Numerical Linear Algebra Lecture notes

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Linear Algebra Jim Hefferon ¡¢ 1 3 ¡¢ 2 1 ¯ ¯ ¯ ¯ 1 2 ¯ ¯ ¯ ¯ 3 1 ¡¢ 1 x ¢ 1 3 ¡¢ 2 1 ¯ ¯ ¯ ¯ x¢1 2 ¯ ¯ ¯ ¯ x¢3 1 ¡¢ 6 8 ¡¢ 2 1 ¯ ¯ ¯ ¯ 6 2 ¯ ¯ ¯ ¯ 8 1Chapter One Linear Systems I Solving Linear Systems Systemsoflinearequationsarecommoninscienceandmathematics. Thesetwo examples from high school science Onan give a sense 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. 40 50 25 50 c c 2 2 h h 15 25 Sincethesumofmomentsontheleftofeachbalanceequalsthesumofmoments 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 C H and nitric acid HNO to produce 7 8 3 trinitrotoluene C H O N along with the byproduct water (conditions have to 7 5 6 3 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 xC H + yHNO ¡ zC H O N + wH O 7 8 3 7 5 6 3 2 must equal the number present afterward. Applying that principle to the ele- 12 Chapter One. Linear Systems ments C, H, N, and O in turn gives this system. 7x=7z 8x+1y =5z+2w 1y =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 Gauss’ Method 1.1 Definition A linear equation in variables x ;x ;:::;x has the form 1 2 n a x +a x +a x +¢¢¢+a x =d 1 1 2 2 3 3 n n where the numbers a ;:::;a 2 R are the equation’s coefficients and d 2 R 1 n n is the constant. An n-tuple (s ;s ;:::;s )2R is a solution of, or satisfies, 1 2 n that equation if substituting the numbers s , ..., s for the variables gives a 1 n true statement: a s +a s +:::+a s =d. 1 1 2 2 n n A system of linear equations a x + a x +¢¢¢+ a x = d 1;1 1 1;2 2 1;n n 1 a x + a x +¢¢¢+ a x = d 2;1 1 2;2 2 2;n n 2 . . . a x +a x +¢¢¢+a x =d m;1 1 m;2 2 m;n n m has the solution (s ;s ;:::;s ) if that n-tuple is a solution of all of the equa- 1 2 n tions in the system. 1.2 Example The ordered pair (¡1;5) is a solution of this system. 3x +2x =7 1 2 ¡x + x =6 1 2 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 3x =9 3 x +5x ¡2x =2 1 2 3 1 x +2x =3 1 2 3 we repeatedly transform it until it is in a form that is easy to solve. 1 x +2x =3 1 2 3 swap row 1 with row 3 ¡ x +5x ¡2x =2 1 2 3 3x =9 3 x +6x =9 1 2 multiply row 1 by 3 x +5x ¡2x =2 ¡ 1 2 3 3x =9 3 x + 6x = 9 1 2 add¡1 times row 1 to row 2 ¡ ¡x ¡2x =¡7 2 3 3x = 9 3 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 x = 3. Substituting 3 for x in the middle equation shows that x = 1. 3 3 2 Substitutingthosetwointothetopequationgivesthatx =3andsothesystem 1 has a unique solution: the solution set is f(3;1;3)g. 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 adding ¡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 Proof. 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. a x + a x +¢¢¢ a x = d a x + a x +¢¢¢ a x = d 1;1 1 1;2 2 1;n n 1 1;1 1 1;2 2 1;n n 1 . . . . . . a x + a x +¢¢¢ a x = d a x + a x +¢¢¢ a x = d i;1 1 i;2 2 i;n n i j;1 1 j;2 2 j;n n j . . . . ¡ . . a x + a x +¢¢¢ a x = d a x + a x +¢¢¢ a x = d j;1 1 j;2 2 j;n n j i;1 1 i;2 2 i;n n i . . . . . . a x +a x +¢¢¢ a x =d a x +a x +¢¢¢ a x =d m;1 1 m;2 2 m;n n m m;1 1 m;2 2 m;n n m The n-tuple (s ;::: ;s ) satisfies the system before the swap if and only if 1 n substituting the values, the s’s, for the variables, the x’s, gives true statements: a s +a s +¢¢¢+a s =d and... a s +a s +¢¢¢+a s =d and... 1;1 1 1;2 2 1;n n 1 i;1 1 i;2 2 i;n n i a s +a s +¢¢¢+a s =d and ... a s +a s +¢¢¢+a s =d . j;1 1 j;2 2 j;n n j m;1 1 m;2 2 m;n n m In a requirement consisting of statements and-ed together we can rearrange theorderofthestatements,sothatthisrequirementismetifandonlyifa s + 1;1 1 a s +¢¢¢+a s = d and ... a s +a s +¢¢¢+a s = d and ... 1;2 2 1;n n 1 j;1 1 j;2 2 j;n n j a s +a s +¢¢¢+a s =d and ... a s +a s +¢¢¢+a s =d . i;1 1 i;2 2 i;n n i m;1 1 m;2 2 m;n n m Thisisexactlytherequirementthat(s ;::: ;s )solvesthesystemaftertherow 1 n swap. QED 1.5 Definition 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 i’ by ‘½ ’. For i instance, we will denote a pivot operation by k½ + ½ , with the row that is i j changed written second. We will also, to save writing, often list pivot steps together when they use the same ½ . i 1.6 Example A typical use of Gauss’ method is to solve this system. x+ y =0 2x¡ y+3z=3 x¡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 2x, 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. x+ y =0 ¡2½ +½ 1 2 ¡ ¡3y+3z=3 ¡½ +½ 1 3 ¡3y¡ z=3Section I. Solving Linear Systems 5 (Note that the two ½ steps¡2½ +½ and¡½ +½ are written as one opera- 1 1 2 1 3 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. x+ y =0 ¡½ +½ 2 3 ¡ ¡3y+ 3z=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. 5=4½ +½ 40h+15c=100 1 2 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 x+ y+ z=9 x+ y+ z= 9 ¡2½ +½ 1 2 2x+4y¡3z=1 ¡ 2y¡5z=¡17 ¡3½ +½ 1 3 3x+6y¡5z=0 3y¡8z=¡27 x+ y+ z= 9 ¡(3=2)½ +½ 2 3 ¡ 2y¡ 5z= ¡17 ¡(1=2)z=¡(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. 1.9Definition Ineachrow,thefirstvariablewithanonzerocoefficientisthe 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. Hereisalinearsystemthatrequirestheoperationofswappingequations. After the first pivot x¡ y =0 x¡y =0 ¡2½ +½ 2x¡2y+ z+2w=4 1 2 z+2w=4 ¡ y + w=0 y + w=0 2z+ w=5 2z+ w=56 Chapter One. Linear Systems the second equation has no leading y. To get one, we look lower down in the system for a row that has a leading y and swap it in. x¡y =0 ½ ½ 2 3 y + w=0 ¡ z+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. x¡y = 0 ¡2½ +½ 3 4 y + w= 0 ¡ z+ 2w= 4 ¡3w=¡3 Back-substitution gives w =1, z =2 , y =¡1, and x=¡1. Strictlyspeaking,theoperationofrescalingrowsisnotneededtosolvelinear 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. Wefinishthissubsectionbyseeingforcontrastsomeotherthingsthat can happen. 1.11 Example Linear systems need not have the same number of equations as unknowns. This system x+3y= 1 2x+ y=¡3 2x+2y=¡2 has more equations than variables. Gauss’ method helps us understand this system also, since this x+ 3y= 1 ¡2½ +½ 1 2 ¡ ¡5y=¡5 ¡2½ +½ 1 3 ¡4y=¡4 shows that one of the equations is redundant. Echelon form x+ 3y= 1 ¡(4=5)½ +½ 2 3 ¡ ¡5y=¡5 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. Many 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. x+3y= 1 x+ 3y= 1 ¡2½ +½ 1 2 2x+ y=¡3 ¡ ¡5y=¡5 ¡2½ +½ 1 3 2x+2y= 0 ¡4y=¡2 Here the system is inconsistent: no pair of numbers satisfies all of the equations simultaneously. Echelon form makes this inconsistency obvious. x+ 3y= 1 ¡(4=5)½ +½ 2 3 ¡ ¡5y=¡5 0= 2 The solution set is empty. 1.13Example Thepriorsystemhasmoreequationsthanunknowns,butthat 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 necessaryforinconsistency, asisillustratedbythisinconsistentsystemwiththe same number of equations as unknowns. ¡2½ +½ x+2y=8 1 2 x+2y= 8 ¡ 2x+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 x+ y=4 2x+2y=8 any pair of numbers satisfying the first equation automatically satisfies the sec- ¯ ¯ ond. The solution set f(x;y) x + y = 4g 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 prior example because we do not get a contradictory equa- tion. ¡2½ +½ x+y=4 1 2 ¡ 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 x+y+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;¡¼;¼). 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 manysolutions—infactithasnone—despitethatwhenitisbroughttoechelon form it has a ‘0=0’ row. 2x ¡2z=6 2x ¡2z=6 y+ z=1 ¡½ +½ y+ z=1 1 3 ¡ 2x+ y¡ z=7 y+ z=1 3y+3z=0 3y+3z=0 2x ¡2z= 6 ¡½ +½ y+ z= 1 2 3 ¡ ¡3½ +½ 0= 0 2 4 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 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 X 1.16 Use Gauss’ method to find the unique solution for each system. x ¡z=0 2x+3y= 13 (a) (b) 3x+y =1 x¡ y=¡1 ¡x+y+z=4 X 1.17 Use Gauss’ method to solve each system or conclude ‘many solutions’ or ‘no solutions’.Section I. Solving Linear Systems 9 (a) 2x+2y=5 (b) ¡x+y=1 (c) x¡3y+ z= 1 x¡4y=0 x+y=2 x+ y+2z=14 (d) ¡x¡ y=1 (e) 4y+z=20 (f) 2x + z+w= 5 ¡3x¡3y=2 2x¡2y+z= 0 y ¡w=¡1 x +z= 5 3x ¡ z¡w= 0 x+ y¡z=10 4x+y+2z+w= 9 X 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 thesolutionisderived,andthenback-substitutioncanbedone. Thismethodboth takes longer than Gauss’ method, since it involves more arithmetic operations and is more likely to lead to errors. To illustrate how it can lead to wrong conclusions, we will use the system x+3y= 1 2x+ y=¡3 2x+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? X 1.19 For which values of k are there no solutions, many solutions, or a unique solution to this system? x¡ y=1 3x¡3y=k X 1.20 This system is not linear, in some sense, 2sin®¡ cos¯+3tan°= 3 4sin®+2cos¯¡2tan°=10 6sin®¡3cos¯+ tan°= 9 and yet we can nonetheless apply Gauss’ method. Do so. Does the system have a solution? X 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) x +2x +3x =b 1 1 2 3 1 3x+ y=b 2x +5x +3x =b 2 1 2 3 2 x+7y=b x +8x =b 3 1 3 3 2x+4y=b 4 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 problem—have infinitely many solutions? 2 X 1.24 Findthecoefficientsa,b,andcsothatthegraphoff(x)=ax +bx+cpasses 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) Cantheequation3x¡2y =5bederived,byasequenceofGaussianreduction steps, from the equations in this system? x+y=1 4x¡y=6 (b) Cantheequation5x¡3y =2bederived,byasequenceofGaussianreduction steps, from the equations in this system? 2x+2y=5 3x+ 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? 2x+ y¡z=4 6x¡3y+z=5 1.26 Prove that, where a;b;:::;e are real numbers and a6=0, if ax+by =c has the same solution set as ax+dy =e then they are the same equation. What if a=0? X 1.27 Show that if ad¡bc6=0 then ax+ by= j cx+dy=k has a unique solution. X 1.28 In the system ax+by= c dx+ey=f each of the equations describes a line in the xy-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. 2 1.30 Is there a two-unknowns linear system whose solution set is all ofR ? X 1.31 Are any of the operations used in Gauss’ method redundant? That is, can any of the operations be synthesized from the others? 1.32 ProvethateachoperationofGauss’methodisreversible. Thatis,showthatif two systems are related by a row operation S S then there is a row operation 1 2 to go back S S . 2 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) 23 (d) 29 (e) 17 Con. Prob. 1955 ?X 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 theother9preferredB overA). Similarly, itwasfoundthat12memberspreferred C overA. Given these results, it was suggested that B should withdraw, to enable a runoff election between A and C. However, B protested, and it was then found that14memberspreferredB overC TheBoardhasnotyetrecoveredfromthere- sulting confusion. Given that every possible order of A, B, C appeared on at least oneballot,howmanymembersvotedforB astheirfirstchoice? 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 6x+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 I.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 +z=3 2x + z= 3 ¡(1=2)½ +½ 1 2 x¡y¡z=1 ¡ ¡y¡(3=2)z=¡1=2 ¡(3=2)½ +½ 1 3 3x¡y =4 ¡y¡(3=2)z=¡1=2 2x + z= 3 ¡½ +½ 2 3 ¡ ¡y¡(3=2)z=¡1=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,thesolutionsetf(x;y;z) 2x+z =3 and x¡y¡z =1 and 3x¡y =4g12 Chapter One. Linear Systems ¯ ¯ can also be described as f(x;y;z) 2x+z =3 and¡y¡3z=2=¡1=2g. 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- ¯ ¯ scribedasf(x;y;z)=((3=2)¡(1=2)z;(1=2)¡(3=2)z;z) z2Rg. Forinstance, (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. Theadvantageofthisdescriptionovertheonesaboveisthattheonlyvariable 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. Intheechelonformsystemderivedintheaboveexample, xandy areleading variables and z is free. 2.3 Example A linear system can end with more than one variable free. This row reduction x+ y+ z¡ w= 1 x+ y+ z¡ w= 1 y¡ z+ w=¡1 ¡3½ +½ y¡ z+ w=¡1 1 3 ¡ 3x +6z¡6w= 6 ¡3y+3z¡3w= 3 ¡y+ z¡ w= 1 ¡y+ z¡ w= 1 x+y+z¡w= 1 3½ +½ y¡z+w=¡1 2 3 ¡ ½ +½ 0= 0 2 4 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 + z ¡ w. Next, moving up to the top equation, substituting for y in the first equation x+(¡1+z¡w)+z¡ w = 1 and solving for x yields x = 2¡ 2z + 2w. Thus, the solution set is ¯ ¯ f2¡2z+2w;¡1+z¡w;z;w) z;w2Rg. 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 2x¡2y =0 2x¡2y =0 ¡(3=2)½ +½ z+3w=2 1 3 z+3w=2 ¡ 3x¡3y =0 0=0 ¡(1=2)½ +½ 1 4 x¡ y+2z+6w=4 2z+6w=4 2x¡2y =0 ¡2½ +½ 2 4 z+3w=2 ¡ 0=0 0=0 ¯ ¯ x andz lead, y andw are free. The solution set isf(y;y;2¡3w;w) y;w2Rg. For instance, (1;1;2;0) satisfies the system—take y = 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 parametersarey andz. Noticethatwecouldnothaveparamatrizedwithxand y, so there is sometimes a restriction on the choice of parameters. The terms ‘parameter’and‘free’arerelatedbecause, asweshallshowlaterinthischapter, 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. x+2y =1 x+ 2y =1 ¡2½ +½ 1 2 2x +z =2 ¡ ¡4y+z =0 ¡3½ +½ 1 3 3x+2y+z¡w=4 ¡4y+z¡w=1 x+ 2y =1 ¡½ +½ 2 3 ¡4y+z =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. 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 isf(1¡(1=2)z;(1=4)z;z;¡1) z2Rg. 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 m£n 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. Each entry is denoted by the corresponding lower-case letter, e.g. a is the number in row i i;j and column j of the array. For instance, µ ¶ 1 2:2 5 A= 3 4 ¡7 has two rows and three columns, and so is a 2£3 matrix. (Read that “two- by-three”; the number of rows is always stated first.) The entry in the second row and first column is a =3. Note that the order of the subscripts matters: 2;1 a 6= a since a = 2:2. (The parentheses around the array are a typo- 1;2 2;1 1;2 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 x +2x =4 1 2 x ¡ x =0 2 3 x +2x =4 1 3 with this matrix. 0 1 1 2 0 4 A 0 1 ¡1 0 1 0 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 parts, we call it an augmented matrix. In this notation, Gauss’ method goes this way. 0 1 0 1 0 1 1 2 0 4 1 2 0 4 1 2 0 4 ¡½ +½ 2½ +½ 1 3 2 3 A A A 0 1 ¡1 0 ¡ 0 1 ¡1 0 ¡ 0 1 ¡1 0 1 0 2 4 0 ¡2 2 0 0 0 0 0 The second row stands for y¡z =0 and the first row stands for x+2y =4 so ¯ ¯ the solution set isf(4¡2z;z;z) z2Rg. One advantageof 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 f(2¡2z+2w;¡1+z¡w;z;w) z;w2Rg from Ex- ample2.3ishardtoread. Wewillrewriteittogroupalltheconstantstogether, all the coefficients of z together, and all the coefficients of w together. We will write them vertically, in one-column wide matrices. 0 1 0 1 0 1 2 ¡2 2 B C B C B C ¯ ¡1 1 ¡1 B C B C B C ¯ f + ¢z+ ¢w z;w2Rg A A A 0 1 0 0 0 1Section I. Solving Linear Systems 15 For instance, the top line says that x = 2¡2z+2w. The next section gives a geometric interpretation that will help us picture the solution sets when they are written in this way. 2.8Definition Avector (orcolumn vector)isamatrixwithasinglecolumn. 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 anarrow: a,b,... or® ,¯,... (boldfaceisalsocommon: aor®). Forinstance, this is a column vector with a third component of 7. 0 1 1 A v = 3 7 2.9 Definition The linear equation a x + a x + ¢¢¢ + a x = d with 1 1 2 2 n n unknowns x ;::: ;x is satisfied by 1 n 0 1 s 1 B C . . s= A . s n if a s +a s +¢¢¢ +a s =d. A vector satisfies a linear system if it satisfies 1 1 2 2 n n 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. 0 1 0 1 0 1 u v u +v 1 1 1 1 B C B C B C . . . u+v = . + . = . A A A . . . u v u +v n n n n In general, two matrices with the same number of rows and the same number of columns add in this way, entry-by-entry. 2.11Definition Thescalarmultiplication oftherealnumberrandthevector v is this. 0 1 0 1 v rv 1 1 B C B C . . r¢v =r¢ . = . A A . . v rv n n 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¢v orv¢r, or without the ‘¢’ symbol: rv. (Do not refer to scalar multiplication as ‘scalar product’ because that name is used for a different operation.) 2.12 Example 0 1 0 1 0 1 0 1 0 1 0 1 1 7 2 3 2+3 5 B C B C 4 28 A A A A B C B C 3 + ¡1 = 3¡1 = 2 7¢ = A A ¡1 ¡7 1 4 1+4 5 ¡3 ¡21 Notice that the definitions of vector addition and scalar multiplication agree where they overlap, for instance, v+v =2v. Withthenotationdefined, wecannowsolvesystemsinthewaythatwewill use throughout this book. 2.13 Example This system 2x+y ¡ w =4 y + w+u=4 x ¡z+2w =0 reduces in this way. 0 1 0 1 2 1 0 ¡1 0 4 2 1 0 ¡1 0 4 ¡(1=2)½ +½ 1 3 A A 0 1 0 1 1 4 ¡ 0 1 0 1 1 4 1 0 ¡1 2 0 0 0 ¡1=2 ¡1 5=2 0 ¡2 0 1 2 1 0 ¡1 0 4 (1=2)½ +½ 2 3 A 0 1 0 1 1 4 ¡ 0 0 ¡1 3 1=2 0 ¯ ¯ The solution set is f(w+(1=2)u;4¡w¡u;3w+(1=2)u;w;u) w;u2Rg. We write that in vector form. 0 1 0 1 0 1 0 1 x 0 1 1=2 B C B C B C B C y 4 ¡1 ¡1 B C B C B C B C ¯ B C B C B C B C ¯ f z = 0 + 3 w+ 1=2 u w;u2Rg B C B C B C B C A A A A 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 0 1 0 1 x 0 B C B C y 4 B C B C B C B C z = 0 B C B C A A w 0 u 0 is a particular solution of the linear system. 2.14 Example In the same way, this system x¡ y+ z=1 3x + z=3 5x¡2y+3z=5 reduces 0 1 0 1 0 1 1 ¡1 1 1 1 ¡1 1 1 1 ¡1 1 1 ¡3½ +½ ¡½ +½ 1 2 2 3 A A A 3 0 1 3 ¡ 0 3 ¡2 0 ¡ 0 3 ¡2 0 ¡5½ +½ 1 3 5 ¡2 3 5 0 3 ¡2 0 0 0 0 0 to a one-parameter solution set. 0 1 0 1 1 ¡1=3 ¯ A A ¯ f 0 + 2=3 z z2Rg 0 1 Before the exercises, we pause 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. Forexample,canwealwaysdescribesolutionsetsasabove,withaparticular 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 2 3 also help us picture the solution sets, inR , or inR , 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 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 X 2.15 Find the indicated entry of the matrix, if it is defined. µ ¶ 1 3 1 A= 2 ¡1 4 (a) a (b) a (c) a (d) a 2;1 1;2 2;2 3;1 X 2.16 Give the size of each matrix. à µ ¶ µ ¶ 1 1 1 0 4 5 10 (a) (b) ¡1 1 (c) 2 1 5 10 5 3 ¡1 X 2.17 Do the indicated vector operation, if it is defined. à à à à µ ¶ µ ¶ µ ¶ 2 3 1 3 4 2 3 (a) 1 + 0 (b) 5 (c) 5 ¡ 1 (d) 7 +9 ¡1 1 5 1 4 1 1 à à à à µ ¶ 1 3 2 1 1 (e) + 2 (f) 6 1 ¡4 0 +2 1 2 3 1 3 5 X 2.18 Solve each system using matrix notation. Express the solution using vec- tors. (a) 3x+6y=18 (b) x+y= 1 (c) x + x = 4 1 3 x+2y= 6 x¡y=¡1 x ¡x +2x = 5 1 2 3 4x ¡x +5x =17 1 2 3 (d) 2a+b¡c=2 (e) x+2y¡z =3 (f) x +z+w=4 2a +c=3 2x+ y +w=4 2x+y ¡w=2 a¡b =0 x¡ y+z+w=1 3x+y+z =7 X 2.19 Solve each system using matrix notation. Give each solution set in vector notation. (a) 2x+y¡z=1 (b) x ¡ z =1 (c) x¡ y+ z =0 4x¡y =3 y+2z¡w=3 y +w=0 x+2y+3z¡w=7 3x¡ 2y+3z+w=0 ¡y ¡w=0 (d) a+2b+3c+d¡e=1 3a¡ b+ c+d+e=3 X 2.20 The vector is in the set. What value of the parameters produces that vec- tor? µ ¶ µ ¶ ¯ 5 1 ¯ (a) ,f k k2Rg ¡5 ¡1 à à à ¡1 ¡2 3 ¯ ¯ (b) 2 ,f 1 i+ 0 j i;j2Rg 1 0 1Section I. Solving Linear Systems 19 à à à 0 1 2 ¯ ¯ (c) ¡4 ,f 1 m+ 0 n m;n2Rg 2 0 1 2.21 Decide if the vector is in the set. µ ¶ µ ¶ ¯ 3 ¡6 ¯ (a) ,f k k2Rg ¡1 2 µ ¶ µ ¶ ¯ 5 5 ¯ (b) ,f j j2Rg 4 ¡4 à à à 2 0 1 ¯ ¯ (c) 1 ,f 3 + ¡1 r r2Rg ¡1 ¡7 3 à à à 1 2 ¡3 ¯ ¯ (d) 0 ,f 0 j+ ¡1 k j;k2Rg 1 1 1 2.22 Paramatrize the solution set of this one-equation system. x +x +¢¢¢+x =0 1 2 n X 2.23 (a) Apply Gauss’ method to the left-hand side to solve x+2y ¡ w=a 2x +z = b x+ 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 +z = 1 x+ y +2w=¡2 X 2.24 Why is the comma needed in the notation ‘a ’ for matrix entries? i;j X 2.25 Give the 4£4 matrix whose i;j-th entry is (a) i+j; (b) ¡1 to the i+j power. trans 2.26 For any matrix A, the transpose of A, written A , is the matrix whose columns are the rows of A. Find the transpose of each of these. à µ ¶ µ ¶ µ ¶ 1 1 2 3 2 ¡3 5 10 (a) (b) (c) (d) 1 4 5 6 1 1 10 5 0 2 X 2.27 (a) Describe all functions f(x) = ax + bx + c such that f(1) = 2 and f(¡1)=6. 2 (b) Describe all functions f(x)=ax +bx+c such that f(1)=2. 2 2.28 Show that any set of five points from the plane R lie on a common conic 2 2 section, that is, they all satisfy some equation of the form ax +by +cxy+dx+ 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. ? 2.30 (a) Solve the system of equations. 2 ax+ y=a x+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?

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