS NT version 4 0 On all of them the preliminary command to load the linear algebra package along with Maple s responses to the Enter key have been omitted Other systems have similar commands Topic Computer Algebra Systems 63 1 Use the computer to solve the two problems that opened this chapter a This is the Statics problem 40h 15c 100 25c 50 50h b This is the Chemistry problem Th Tj 8h li 5j 2k li 33 3i 6j 1k 2 Use the computer to solve these systems from the first subsection many solutions or no solutions or conclude a 2a 2y 5 b a y 1 c x 3y z 1 x 4Ay 0 x y 2 e y 2z 14
42. a a ay 1 For what values of a does the system fail to have solutions and for what values of a are there infinitely many solutions 2 20 Chapter One Linear Systems b Answer the above question for the system az y a x ay 1 USSR Olympiad no 174 2 31 In air a gold surfaced sphere weighs 7588 grams It is known that it may contain one or more of the metals aluminum copper silver or lead When weighed successively under standard conditions in water benzene alcohol and glycerine its respective weights are 6588 6688 6778 and 6328 grams How much if any of the forenamed metals does it contain if the specific gravities of the designated substances are taken to be as follows Aluminum 2 1 Alcohol 0 81 Copper 8 9 Benzene 0 90 Gold 19 3 Glycerine 1 26 Lead 11 3 Water 1 00 Silver 10 8 Math Mag Sept 1952 1 3 General Particular Homogeneous The prior subsection has many descriptions of solution sets They all fit a pattern They have a vector that is a particular solution of the system added to an unrestricted combination of some other vectors The solution set from Example 2 13 illustrates 0 1 1 2 4 1 1 0 w 3 u 1 2 wucR 0 1 0 0 0 1 particular unrestricted solution combination The combination is unrestricted in that w and u can be any real numbers there is no condition like such that 2w u 0 that would restrict which pairs w u can be used
43. a contradictory equation and each variable is a leading variable in its row then the system has a unique solution and we find it by back substitution Finally if we reach echelon form without a contradictory equation and there is not a unique solution at least one variable is not a leading variable then the system has many solutions The next subsection deals with the third case we will see how to describe the solution set of a system with many solutions Exercises v 1 16 Use Gauss method to find the unique solution for each system x z 0 a ere a 13 b doy x y l 2 y z 4 v 1 17 Use Gauss method to solve each system or conclude many solutions or no solutions Section I Solving Linear Systems 9 a 22 2y 5 b y 1 c v 3y4 z 1 x 4dy 0 y 2 y 2z 14 d z y 1 e Ay z 20 f 2a z w 5 3z 3y 2 22 2y z 0 y w 1 x z 5 3x z w 0 a y z 10 di y 224w 9 Y 1 18 There are methods for solving linear systems other than Gauss method One often taught in high school is to solve one of the equations for a variable then substitute the resulting expression into other equations That step is repeated until there is an equation with only one variable From that the first number in the solution is derived and then back substitution can be done This method both takes longer than Gauss method since it involves more arithmetic operations and is more likely to
44. and k m R kt mu k t v m u w v 2 38 The geometric mean of two positive reals x y is y TY It is analogous to the arithmetic mean x y 2 Use the Cauchy Schwartz inequality to show that the geometric mean of any x y R is less than or equal to the arithmetic mean 2 39 A ship is sailing with speed and direction v1 the wind blows apparently judging by the vane on the mast in the direction of a vector d on changing the direction and speed of the ship from v1 to U2 the apparent wind is in the direction of a vector b Find the vector velocity of the wind Am Math Mon Feb 1933 2 40 Verify the Cauchy Schwartz inequality by first proving Lagrange s identity 2 Ne x X u X om 1 lt j lt n 1 lt j lt n 1 lt j lt n 1 lt k lt j lt n and then noting that the final term is positive Recall the meaning 5 ajbj a b agbe Feed Anbn 1 lt j lt n 2 2 2 Lp 2 aj a a2 Tr Tn 1 lt j lt n and Section II Linear Geometry of n Space 45 of the X notation This result is an improvement over Cauchy Schwartz because it gives a formula for the difference between the two sides Interpret that difference in R 46 Chapter One Linear Systems III Reduced Echelon Form After developing the mechanics of Gauss method we observed that it can be done in more than one way One example is that we sometimes have to swap rows and there can be more than one row to choose from Another
45. and slopes if b a1 di c and 6 a d c2 Is that only if 1 10 How should R be defined y 1 11 A person traveling eastward at a rate of 3 miles per hour finds that the wind appears to blow directly from the north On doubling his speed it appears to come from the north east What was the wind s velocity Math Mag Jan 1957 1 12 Euclid describes a plane as a surface which lies evenly with the straight lines on itself Commentators e g Heron have interpreted this to mean A plane surface is such that if a straight line pass through two points on it the line coincides wholly with it at every spot all ways Translations from Heath pp 171 172 Do planes as described in this section have that property Does this description adequately define planes Section II Linear Geometry of n Space 39 II 2 Length and Angle Measures We ve translated the first section s results about solution sets into geometric terms for insight into how those sets look But we must watch out not to be mislead by our own terms labeling subsets of R of the forms p tU te R and p tU sw t s R as lines and planes doesn t make them act like the lines and planes of our prior experience Rather we must ensure that the names suit the sets While we can t prove that the sets satisfy our intuition we can t prove anything about intuition in this subsection we ll observe that a result
46. any examples are in the next subsection Section I Solving Linear Systems 7 Another way that linear systems can differ from the examples shown earlier is that some linear systems do not have a unique solution This can happen in two ways The first is that it can fail to have any solution at all 1 12 Example Contrast the system in the last example with this one a 3y 1 ER a 3y 1 2r ya a EE 5y 5 2r 2y 0 res 4y 2 Here the system is inconsistent no pair of numbers satisfies all of the equations simultaneously Echelon form makes this inconsistency obvious 3y 1 4 5 p2 p3 di Es 5 0 2 The solution set is empty 1 13 Example The prior system has more equations than unknowns but that is not what causes the inconsistency Example 1 11 has more equations than unknowns and yet is consistent Nor is having more equations than unknowns necessary for inconsistency as is illustrated by this inconsistent system with the same number of equations as unknowns x 2y 8 2p p2 r 2y 8 2r 4y 8 0 8 The other way that a linear system can fail to have a unique solution is to have many solutions 1 14 Example In this system y 4 2r 2y 8 any pair of numbers satisfying the first equation automatically satisfies the sec ond The solution set z y z y 4 is infinite some of its members are 0 4 1 5 and 2 5 1 5 The result of applying Gauss method here contrasts with the
47. apter One Linear Systems T o il l i2 The Current Law applied to the point in the upper right where the flow io meets i and ia gives that iy 11 i2 Applied to the lower right it gives 11 i2 ig In the circuit that loops out of the top of the battery down the left branch of the parallel portion and back into the bottom of the battery the voltage rise is 20 while the voltage drop is i4 12 so the Voltage Law gives that 124 20 Similarly the circuit from the battery to the right branch and back to the battery gives that 8i2 20 And in the circuit that simply loops around in the left and right branches of the parallel portion arbitrarily taken clockwise there is a voltage rise of 0 and a voltage drop of 8i2 12 so the Voltage Law gives that 822 1211 0 lo 4q 0 to i t2 0 8i2 20 127 8i2 0 The solution is io 25 6 i 5 3 and ia 5 2 all in amperes Incidentally this illustrates that redundant equations do indeed arise in practice Kirchhoff s laws can be used to establish the electrical properties of networks of great complexity The next diagram shows five resistors wired in a series parallel way 5 ohm A 2 ohm 10 volt 10 ohm ES 4 ohm This network is a Wheatstone bridge see Exercise 4 To analyze it we can place the arrows in this way is gt Topic Analyzing Networks 75 Kirchoff s Current Law applied to the top node the left node
48. arrow d b e Ord B boldface is also common a or For instance this is a column vector with a third component of 7 1 v 3 7 2 9 Definition The linear equation az a319 Apt d with unknowns z4 n is satisfied by 1 s Sn if a181 a252 anSn d A vector satisfies a linear system if it satisfies each equation in the system The style of description of solution sets that we use involves adding the vectors and also multiplying them by real numbers such as the z and w We need to define these operations 2 10 Definition The vector sum of u and v is this u Ui U1 T U1 a v 4 Un Un Un T Un In general two matrices with the same number of rows and the same number of columns add in this way entry by entry 2 11 Definition The scalar multiplication of the real number r and the vector U is this vi TU Dre Bun In general any matrix is multiplied by a real number in this entry by entry way 16 Chapter One Linear Systems Scalar multiplication can be written in either order r u or U r or without the symbol rd Do not refer to scalar multiplication as scalar product because that name is used for a different operation 2 12 Example 2 3 24 3 5 F p 3 1 5 3 1 2 2 7 2 1 4 1 4 5 La Notice that the definitions of vector addition and scalar multiplication agree where they overlap for instance Y 0 20 Wi
49. arts we call it an augmented matrix In this notation Gauss method goes this way 12 0 4 fa 014 fi 2 0 4 ai alo Seo 1 Ale b dom 10 2 4 0 2 2 0 00 010 The second row stands for y z 0 and the first row stands for x 2y 4 so the solution set is 4 2z z 2 z R One advantage of the new notation is that the clerical load of Gauss method the copying of variables the writing of s and s etc is lighter We will also use the array notation to clarify the descriptions of solution sets A description like 2 2z 2 0 1 z w z w z w E R from Ex ample 2 3 is hard to read We will rewrite it to group all the constants together all the coefficients of z together and all the coefficients of w together We will write them vertically in one column wide matrices 2 2 2 1 1 1 a lela Pele le z w E R 0 0 1 Section I Solving Linear Systems 15 For instance the top line says that x 2 22 2w The next section gives a geometric interpretation that will help us picture the solution sets when they are written in this way 2 8 Definition A vector or column vector is a matrix with a single column A matrix with a single row is a row vector The entries of a vector are its components Vectors are an exception to the convention of representing matrices with capital roman letters We use lower case roman or greek letters overlined with an
50. bottom most non 0 0 equation the case where all the equations are 0 0 is trivial We call that the m th row Am bm Ulm Am u ET 1 ee Gm nin 0 where am 0 The notation here has stand for leading so a means the coefficient from the row m of the variable leading row m Either there are variables in this equation other than the leading one x or else there are not If there are other variables xp 1 etc then they must be free variables because this is the bottom non 0 0 row Move them to the right and divide by am lm Tp T am tm 1 Am tu Elm 1 art Ann Om bin Bn to expresses this leading variable in terms of free variables If there are no free variables in this equation then z 0 see the tricky point noted following this proof For the inductive step we assume that for the m th equation and for the m 1 th equation and for the m t th equation we can express the leading variable in terms of free variables where 0 lt t lt m To prove that the same is true for the next equation up the m t 1 th equation we take each variable that leads in a lower down equation z r5 and substitute its expression in terms of free variables The result has the form Om t 1 lm a41 T Es iii SUMS of multiples of free variables 0 z 0 We move the free variables to the right hand side expressed in terms of free where y
51. classes There are infinitely many matrices in the pictured class but we ve only got room to show two We have proved there is one and only one reduced echelon form matrix in each row equivalence class So the reduced echelon form is a canonical form for row equivalence the reduced echelon form matrices are representatives of the classes We can answer questions about the classes by translating them into questions about the representatives 2 8 Example We can decide if matrices are interreducible by seeing if Gauss Jordan reduction produces the same reduced echelon form result Thus these are not row equivalent 1 3 1 3 2 6 2 5 More information on canonical representatives is in the appendix Section III Reduced Echelon Form 59 because their reduced echelon forms are not equal 6 v 1 2 9 Example Any nonsingular 3x3 matrix Gauss Jordan reduces to this QUO B Q ERO 0 0 1 2 10 Example We can describe the classes by listing all possible reduced echelon form matrices Any 2x2 matrix lies in one of these the class of matrices row equivalent to this 0 0 up the infinitely many classes of matrices row equivalent to one of this type la 0 0 where a R including a 0 the class of matrices row equivalent to this 0 1 0 0 and the class of matrices row equivalent to this 1 0 0 1 this the class of nonsingular 2x2 matrices Exercises Y 2 11 Decide if the mat
52. d in the pivot_row pivot_row position for use as the pivot entry To make it practi cal one way in which this code needs to be reworked is to cover the case where finding a zero in that location leads to a row swap or to the conclusion that the matrix is singular Adding some if statements to cover those cases is not hard but we will instead consider some more subtle ways in which the code is naive There are pitfalls arising from the computer s reliance on finite precision floating point arithmetic For example we have seen above that we must handle as a separate case a system that is singular But systems that are nearly singular also require care Consider this one r 2y 3 1 000 000 01x 2y 3 000 000 01 By eye we get the solution x 1 and y 1 But a computer has more trouble A computer that represents real numbers to eight significant places as is common usually called single precision will represent the second equation internally as 1 000 000 Ox 2y 3 000 0000 losing the digits in the ninth place Instead of reporting the correct solution this computer will report something that is not even close this computer thinks that the system is singular because the two equations are represented internally as equal For some intuition about how the computer could think something that is so far off we can graph the system Topic Accuracy of Computations 69 At the scale of this graph the two lines can
53. d then to a third stage that uses the leading entries to eliminate all of the other entries in each column by pivoting upwards 11 012 10 0l 1 rdg 1 oli S949 1 0 1 est A qe 112 00 11 2 The answer is x 1 y 1 and z 2 Note that the pivot operations in the first stage proceed from column one to column three while the pivot operations in the third stage proceed from column three to column one 1 2 Example We often combine the operations of the middle stage into a single step even though they are operations on different rows 2 1 7 tm 2 1 7 4 2 6 0 4 8 1 2 t 1 2 71 4 p2 0 1 2 2e m 1 0 5 2 0 1 2 The answer is x 5 2 and y 2 This extension of Gauss method is Gauss Jordan reduction It goes past echelon form to a more refined more specialized matrix form 1 3 Definition A matrix is in reduced echelon form if in addition to being in echelon form each leading entry is a one and is the only nonzero entry in its column The disadvantage of using Gauss Jordan reduction to solve a system is that the additional row operations mean additional arithmetic The advantage is that the solution set can just be read off In any echelon form plain or reduced we can read off when a system has an empty solution set because there is a contradictory equation we can read off when a system has a one element solution set because there is no contradiction and every variable is the leading variable
54. ding 1 times the row to itself has the effect of multiplying the row by 0 Finally swap ping a row with itself is disallowed to make some results in the fourth chapter easier to state and remember and besides self swapping doesn t accomplish anything 4 Chapter One Linear Systems Proor We will cover the equation swap operation here and save the other two cases for Exercise 29 Consider this swap of row i with row j 01 181 T 01 222 T Al nin d a1 1 41272 t AL nin d 05181 T 272 Tr Qinin di Qj 11 T 45272 r Qj nin dj 05181 T Qj 22 Ttt QjnXn dj 05 181 W2X2 Tr Qinin di Am 1 81 T Am 2 82 7 Am nin dm Am 1 1 T Am 2 2 T Am nn dm The n tuple 51 Sn satisfies the system before the swap if and only if substituting the values the s s for the variables the x s gives true statements a1 151 01 282 01n8n d and 0 151 05282 Qj 84 d and 45 181 052582 4jn8n dj and am 181 am 282 am Sn dm In a requirement consisting of statements and ed together we can rearrange the order of the statements so that this requirement is met if and only if a s 01 282 A1 nSn d and 05181 05 282 Aj nSn dj and Qi 151 04 282 nsSn di and am 181 Fam 282 Am nSn Im This is exactly the requirement that s Sn solves the system after the row swap QED 1 5 Definition
55. e unit cube in R and one of the axes 2 16 Is any vector perpendicular to itself Y 2 17 Describe the algebraic properties of dot product a Is it right distributive over addition v b Is is left distributive over addition c Does it commute d Associate e How does it interact with scalar multiplication As always any assertion must be backed by either a proof or an example 2 18 Verify the equality condition in Corollary 2 6 the Cauchy Schwartz Inequal ity a Show that if 4 is a negative scalar multiple of Y then t Y and 4 U are less than or equal to zero b Show that u v u if and only if one vector is a scalar multiple of the other 2 19 Suppose that i Y i 1 and 0 Must d di v 2 20 Does any vector have length zero except a zero vector If yes produce an example If no prove it v 2 21 Find the midpoint of the line segment connecting 11 y1 with x2 y2 in R Generalize to R 2 22 Show that if y 0 then 5 7 has length one What if 0 2 23 Show that if r gt O then rv is r times as long as Y What if r lt 0 w U w 44 Chapter One Linear Systems v 2 24 A vector Y R of length one is a unit vector Show that the dot product of two unit vectors has absolute value less than or equal to one Can less than happen Can equal to 2 25 Prove that 7 Ju v 2 u 2 15 2 26 Show that if 7 y 0
56. e genuinely different sets Here is an informal argument that the reduced echelon form version of a matrix is unique Consider again the example that started this section of a matrix that reduces to three different echelon form matrices The first matrix of the three is the natural echelon form version The second matrix is the same as the first except that a row has been halved The third matrix too is just a cosmetic variant of the first The definition of reduced echelon form outlaws this kind of fooling around In reduced echelon form halving a row is not possible because that would change the row s leading entry away from one and neither is combining rows possible because then a leading entry would no longer be alone in its column This informal justification is not a proof we have argued that no two different reduced echelon form matrices are related by a single row operation step but we have not ruled out the possibility that multiple steps might do Before we go to that proof we finish this subsection by rephrasing our work in a terminology that will be enlightening Many different matrices yield the same reduced echelon form matrix The three echelon form matrices from the start of this section and the matrix they were derived from all give this reduced echelon form matrix t We think of these matrices as related to each other The next result speaks to 50 Chapter One Linear Systems this relationship 1 4 Lemma Eleme
57. es to B by some row operations then also B reduces to A by reversing those operations For transitivity suppose that A reduces to B and that B reduces to C Linking the reduction steps from A B with those from B gt gt C gives a reduction from A to C QED 1 6 Definition Two matrices that are interreducible by the elementary row operations are row equivalent More information on equivalence relations is in the appendix Section III Reduced Echelon Form 51 The diagram below shows the collection of all matrices as a box Inside that box each matrix lies in some class Matrices are in the same class if and only if they are interreducible The classes are disjoint no matrix is in two distinct classes The collection of matrices has been partitioned into row equivalence classes One of the classes in this partition is the cluster of matrices shown above expanded to include all of the nonsingular 2x2 matrices The next subsection proves that the reduced echelon form of a matrix is unique that every matrix reduces to one and only one reduced echelon form matrix Rephrased in terms of the row equivalence relationship we shall prove that every matrix is row equivalent to one and only one reduced echelon form matrix In terms of the partition what we shall prove is every equivalence class contains one and only one reduced echelon form matrix So each reduced echelon form matrix serves a
58. familiar from R and R when generalized to arbitrary R supports the idea that a line is straight and a plane is flat Specifically we ll see how to do Euclidean geometry in a plane by giving a definition of the angle between two R vectors in the plane that they generate 2 1 Definition The length of a vector U R is this Ig fot uv2 2 2 Remark This is a natural generalization of the Pythagorean Theorem A classic discussion is in Polya We can use that definition to derive a formula for the angle between two vectors For a model of what to do consider two vectors in R a Put them in canonical position and in the plane that they determine consider the triangle formed by 4 v and d v ia eS Apply the Law of Cosines i 7 u IIg 2 cos 0 where 0 is the angle between the vectors Expand both sides uf u2 u2 v2 v 4 v3 2 ui v cos 8 and simplify u1Uj U2V2 UZU3 0 arccos ja l l 40 Chapter One Linear Systems In higher dimensions no picture suffices but we can make the same argument analytically First the form of the numerator is clear it comes from the middle terms of the squares u1 v1 ug v2 etc 2 3 Definition The dot product or inner product or scalar product of two n component real vectors is the linear combination of their components UV
59. for every y then 7 0 2 27 Is dnl lt ld J If it is true then it would generalize the Triangle Inequality 2 28 What is the ratio between the sides in the Cauchy Schwartz inequality 2 29 Why is the zero vector defined to be perpendicular to every vector 2 30 Describe the angle between two vectors in R 2 31 Give a simple necessary and sufficient condition to determine whether the angle between two vectors is acute right or obtuse v 2 32 Generalize to R the converse of the Pythagorean Theorem that if and v are perpendicular then v 2 33 Show that if and only if 4 and 4 v are perpendicular Give an example in R 2 34 Show that if a vector is perpendicular to each of two others then it is perpen dicular to each vector in the plane they generate Remark They could generate a degenerate plane a line or a point but the statement remains true 2 35 Prove that where 4 Y R are nonzero vectors the vector a M v lall all bisects the angle between them Illustrate in R 2 36 Verify that the definition of angle is dimensionally correct 1 if k gt 0 then the cosine of the angle between ku and Y equals the cosine of the angle between 4 and v and 2 if k lt 0 then the cosine of the angle between k and Y is the negative of the cosine of the angle between 4 and v vV 2 37 Show that the inner product operation is linear for t Y w R
60. g this year s external demands and reducing them 7 The study from which these numbers are taken concluded that because of the closing of a local military facility overall personal income in the area would fall 7 so this might be a first guess at what would actually happen 68 Chapter One Linear Systems Topic Accuracy of Computations Gauss method lends itself nicely to computerization The code below illustrates It operates on an nxn matrix a pivoting with the first row then with the second row etc for pivot_row 1 pivot_row lt n 1 pivot_row for row_below pivot_row 1 row_below lt n row_below 4 multiplier a row_below pivot_row a pivot_row pivot_row for col pivot_row col lt n co1 4 a row belovw col multiplier a pivot row col This code is in the C language Here is a brief translation The loop con struct for pivot_row 1 pivot_row lt n 1 pivot_rowt sets pivot row to 1 and then iterates while pivot row is less than or equal to n 1 each time through incrementing pivot_row by one with the operation The other non obvious construct is that the in the innermost loop amounts to the alrow_below col multiplier alpivot_row col alrow_below col operation While this code provides a quick take on how Gauss method can be mech anized it is not ready to use It is naive in many ways The most glar ing way is that it assumes that a nonzero number is always foun
61. ges of having a mathematical model is that we can ask What if questions For instance we can ask What if the estimates for 66 Chapter One Linear Systems next year s external demands are somewhat off To try to understand how much the model s predictions change in reaction to changes in our estimates we can try revising our estimate of next year s external steel demand from 17 589 down to 17 489 while keeping the assumption of next year s external demand for autos fixed at 21 243 The resulting system 0 7880s 0 0879a 17 489 0 0019s 0 7024a 21 243 when solved gives s 25577 and a 30314 This kind of exploration of the model is sensitivity analysis We are seeing how sensitive the predictions of our model are to the accuracy of the assumptions Obviously we can consider larger models that detail the interactions among more sectors of an economy These models are typically solved on a computer using the techniques of matrix algebra that we will develop in Chapter Three Some examples are given in the exercises Obviously also a single model does not suit every case expert judgment is needed to see if the assumptions under lying the model can are reasonable ones to apply to a particular case With those caveats however this model has proven in practice to be a useful and ac curate tool for economic analysis For further reading try Leontief 1951 and Leontief 1965 Exercises Hint these sys
62. gs When components are wired one after another as here they are said to be in series 2 ohm resistance 20 volt 5 ohm potential resistance 3 ohm resistance By Kirchhoff s Voltage Law because the voltage rise is 20 volts the total voltage drop must also be 20 volts Since the resistance from start to finish is 10 ohms the resistance of the wires is negligible we get that the current is 20 10 2 amperes Now by Kirchhoff s Current Law there are 2 amperes through each resistor And therefore the voltage drops are 4 volts across the 2 oh m resistor 10 volts across the 5 ohm resistor and 6 volts across the 3 ohm resistor The prior network is so simple that we didn t use a linear system but the next network is more complicated In this one the resistors are in parallel This network is more like the car lighting diagram shown earlier 20 volt 12 ohm 8 ohm We begin by labeling the branches shown below Let the current through the left branch of the parallel portion be 1 and that through the right branch be i5 and also let the current through the battery be ip We are following Kirchoff s Current Law for instance all points in the right branch have the same current which we call i2 Note that we don t need to know the actual direction of flow if current flows in the direction opposite to our arrow then we will simply get a negative number in the solution 74 Ch
63. han one variable free This row reduction t yt z w y z w 3x 6z 6w y 2 w ct y z w 1 3pitps y z w 1 3y 3z 3w 3 y z w 1 oeme r ytz w 1 3p2 p3 y zt tw l p2tpa 0 0 0 0 ends with x and y leading and with both z and w free To get the description that we prefer we will start at the bottom We first express y in terms of the free variables z and w with y 1 4 z w Next moving up to the top equation substituting for y in the first equation z 1 z w z w 1 and solving for x yields x 2 2z 2w Thus the solution set is 2 2z 2w 1 z w z w z w ER We prefer this description because the only variables that appear z and w are unrestricted This makes the job of deciding which four tuples are system solutions into an easy one For instance taking z 1 and w 2 gives the solution 4 2 1 2 In contrast 3 2 1 2 is not a solution since the first component of any solution must be 2 minus twice the third component plus twice the fourth Section I Solving Linear Systems 13 2 4 Example After this reduction 2r 2y 0 22 2y 0 z 3w 2 8 2 p ps z 3w 2 3x 3y a 2 1i p 0 0 X y 22 6w 4 2z 6w 4 2x 2y 0 2patpa z 3w 2 0 0 0 0 x and z lead y and w are free The solution set is y y 2 3w w y w ER For instance 1 1 2 0 satisfies the system take y
64. has a constant of zero that is if it can be put in the form a474 a215 0 tp 0 These are homogeneous because all of the terms involve the same power of their variable the first power including a Oz that we can imagine is on the right side 3 3 Example With any linear system like 3a 4y 3 22 y 1 we associate a system of homogeneous equations by setting the right side to Zeros 3a 4y 0 2r y 0 Our interest in the homogeneous system associated with a linear system can be understood by comparing the reduction of the system 3r 4y 3 2 3 p1 p2 3x 4y 3 2y y 1 11 3 y 1 with the reduction of the associated homogeneous system 3a 4y 0 2 3 p1 p2 3a 4 4y 0 2y y 0 11 3 y 0 Obviously the two reductions go in the same way We can study how linear sys tems are reduced by instead studying how the associated homogeneous systems are reduced Studying the associated homogeneous system has a great advantage over studying the original system Nonhomogeneous systems can be inconsistent But a homogeneous system must be consistent since there is always at least one solution the vector of zeros 22 Chapter One Linear Systems 3 4 Definition A column or row vector of all zeros is a zero vector denoted 0 There are many different zero vectors e g the one tall zero vector the two tall zero vector etc Nonetheless people often refer t
65. he operations be synthesized from the others 1 32 Prove that each operation of Gauss method is reversible That is show that if two systems are related by a row operation S S2 then there is a row operation to go back 5 1 1 33 A box holding pennies nickels and dimes contains thirteen coins with a total value of 83 cents How many coins of each type are in the box Anton 1 34 Four positive integers are given Select any three of the integers find their arithmetic average and add this result to the fourth integer Thus the numbers 29 23 21 and 17 are obtained One of the original integers is Section I Solving Linear Systems 11 a 19 b 21 c 3 d 29 e 17 Con Prob 1955 Y 1 35 Laugh at this AHAHA TEHE TEHAW It resulted from substituting a code letter for each digit of a simple example in addition and it is required to identify the letters and prove the solution unique Am Math Mon Jan 1935 1 36 The Wohascum County Board of Commissioners which has 20 members re cently had to elect a President There were three candidates A B and C on each ballot the three candidates were to be listed in order of preference with no abstentions It was found that 11 members a majority preferred A over B thus the other 9 preferred B over A Similarly it was found that 12 members preferred C over A Given these results it was suggested that B should withdraw to enable a runoff election betwee
66. he relationship to include just the nonzero rows Bi C4101 Cindy The preliminary result also says that for each row j between 1 and r the leading entries of the j th row of B and D appear in the same column denoted Lj Rewriting the above relationship to focus on the entries in the th column Die ce Jes l din c ci 2 e dae c0 Cir e deg ce gives this set of equations for i 1 up to i r bie c1 1d e a 1 545 0 ap Pete Te Cir dr bje hr o Cjrdre bre Cra diu ten o Onde 58 Chapter One Linear Systems Since D is in reduced echelon form all of the d s in column are zero except for d e which is 1 Thus each equation above simplifies to b e Ci d je ci j 1 But B is also in reduced echelon form and so all of the b s in column are zero except for b which is 1 Therefore each c j is zero except that c 1 and c22 1 andc 1 We have shown that the only nonzero coefficient in the linear combination labelled is cj j which is 1 Therefore 6 Because this holds for all nonzero rows B D QED We end with a recap In Gauss method we start with a matrix and then derive a sequence of other matrices We defined two matrices to be related if one can be derived from the other That relation is an equivalence relation called row equivalence and so partitions the set of all matrices into row equivalence
67. ht that Gauss method works by taking linear combinations of the rows But to what end why do we go to echelon form as a particularly simple or basic version of a linear system The answer of course is that echelon form is suitable for back substitution because we have isolated the variables For instance in this matrix ooo bl oO CoO OW oorN O w OU OO Noe Momo z has been removed from z5 s equation That is Gauss method has made x5 s row independent of z4 s row Independence of a collection of row vectors or of any kind of vectors will be precisely defined and explored in the next chapter But a first take on it is that we can show that say the third row above is not comprised of the other Section III Reduced Echelon Form 55 rows that p3 C1p1 Capa Capa For suppose that there are scalars c co and c4 such that this relationship holds 0 0 0 3 3 Deal 37 8 0 0 amp 0 0 1 5 1 1 40 0 0 0 2 1 The first row s leading entry is in the first column and narrowing our considera tion of the above relationship to consideration only of the entries from the first column 0 2c 0c2 0c gives that c 0 The second row s leading entry is in the third column and the equation of entries in that column 0 7c 1c2 0ca along with the knowledge that c 0 gives that ca 0 Now to finish the third row s leading entry is in the fourth column and the equation of entries in that column 3 8c 5ca
68. in some row and we can read off when a system has an infinite solution set because there is no contradiction and at least one variable is free In reduced echelon form we can read off not just what kind of solution set the system has but also its description Whether or not the echelon form is reduced we have no trouble describing the solution set when it is empty of course The two examples above show that when the system has a single solution then the solution can be read off from the right hand column In the case when the solution set is infinite its parametrization can also be read off 48 Chapter One Linear Systems of the reduced echelon form Consider for example this system that is shown brought to echelon form and then to reduced echelon form 2 6 1215 261 215 o0 3 1 41 SEPTEO3S Aa 031215 000 4 2 1 0 1 2 0 9 2 1 2 p 4 3 p3 p2 3p2 P1 0 1 1 3 0 3 1 3 p padpi 1 2 ps 00 0 1 2 Starting with the middle matrix the echelon form version back substitution produces 2x 4 4 so that x4 2 then another back substitution gives 3z9 z3 4 2 1 implying that z 3 1 3 x3 and then the final back substitution gives 2x1 6 3 1 3 z3 z3 2 2 5 implying that z 9 2 1 2 x3 Thus the solution set is this 31 9 2 1 2 S Er Kar 0 1 a3 3 R XA 2 0 Now considering the final matrix the reduced echelon form version note that adjusting the parametriza
69. inear system 1 Produce an equation not implied by this system 3x 4y 8 224 y 3 Section III Reduced Echelon Form 61 2 Can any equation be derived from an inconsistent system 2 27 Extend the definition of row equivalence to linear systems Under your defi nition do equivalent systems have the same solution set Hoffman amp Kunze X 2 28 In this matrix 12 3 3 0 3 145 the first and second columns add to the third a Show that remains true under any row operation b Make a conjecture c Prove that it holds 62 Chapter One Linear Systems Topic Computer Algebra Systems The linear systems in this chapter are small enough that their solution by hand is easy But large systems are easiest and safest to do on a computer There are special purpose programs such as LINPACK for this job Another popular tool is a general purpose computer algebra system including both commercial packages such as Maple Mathematica or MATLAB or free packages such as SciLab MuPAD or Octave For example in the Topic on Networks we need to solve this o t1 la 0 ii da i 0 19 m i5 0 ig la ig 0 5i 1033 10 2i 4i4 10 511 2ig 5015 0 It can be done by hand but it would take a while and be error prone Using a computer is better We illustrate by solving that system under Maple for another system a user s manual would obviously detail the exact syntax needed
70. inishes this proof is saved for Exercise 21 QED 56 Chapter One Linear Systems We can now prove that each matrix is row equivalent to one and only one reduced echelon form matrix We will find it convenient to break the first half of the argument off as a preliminary lemma For one thing it holds for any echelon form whatever not just reduced echelon form 2 6 Lemma If two echelon form matrices are row equivalent then the leading entries in their first rows lie in the same column The same is true of all the nonzero rows the leading entries in their second rows lie in the same column etc For the proof we rephrase the result in more technical terms Define the form of an m x n matrix to be the sequence 4 5 m where 4 is the column number of the leading entry in row and 4 oo if there is no leading entry in that column The lemma says that if two echelon form matrices are row equivalent then their forms are equal sequences Proor Let Band D be echelon form matrices that are row equivalent Because they are row equivalent they must be the same size say nx m Let the column number of the leading entry in row of B be and let the column number of the leading entry in row j of D be kj We will show that 4 ky that 5 k2 etc by induction This induction argument relies on the fact that the matrices are row equiv alent because the Linear Combination Lemma and its corollary therefore give that each row
71. ivot_row a pivot_row pivot_row for col pivot_row col lt n col a row_below col multiplier a pivot_row col A full analysis of the best way to implement Gauss method is outside the scope of the book see Wilkinson 1965 but the method recommended by most experts is a variation on the code above that first finds the best pivot among the candidates and then scales it to a number that is less likely to give trouble This is scaled partial pivoting Topic Accuracy of Computations 71 In addition to returning a result that is likely to be reliable most well done code will return a number called the conditioning number that describes the factor by which uncertainties in the input numbers could be magnified to become inaccuracies in the results returned see Rice The lesson of this discussion is that just because Gauss method always works in theory and just because computer code correctly implements that method and just because the answer appears on green bar paper doesn t mean that the answer is reliable In practice always use a package where experts have worked hard to counter what can go wrong Exercises 1 Using two decimal places add 253 and 2 3 2 This intersect the lines problem contrasts with the example discussed above e 2y 3 3r 2y 1 Illustrate that in this system some small change in the numbers will produce only a small change in the solution by changing the constant in the bot
72. lead to errors To illustrate how it can lead to wrong conclusions we will use the system r 3y 1 27 y 3 2r 2y 0 from Example 1 12 a Solve the first equation for x and substitute that expression into the second equation Find the resulting y b Again solve the first equation for x but this time substitute that expression into the third equation Find this y What extra step must a user of this method take to avoid erroneously concluding a system has a solution Y 1 19 For which values of k are there no solutions many solutions or a unique solution to this system z y 1 3x 3y k v 1 20 This system is not linear in some sense 2sina cosB 3tany 3 4sina 2cos 2tan y 10 6sina 3 cos 8 tany 9 and yet we can nonetheless apply Gauss method Do so Does the system have a solution v 1 21 What conditions must the constants the b s satisfy so that each of these systems has a solution Hint Apply Gauss method and see what happens to the right side Anton a x 3y b b 1 2r2 323 b 3x y ba 2x1 5x2 3x3 ba x Ty b3 X1 823 b3 2r 4y ba 1 22 True or false a system with more unknowns than equations has at least one solution As always to say true you must prove it while to say false you must produce a counterexample 1 23 Must any Chemistry problem like the one that starts this subsection a bal ance the reaction p
73. n the intuition is that a line per mits motion in one direction a plane permits motion in combinations of two directions etc A linear surface description can be misleading about the dimension this 1 1 2 0 1 2 L i tt o 1 81 6 t s R 2 1 2 is a degenerate plane because it is actually a line 1 1 0 1 L 4 altri o reR 2 1 We shall see in the Linear Independence section of Chapter Two what relation ships among vectors causes the linear surface they generate to be degenerate We finish this subsection by restating our conclusions from the first section in geometric terms First the solution set of a linear system with n unknowns is a linear surface in R Specifically it is a k dimensional linear surface where k is the number of free variables in an echelon form version of the system Second the solution set of a homogeneous linear system is a linear surface passing through the origin Finally we can view the general solution set of any linear system as being the solution set of its associated homogeneous system offset from the origin by a vector namely by any particular solution Exercises v 1 1 Find the canonical name for each vector a the vector from 2 1 to 4 2 in R b the vector from 3 3 to 2 5 in R c the vector from 1 0 6 to 5 0 3 in R d the vector from 6 8 8 to 6 8 8 in R Y 1 2 Decide if the two vectors are equal a the vector from 5 3 to 6 2 and the vector from
74. n A and C However B protested and it was then found that 14 members preferred B over C The Board has not yet recovered from the re sulting confusion Given that every possible order of A B C appeared on at least one ballot how many members voted for B as their first choice Wohascum no 2 1 37 This system of n linear equations with n unknowns said the Great Math ematician has a curious property Good heavens said the Poor Nut What is it Note said the Great Mathematician that the constants are in arithmetic progression It s all so clear when you explain it said the Poor Nut Do you mean like 6x2 9y 12 and 15x 18y 21 Quite so said the Great Mathematician pulling out his bassoon Indeed the system has a unique solution Can you find it Good heavens cried the Poor Nut I am baffled Are you Am Math Mon Jan 1963 1 2 Describing the Solution Set A linear system with a unique solution has a solution set with one element A linear system with no solution has a solution set that is empty In these cases the solution set is easy to describe Solution sets are a challenge to describe only when they contain many elements 2 1 Example This system has many solutions because in echelon form 2x 2 3 x 2r 2 3 xy z 1 y 3 2 z 1 2 31 y 4 3 2 p1 es y 3 2 z 1 2 p2 Pp3 u i z y 3 2 z 1
75. n above in the discussion for the 1958 economy d Solve these equations with the 1958 external demands note the difference in units a 1947 dollar buys about what 1 30 in 1958 dollars buys How far off are the predictions for total output 4 Predict next year s total productions of each of the three sectors of the hypothet ical economy shown below used by used by used by used by farm rail shipping others total value of farm 25 50 100 800 value oj 25 50 50 300 rail value oj 15 10 0 500 shipping if next year s external demands are as stated a 625 for farm 200 for rail 475 for shipping b 650 for farm 150 for rail 450 for shipping 5 This table gives the interrelationships among three segments of an economy see Clark amp Coupe used by used by used by used by total food wholesale retail others en value of food 0 2318 4679 11869 value of wholesale 393 1089 22 459 122 242 value of retail 3 53 75 116041 We will do an Input Output analysis on this system a Fill in the numbers for this year s external demands b Set up the linear system leaving next year s external demands blank c Solve the system where next year s external demands are calculated by tak ing this year s external demands and inflating them 10 Do all three sectors increase their total business by 10 Do they all even increase at the same rate d Solve the system where next year s external demands are calculated by takin
76. not be resolved apart This system is nearly singular in the sense that the two lines are nearly the same line Near singularity gives this system the property that a small change in the system can cause a large change in its solution for instance changing the 3 000 000 01 to 3 00000003 changes the intersection point from 1 1 to 3 0 This system changes radically depending on a ninth digit which explains why the eight place computer has trouble A problem that is very sensitive to inaccuracy or uncertainties in the input values is ill conditioned The above example gives one way in which a system can be difficult to solve on a computer It has the advantage that the picture of nearly equal lines gives a memorable insight into one way that numerical difficulties can arise Unfortunately this insight isn t very useful when we wish to solve some large system We cannot typically hope to understand the geometry of an arbitrary large system In addition there are ways that a computer s results may be unreliable other than that the angle between some of the linear surfaces is quite small For an example consider the system below from Hamming 0 001z y 1 z y 0 The second equation gives x y so x y 1 1 001 and thus both variables have values that are just less than 1 A computer using two digits represents the system internally in this way we will do this example in two digit floating point arithmetic but a similar one with
77. ntary row operations are reversible Proor For any matrix A the effect of swapping rows is reversed by swapping them back multiplying a row by a nonzero k is undone by multiplying by 1 k and adding a multiple of row i to row j with 4 4 7 is undone by subtracting the same multiple of row i from row j ARES PIO 4 A AE 1 k pi A gree hpet pj A The i Z j conditions is needed See Exercise 13 QED This lemma suggests that reduces to is misleading where A B we shouldn t think of B as after A or simpler than A Instead we should think of them as interreducible or interrelated Below is a picture of the idea The matrices from the start of this section and their reduced echelon form version are shown in a cluster They are all interreducible these relationships are shown also We say that matrices that reduce to each other are equivalent with respect to the relationship of row reducibility The next result verifies this statement using the definition of an equivalence 1 5 Lemma Between matrices reduces to is an equivalence relation Proor We must check the conditions i reflexivity that any matrix reduces to itself ii symmetry that if A reduces to B then B reduces to A and iii tran sitivity that if A reduces to B and B reduces to C then A reduces to C Reflexivity is easy any matrix reduces to itself in zero row operations That the relationship is symmetric is Lemma 1 4 if A reduc
78. o the zero vector expecting that the size of the one being discussed will be clear from the context 3 5 Example Some homogeneous systems have the zero vector as their only solution 32 2y z 0 N 3T 2y z 0 3T 2y z 0 6r 4y 0 5 22 0 f y 2 0 y z 0 y z 0 2z 0 3 6 Example Some homogeneous systems have many solutions One example is the Chemistry problem from the first page of this book Tx Tz 0 Tx 7z 0 8rd y 5z 2k 0 8 Np P2 y 32 2w 0 y 32 0 y 32 0 3y 6z k 0 3y 6z w 0 7x Tz 0 patps y 32 2w 0 3p2 pa 62 2w 0 15z 5w 0 Tx Tz 0 5 2 p3 pa y 3z2 2w 0 6z 2w 0 0 0 The solution set 1 3 Qu w we R 1 3 w w 1 has many vectors besides the zero vector if we interpret w as a number of molecules then solutions make sense only when w is a nonnegative multiple of 3 We now have the terminology to prove the two parts of Theorem 3 1 The first lemma deals with unrestricted combinations 3 7 Lemma For any homogeneous linear system there exist vectors A ds gt By such that the solution set of the system is ME ef c1 cx ER where k is the number of free variables in an echelon form version of the system Section I Solving Linear Systems 23 Before the proof we will recall the back substitution calculations that were done in the prior subsection Imagine that we have brought a system to this echelon form
79. o echelon form versions of a matrix have the same free variables and consequently ii any two echelon form versions have the same number of free variables There is no linear system and no combination of row operations such that say we could solve the system one way and get y and z free but solve it another way and get y and w free or solve it one way and get two free variables while solving it another way yields three We finish now by specializing to the case of reduced echelon form matrices 2 7 Theorem Each matrix is row equivalent to a unique reduced echelon form matrix Proor Clearly any matrix is row equivalent to at least one reduced echelon form matrix via Gauss Jordan reduction For the other half that any matrix is equivalent to at most one reduced echelon form matrix we will show that if a matrix Gauss Jordan reduces to each of two others then those two are equal Suppose that a matrix is row equivalent the two reduced echelon form ma trices B and D which are therefore row equivalent to each other The Linear Combination Lemma and its corollary allow us to write the rows of one say B as a linear combination of the rows of the other B c j 161 Cimdm The preliminary result Lemma 2 6 says that in the two matrices the same collection of rows are nonzero Thus if 61 through 6 are the nonzero rows of B then the nonzero rows of D are 6 through 6 Zero rows don t contribute to the sum so we can rewrite t
80. o master this one and the ones later in this chapter before going on to the second chapter Proor First use Gauss method to reduce the homogeneous system to echelon form We will show that each leading variable can be expressed in terms of free variables That will finish the argument because then we can use those free variables as the parameters That is the 3 s are the vectors of coefficients of the free variables as in Example 3 6 where the solution is x 1 3 w y w z 1 3 w and w w We will proceed by mathematical induction which has two steps The base step of the argument will be to focus on the bottom most non 0 0 equation and write its leading variable in terms of the free variables The inductive step of the argument will be to argue that if we can express the leading variables from the bottom t rows in terms of free variables then we can express the leading variable of the next row up the t 1 th row up from the bottom in terms of free variables With those two steps the theorem will be proved because by the base step it is true for the bottom equation and by the inductive step the fact that it is true for the bottom equation shows that it is true for the next one up and then another application of the inductive step implies it is true for third equation up etc More information on mathematical induction is in the appendix 24 Chapter One Linear Systems For the base step consider the
81. of how they arise The first example is from Physics Suppose that we are given three objects one with a mass known to be 2 kg and are asked to find the unknown masses Suppose further that experimentation with a meter stick produces these two balances He 40 gt ja 50 gt me 25 ee 50 15 m we 25 Since the sum of moments on the left of each balance equals the sum of moments on the right the moment of an object is its mass times its distance from the balance point the two balances give this system of two equations 40h 15c 100 25c 50 50h The second example of a linear system is from Chemistry We can mix under controlled conditions toluene C7Hg and nitric acid HNO3 to produce trinitrotoluene CzH5OgNs along with the byproduct water conditions have to be controlled very well indeed trinitrotoluene is better known as TNT In what proportion should those components be mixed The number of atoms of each element present before the reaction xC7Hg yHNOz3 2zC7H506N3 wH20 must equal the number present afterward Applying that principle to the ele 1 2 Chapter One Linear Systems ments C H N and O in turn gives this system a 12 8x ly 5z 2w ly 3z 3y 6z 1w To finish each of these examples requires solving a system of equations In each the equations involve only the first power of the variables This chapter shows how to solve any such system I 1 Gau
82. own in the system for a row that has a leading y and swap it in TY 0 paps y w 0 2 2w 4 2z w 5 Had there been more than one row below the second with a leading y then we could have swapped in any one The rest of Gauss method goes as before TY 0 2ps pa y w 0 z 2w 4 3w 3 Back substitution gives w 1 z 2 y 1 and z 1 Strictly speaking the operation of rescaling rows is not needed to solve linear systems We have included it because we will use it later in this chapter as part of a variation on Gauss method the Gauss Jordan method All of the systems seen so far have the same number of equations as un knowns All of them have a solution and for all of them there is only one solution We finish this subsection by seeing for contrast some other things that can happen 1 11 Example Linear systems need not have the same number of equations as unknowns This system x 3y 1 2r y 3 2x 2y 2 has more equations than variables Gauss method helps us understand this system also since this t 3y 1 2p p2 om 5y 5 2p1TPp3 dy 4 shows that one of the equations is redundant Echelon form 3y 1 4 5 pa ps i cue 0 0 gives y 1 and x 2 The 0 0 is derived from the redundancy That example s system has more equations than variables Gauss method is also useful on systems with more variables than equations M
83. prior example because we do not get a contradictory equa tion 2pitp2 z y 4 0 0 Don t be fooled by the 0 0 equation in that example It is not the signal that a system has many solutions 8 Chapter One Linear Systems 1 15 Example The absence of a 0 0 does not keep a system from having many different solutions This system is in echelon form ys z 0 y z 0 has no 0 0 and yet has infinitely many solutions For instance each of these is a solution 0 1 1 0 1 2 1 2 0 0 0 and 0 7 7 There are infinitely many solutions because any triple whose first component is 0 and whose second component is the negative of the third is a solution Nor does the presence of a 0 0 mean that the system must have many solutions Example 1 11 shows that So does this system which does not have many solutions in fact it has none despite that when it is brought to echelon form it has a 0 0 row 2x 2z 6 2x 2z 6 y z 1 Pp tps y z 1 202 y z 7 mE y z 1 3y 132 0 3y 1 32 20 2r 2z 6 patps y z 1 3p2 pa 0 0 0 3 We will finish this subsection with a summary of what we ve seen so far about Gauss method Gauss method uses the three row operations to set a system up for back substitution If any step shows a contradictory equation then we can stop with the conclusion that the system has no solutions If we reach echelon form without
84. rem states and as discussed at the start of this subsection in this single solution case the general solution results from taking the particular solu tion and adding to it the unique solution of the associated homogeneous system 3 10 Example Also discussed there is that the case where the general solution set is empty fits the General Particular Homogeneous pattern This system illustrates Gauss method x z w 1 5 x z w l 22 y peus a e 22 w 5 r y 3z 2w 1 t 2z 2 shows that it has no solutions The associated homogeneous system of course has a solution x zZ w 0 A 2 z w 0 2r y Be Ey OS p1 P2 p2tps y 2z w 0 cty 3z2 2w 0 76 0 0 In fact the solution set of the homogeneous system is infinite 1 2 1 0 z F 1 1 0 1 w z w R However because no particular solution of the original system exists the general solution set is empty there are no vectors of the form p h because there are no p s 3 11 Corollary Solution sets of linear systems are either empty have one element or have infinitely many elements Proor We ve seen examples of all three happening so we need only prove that those are the only possibilities First notice a homogeneous system with at least one non 0 solution 7 has infinitely many solutions because the set of multiples sU is infinite if s 4 1 then si 7 s 1 is easily seen to be non 0 and
85. rices are row equivalent 1 0 2 1 2 a E a J b E i jJ 0 5 1 5 2 4 2 1 1 10 2 1 1 0 3 1 e 1 am a C 43 1 0 2 10 1 2 225 e 1 isle A 0 1 2 ee lo o xe E oed 1 2 12 Describe the matrices in each of the classes represented in Example 2 10 oO r c DH 2 13 Describe all matrices in the row equivalence class of these 60 Chapter One Linear Systems a t b E J c t 2 14 How many row equivalence classes are there 2 15 Can row equivalence classes contain different sized matrices 2 16 How big are the row equivalence classes a Show that the class of any zero matrix is finite b Do any other classes contain only finitely many members v 2 17 Give two reduced echelon form matrices that have their leading entries in the same columns but that are not row equivalent v 2 18 Show that any two n x n nonsingular matrices are row equivalent Are any two singular matrices row equivalent v 2 19 Describe all of the row equivalence classes containing these a 2x2 matrices b 2x3 matrices c 3x 2 matrices d 3x3 matrices 2 20 a Show that a vector Bo is a linear combination of members of the set B E Ba if and only there is a linear relationship 0 co Bo Tec CnBn where co is not zero Watch out for the o 0 case b Derive Lemma 2 5 v 2 21 Finish the proof of Lemma 2 5 a First illustrate the inductive step by showing that f2 ka b Do the full inductive step assume that cr i
86. roblem have infinitely many solutions v 1 24 Find the coefficients a b and c so that the graph of f x ax ba c passes through the points 1 2 1 6 and 2 3 10 Chapter One Linear Systems 1 25 Gauss method works by combining the equations in a system to make new equations a Can the equation 3x 2y 5 be derived by a sequence of Gaussian reduction steps from the equations in this system r y 1 4x y 6 b Can the equation 5x 3y 2 be derived by a sequence of Gaussian reduction steps from the equations in this system 2r 2y 5 3c y 4 c Can the equation 6x 9y 5z 2 be derived by a sequence of Gaussian reduction steps from the equations in the system 2r y z 4 62 3y 2 5 1 26 Prove that where a b e are real numbers and a 4 0 if ax by c has the same solution set as ax dy e then they are the same equation What if a 0 v 1 27 Show that if ad bc 4 0 then ax by j cx dy k has a unique solution v 1 28 In the system ax by c dx ey f each of the equations describes a line in the zy plane By geometrical reasoning show that there are three possibilities there is a unique solution there is no solution and there are infinitely many solutions 1 29 Finish the proof of Theorem 1 4 1 30 Is there a two unknowns linear system whose solution set is all of R Y 1 31 Are any of the operations used in Gauss method redundant That is can any of t
87. s a representative of its class After that proof we shall as mentioned in the introduction to this section have a way to decide if one matrix can be derived from another by row reduction We just apply the Gauss Jordan procedure to both and see whether or not they come to the same reduced echelon form Exercises Y 1 7 Use Gauss Jordan reduction to solve each system a r y 2 b z z 4 c 3x 2y 1 u y 0 2x 2y 1 62 y 1 2 d 22 y 1 x 3y z 5 y 2z 5 v 1 8 Find the reduced echelon form of each matrix E i1 3 1 1 0 3 1 2 a J b 0 1 c 421 5 f 3 3 34812 0132 d 0 0 5 6 1515 v 1 9 Find each solution set by using Gauss Jordan reduction then reading off the parametrization a 22 y z 1 b x z 1 c z y 2 0 4x y 3 y 22 w 3 y w 0 xz 2y 32 w 7 3z 2y 3z w 0 y w 0 d a 2b 3c d e 1 3a b c d e 3 More information on partitions and class representatives is in the appendix 52 Chapter One Linear Systems 1 10 Give two distinct echelon form versions of this matrix 2 1 1 3 6 4 1 2 151 5 v 1 11 List the reduced echelon forms possible for each size a 2x2 b 2x3 c 3x2 d 3x3 v 1 12 What results from applying Gauss Jordan reduction to a nonsingular matrix 1 13 The proof of Lemma 1 4 contains a reference to the i j condition on the row pivoting operation a The definition of row operations has an j condition on the swap operation
88. s act to transform one matrix into another we consider the effect that they have on the parts of a matrix The crucial observation is that row operations combine the rows linearly 2 1 Definition A linear combination of z4 c 4 is an expression of the form c121 C2 2 ct Cm amp m where the cs are scalars We have already used the phrase linear combination in this book The mean ing is unchanged but the next result s statement makes a more formal definition in order Section III Reduced Echelon Form 53 2 2 Lemma Linear Combination Lemma A linear combination of linear combinations is a linear combination PROOF Given the linear combinations c 1 1 01 425 through cm z Cm n Tn consider a combination of those d c 181 i 24 day emi dod CmnEn where the d s are scalars along with the c s Distributing those d s and regroup ing gives d101 1 1 d c n amp n data 11 dC 101 dmeintn dicia ers do Cm 1 21 dore d c n dmCm n En which is indeed a linear combination of the z s QED In this subsection we will use the convention that where a matrix is named with an upper case roman letter the matching lower case greek letter names the rows Ayres ree By eee Agee Bg ves A B 2 3 Corollary Where one matrix row reduces to another each row of the second is a linear combination of the rows of the first The proof below uses
89. s for neatness Each light is a circle enclosing a loop Light Dome A Brake Switch Light Actuated Offo Switch Door Dimmer CL Actuated E cud Loe N Hi Switch Brake Parking Rear Headlights Lights Lights Lights The designer of such a network needs to answer questions like How much electricity flows when both the hi beam headlights and the brake lights are on Below we will use linear systems to analyze simpler versions of electrical networks For the analysis we need two facts about electricity and two facts about electrical networks The first fact about electricity is that a battery is like a pump it provides a force impelling the electricity to flow through the circuits connecting the bat tery s ends if there are any such circuits We say that the battery provides a potential to flow Of course this network accomplishes its function when as the electricity flows through a circuit it goes through a light For instance when the driver steps on the brake then the switch makes contact and a cir cuit is formed on the left side of the diagram and the electrical current flowing through that circuit will make the brake lights go on warning drivers behind The second electrical fact is that in some kinds of network components the amount of flow is proportional to the force provided by the battery That is for each such component there is a number it
90. s not correct Compare these two systems 32 2y 5 3x 2y 5 4x 2y 4 3r 2y 4 with the same right sides but different left sides The first has a solution but the second does not Thus the constants on the right side of a system don t decide alone whether a solution exists rather it depends on some interaction between the left and right sides For some intuition about that interaction consider this system with one of the coefficients left as the parameter c xz 2y 3z 1 r y z 1 cx 3y 4z 0 If c 2 this system has no solution because the left hand side has the third row as a sum of the first two while the right hand does not If c Z 2 this system has a unique solution try it with c 1 For a system to have a solution if one row of the matrix of coefficients on the left is a linear combination of other rows then on the right the constant from that row must be the same combination of constants from the same rows More intuition about the interaction comes from studying linear combina tions That will be our focus in the second chapter after we finish the study of Gauss method itself in the rest of this chapter Exercises v 3 15 Solve each system Express the solution set using vectors Identify the par ticular solution and the solution set of the homogeneous system a 3r 6y 18 b t y 1 c m 23 4 x 2y 6 z y l 1 2 2x 3 5 Ari T2 5x3 17 d 2a b c 2 e r 2y z
91. s rows Bi 0 4 1 40 Gm For the inductive step assume the inductive hypothesis with t gt 0 if a matrix can be derived from A in or fewer operations then its rows are linear combinations of the A s rows Consider a B that takes t 1 operations Because there are more than zero operations there must be a next to last matrix G so that A ee G B This G is only t operations away from A and so the inductive hypothesis applies to it that is each row of G is a linear combination of the rows of A If the last operation the one from G to B is a row swap then the rows of B are just the rows of G reordered and thus each row of B is also a linear combination of the rows of A The other two possibilities for this last operation that it multiplies a row by a scalar and that it adds a multiple of one row to another both result in the rows of B being linear combinations of the rows of G But therefore by the Linear Combination Lemma each row of B is a linear combination of the rows of A With that we have both the base step and the inductive step and so the proposition follows QED 2 4 Example In the reduction 0 2Y ee 1 lY G 2p 1 1Y e m amp 1 0 1 1 0 2 0 1 0 1 call the matrices A D G and B The methods of the proof show that there are three sets of linear relationships 6 0 a 4 1 a9 y 0 a 4 1 a2 81 1 2 a 1 a2 1 a 0 a2 ya 1 2 a 0 as Bq 1 2 a 0 ag The prior result gives us the insig
92. s zero for 1 lt k lt i 1 and deduce that cx 1 is also zero c Find the contradiction 2 22 Finish the induction argument in Lemma 2 6 a State the inductive hypothesis Also state what must be shown to follow from that hypothesis b Check that the inductive hypothesis implies that in the relationship 6 41 Sr 1 101 Sr42 202 Sr 1 m0m the coefficients Sr 1 1 741 are each zero c Finish the inductive step by arguing as in the base case that r41 lt kr 1 and kr lt 41 are impossible 2 23 Why in the proof of Theorem 2 7 do we bother to restrict to the nonzero rows Why not just stick to the relationship that we began with B ci 1d1 Cijm6m with m instead of r and argue using it that the only nonzero coefficient is Cii which is 1 v 2 24 Three truck drivers went into a roadside cafe One truck driver purchased four sandwiches a cup of coffee and ten doughnuts for 8 45 Another driver purchased three sandwiches a cup of coffee and seven doughnuts for 6 30 What did the third truck driver pay for a sandwich a cup of coffee and a doughnut Trono 2 25 The fact that Gaussian reduction disallows multiplication of a row by zero is needed for the proof of uniqueness of reduced echelon form or else every matrix would be row equivalent to a matrix of all zeros Where is it used v 2 26 The Linear Combination Lemma says which equations can be gotten from Gaussian reduction from a given l
93. ss Method 1 1 Definition A linear equation in variables 11 2 1 has the form 0111 a2 2 a3 3 antn d where the numbers aj R are the equation s coefficients and d R is the constant An n tuple s1 52 Sn IR is a solution of or satisfies that equation if substituting the numbers s1 Sn for the variables gives a true statement a151 4259 Ay Sy d A system of linear equations Q1 1 1 T 41272 T alnn di 02 4311 T 42272 T Q2 nTn da Am 1 1 T Adm 2 2 UE Um nTn Om has the solution s1 s2 Sn if that n tuple is a solution of all of the equa tions in the system 1 2 Example The ordered pair 1 5 is a solution of this system 311 a 2x9 7 21 T2 6 In contrast 5 1 is not a solution Finding the set of all solutions is solving the system No guesswork or good fortune is needed to solve a linear system There is an algorithm that always works The next example introduces that algorithm called Gauss method It transforms the system step by step into one with a form that is easily solved Section I Solving Linear Systems 3 1 3 Example To solve this system 323 9 1 512 2x3 301 IE 2x9 3 we repeatedly transform it until it is in a form that is easy to solve 321 E 2x2 3 swap row 1 with row 3 z1 4 529 223 2 3x3 9 dq 6x2 9 multiply row 1 by 3 21 519 213 2
94. stem will have a solution and the solution will be unique if and only if it reduces to an echelon form system where every variable leads its row which will happen if and only if the associated homogeneous system has a unique solution Thus the question of uniqueness of solution is especially interesting when the system has the same number of equations as variables 3 12 Definition A square matrix is nonsingular if it is the matrix of coeffi cients of a homogeneous system with a unique solution It is singular otherwise that is if it is the matrix of coefficients of a homogeneous system with infinitely many solutions 3 13 Example The systems from Example 3 3 Example 3 5 and Example 3 9 each have an associated homogeneous system with a unique solution Thus these matrices are nonsingular 3 2 1 1 2 1 E X 6 4 0 2 4 0 01 3 28 Chapter One Linear Systems The Chemistry problem from Example 3 6 is a homogeneous system with more than one solution so its matrix is singular 7 0 7 0 8 1 5 2 01 3 0 0 3 6 1 3 14 Example The first of these matrices is nonsingular while the second is singular 1 2 1 2 3 4 36 because the first of these homogeneous systems has a unique solution while the second has infinitely many solutions z 2y 0 zx 2 y 0 3x2 4y 0 3r by 0 We have made the distinction in the definition because a system with the same number of equations as variables behaves in one of two
95. t s solves the system Then s p solves the associated homogeneous system since for each equation index i between 1 and n Gii s1 Pi Gin Sn Pn 05 181 ai 1P1 Gj aD di di 0 where p and s are the j th components of p and s We can write s p as h where h solves the associated homogeneous system to express s in the required pt h form For set inclusion the other way take a vector of the form p h where p solves the system and h solves the associated homogeneous system and note that it solves the given system for any equation index 2 Gi1 pi hi Gin Pn hn 0 1P1 Qi nPn a ihi Ginn d where hy is the j th component of h QED The two lemmas above together establish Theorem 3 1 We remember that theorem with the slogan General Particular Homogeneous 3 9 Example This system illustrates Theorem 3 1 z2 2y z 1 22 4y 2 y 3z 0 Gauss method P z 2y z 1 z 2y z 1 en 22 0 7 y 3z 0 y 3z 0 2z 0 shows that the general solution is a singleton set 1 0 0 More information on equality of sets is in the appendix 26 Chapter One Linear Systems That single vector is of course a particular solution The associated homoge neous system reduces via the same row operations z 2y z 0 cu z c 2y z 0 2z Ay 0 ee y 3z 0 y 3z 0 2z 0 to also give a singleton set 0 roy 0 As the theo
96. tems are easiest to solve on a computer 1 With the steel auto system given above estimate next year s total productions in these cases a Next year s external demands are up 200 from this year for steel and un changed for autos b Next year s external demands are up 100 for steel and up 200 for autos c Next year s external demands are up 200 for steel and up 200 for autos 2 Imagine that a new process for making autos is pioneered The ratio for use of steel by the auto industry falls to 0500 that is the new process is more efficient in its use of steel a How will the predictions for next year s total productions change compared to the first example discussed above i e taking next year s external demands to be 17 589 for steel and 21 243 for autos b Predict next year s totals if in addition the external demand for autos rises to be 21 500 because the new cars are cheaper 3 This table gives the numbers for the auto steel system from a different year 1947 see Leontief 1951 The units here are billions of 1947 dollars used by used by used by steel auto others total value of steal 6 90 1 28 18 69 value of 0 4 40 14 27 autos a Fill in the missing external demands and compute the ratios b Solve for total output if next year s external demands are steel s demand up 10 and auto s demand up 15 Topic Input Output Analysis 67 c How do the ratios compare to those give
97. th the notation defined we can now solve systems in the way that we will use throughout this book 2 13 Example This system y w u 4 z 2w reduces in this way 2 1 0 1 OA A 1 0 Sh 0 4 eck ge ee AA 1 Oe 1 1 4 1 0 1 2 0 0 1 2 1 5 2 0 D 1 0 1 0 d 1 0 1 1 l4 0 1 A 1 2 p2 p3 oOoON ccr The solution set is w 1 2 u 4 w u 3w 1 2 u w u w u E R We write that in vector form x 0 1 1 2 y 4 zi Si Jz 0 3 we 1 2 u wu R w 0 1 0 u 0 0 1 Note again how well vector notation sets off the coefficients of each parameter For instance the third row of the vector form shows plainly that if u is held fixed then z increases three times as fast as w That format also shows plainly that there are infinitely many solutions For example we can fix u as 0 let w range over the real numbers and consider the first component x We get infinitely many first components and hence infinitely many solutions Section I Solving Linear Systems 17 Another thing shown plainly is that setting both w and u to zero gives that this e E ue sg Il oo 0 Ff o is a particular solution of the linear system 2 14 Example In the same way this system u y z 1 3X 2 3 5x 2y 3z 5 reduces Lot Tii ll A 1 E foe do 3 0 113 9 3 cwm 5 0 3 cw Bo 3168 eig s 0 0 39 to a one parameter solution set 1 1 3 to 2 3 z ze R 0 1 Before the exercises we pa
98. the first determinant Contents Chapter One Linear Systems 1 I Solving Linear Systems 2 eee 1 Ll Gauss Method Dou t Red Rd eh Bed 2 Describing the Solution Set lt i e s eera sesadit esa 11 3 General Particular Homogeneous 20 II Linear Geometry of n Space 2 2 2 nn nennen 32 l Venir Space so wu au X 9 aaa A Semen 32 2 Length and Angle Measures o o 38 111 Reduced Behelon Formi c os 6650202445 a sun 46 1 Gaussslordan Reduetib 222 220 x RR n 46 2 Row Equivele eB 2 2 20 oh b oom RS WIR 52 Topic Computer Algebra Systems ll 62 Topic Input Output Analysis oa s coo V XO E es 64 Topic Accuracy of Computations 2 Cm ao e eee 68 Topic Analyzing Networks ooa e e 72 Chapter Two Vector Spaces 79 I Definition of Vector Space o 80 1 Detinition and Examples 22222 2 9395 80 2 Subspaces and Spanning Sets 0 4 91 11 Linear Independence 2 ke eee ee ee e esua 102 1 Definition and Examples 224 102 HL Basis and Dimensi n 222 2 20 20 Kaas ehe 113 Ben ee Seth oh By Bee ane 113 XE or a Sk ae ea WE OE cee aes 119 3 Vector Spaces and Linear Systems 124 4 Combining Subspaces or 2 4 be eee nna EXE os 131 epi Feds nn ee en EEE BY 141 opie oryetele cos nos nn 6 ee a GO ee 143 Topic Voting Paradoxes 2 25229 win bbb E Wo XR aS 147
99. tion by moving the x3 terms to the other side does indeed give the description of this infinite solution set Part of the reason that this works is straightforward While a set can have many parametrizations that describe it e g both of these also describe the above set S take t to be z3 6 and s to be x3 1 9 2 3 4 1 2 3 2 8 3 1 3 U a EE t tER elle s seR 2 0 2 0 nonetheless we have in this book stuck to a convention of parametrizing using the unmodified free variables that is x3 x3 instead of x3 6t We can easily see that a reduced echelon form version of a system is equivalent to a parametrization in terms of unmodified free variables For instance ei 1 0 2 4 lt 0 1 113 T2 3 23 000 0 to move from left to right we also need to know how many equations are in the system So the convention of parametrizing with the free variables by solving each equation for its leading variable and then eliminating that leading variable from every other equation is exactly equivalent to the reduced echelon form conditions that each leading entry must be a one and must be the only nonzero entry in its column Section III Reduced Echelon Form 49 Not as straightforward is the other part of the reason that the reduced echelon form version allows us to read off the parametrization that we would have gotten had we stopped at echelon form and then done back substitution The prior paragraph shows that reduced echelon
100. to form combinations That example shows an infinite solution set conforming to the pattern We can think of the other two kinds of solution sets as also fitting the same pat tern A one element solution set fits in that it has a particular solution and the unrestricted combination part is a trivial sum that is instead of being a combination of two vectors as above or a combination of one vector it is a combination of no vectors A zero element solution set fits the pattern since there is no particular solution and so the set of sums of that form is empty We will show that the examples from the prior subsection are representative in that the description pattern discussed above holds for every solution set Section I Solving Linear Systems 21 3 1 Theorem For any linear system there are vectors f Am Br such that the solution set can be described as P eB eu 01 cr ER where p is any particular solution and where the system has k free variables This description has two parts the particular solution p and also the un restricted linear combination of the Ps We shall prove the theorem in two corresponding parts with two lemmas We will focus first on the unrestricted combination part To do that we consider systems that have the vector of zeroes as one of the particular solutions so that p eh JT c Ckbk can be shortened to c104 exi 3 2 Definition A linear equation is homogeneous if it
101. tom equation to 1 008 and solving Compare it to the solution of the unchanged system 3 Solve this system by hand Rice 0 000 3x 1 556y 1 569 0 345 4x 2 346y 1 018 a Solve it accurately by hand b Solve it by rounding at each step to four significant digits 4 Rounding inside the computer often has an effect on the result Assume that your machine has eight significant digits a Show that the machine will compute 2 3 4 2 3 1 3 as unequal to 2 3 4 2 3 1 3 Thus computer arithmetic is not associative b Compare the computer s version of 1 3 x y 0 and 2 3 x 2y 0 Is twice the first equation the same as the second 5 Ill conditioning is not only dependent on the matrix of coefficients This example Hamming shows that it can arise from an interaction between the left and right sides of the system Let e be a small real 3c 2y z 6 2x 2ey 282 2 4e x 2ey ez l e a Solve the system by hand Notice that the e s divide out only because there is an exact cancelation of the integer parts on the right side as well as on the left b Solve the system by hand rounding to two decimal places and with e 0 001 72 Chapter One Linear Systems Topic Analyzing Networks The diagram below shows some of a car s electrical network The battery is on the left drawn as stacked line segments The wires are drawn as lines shown straight and with sharp right angle
102. tors v Uz and then the sum v V2 extends along the diagonal to the far corner Section II Linear Geometry of n Space 35 The above drawings show how vectors and vector operations behave in R We can extend to R or to even higher dimensional spaces where we have no pictures with the obvious generalization the free vector that if it starts at 1 n ends at b1 b is represented by this column bi ay b an vectors are equal if they have the same representation we aren t too careful to distinguish between a point and the vector whose canonical representation ends at that point Y R v meR Un and addition and scalar multiplication are component wise Having considered points we now turn to the lines In R the line through 1 2 and 3 1 is comprised of the endpoints of the vectors in this set rien That description expresses this picture 9 00 The vector associated with the parameter t has its whole body in the line it is a direction vector for the line Note that points on the line to the left of x 1 are described using negative values of t In R the line through 1 2 1 and 2 3 2 is the set of endpoints of vectors of this form 1 1 de 1 rem 1 1 ah ES 36 Chapter One Linear Systems and lines in even higher dimensional spaces work in the same way If a line uses one parameter so that there is freedom to move back and forth in
103. u 2 0 v v 24 0 lt 2l 7 Section II Linear Geometry of n Space 41 That in turn holds if and only if the relationship obtained by multiplying both sides by the nonnegative numbers and 17 2 el 2 lglg lglg lt 2 El and rewriting 0x al pep 29 8 1315 13141511 is true But factoring 0 lt ala olla ell e viu shows that this certainly is true since it only says that the square of the length of the vector a v is not negative As for equality it holds when and only when 4 is 0 The check that 1d v 14 if and only if one vector is a nonnegative real scalar multiple of the other is easy QED This result supports the intuition that even in higher dimensional spaces lines are straight and planes are flat For any two points in a linear surface the line segment connecting them is contained in that surface this is easily checked from the definition But if the surface has a bend then that would allow for a shortcut shown here grayed while the segment from P to Q that is contained in the surface is solid pagus Because the Triangle Inequality says that in any IR the shortest cut between two endpoints is simply the line segment connecting them linear surfaces have no such bends Back to the definition of angle measure The heart of the Triangle Inequal ity s proof is the g e u v line At first glance
104. use to point out some things that we have yet to do The first two subsections have been on the mechanics of Gauss method Except for one result Theorem 1 4 without which developing the method doesn t make sense since it says that the method gives the right answers we have not stopped to consider any of the interesting questions that arise For example can we always describe solution sets as above with a particular solution vector added to an unrestricted linear combination of some other vec tors The solution sets we described with unrestricted parameters were easily seen to have infinitely many solutions so an answer to this question could tell us something about the size of solution sets An answer to that question could also help us picture the solution sets in R or in R3 etc Many questions arise from the observation that Gauss method can be done in more than one way for instance when swapping rows we may have a choice of which row to swap with Theorem 1 4 says that we must get the same solution set no matter how we proceed but if we do Gauss method in two different ways must we get the same number of free variables both times so that any two solution set descriptions have the same number of parameters Must those be the same variables e g is it impossible to solve a problem one way and get y and w free or solve it another way and get y and z free 18 Chapter One Linear Systems In the rest of this chapter we
105. vious section because we shall show below how to describe such an object with a column vector We can draw a vector as having some length and pointing somewhere p There is a subtlety here these vectors are equal even though they start in different places because they have equal lengths and equal directions Again those vectors are not just alike they are equal How can things that are in different places be equal Think of a vector as representing a displacement vector is Latin for carrier or traveler These squares undergo the same displacement despite that those displacements start in different places BE ne Sometimes to emphasize this property vectors have of not being anchored they are referred to as free vectors Thus these free vectors are equal as each is a displacement of one over and two up YA More generally vectors in the plane are the same if and only if they have the same change in first components and the same change in second components the vector extending from a4 a2 to b1 b2 equals the vector from c1 c9 to di d2 if and only if bj a1 dy c and bz ag da cs An expression like the vector that were it to start at a1 a2 would extend to b1 b2 is awkward We instead describe such a vector as by ay bz aa so that for instance the one over and two up arrows shown above picture this vector 1 2 34
106. ways depending on whether its matrix of coefficients is nonsingular or singular A system where the matrix of coefficients is nonsingular has a unique solution for any constants on the right side for instance Gauss method shows that this system r 2y a 3a 4y b has the unique solution x b 2a and y 3a b 2 On the other hand a system where the matrix of coefficients is singular never has a unique solutions it has either no solutions or else has infinitely many as with these z 2y 1 a 2y 1 3x 6y 2 32 6y 3 Thus singular can be thought of as connoting troublesome or at least not ideal The above table has two factors We have already considered the factor along the top we can tell which column a given linear system goes in solely by considering the system s left hand side the constants on the right hand side play no role in this factor The table s other factor determining whether a particular solution exists is tougher Consider these two 3r 2y 5 3r 2y 5 3r 2y 5 3x 2y 4 with the same left sides but different right sides Obviously the first has a solution while the second does not so here the constants on the right side Section I Solving Linear Systems 29 decide if the system has a solution We could conjecture that the left side of a linear system determines the number of solutions while the right side determines if solutions exist but that guess i

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