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# Linear Algebra MAT 022A

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This 58 page Class Notes was uploaded by Otilia Murray I on Tuesday September 8, 2015. The Class Notes belongs to MAT 022A at University of California - Davis taught by Staff in Fall. Since its upload, it has received 96 views. For similar materials see /class/187425/mat-022a-university-of-california-davis in Mathematics (M) at University of California - Davis.

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SOLUTIONS OF THEORETICAL EXERCISES selected from INTRODUCTORY LINEAR ALGEBRA WITH APPLICATIONS B KOLMAN D R HILL Eighth Edition7 Prentice Hall7 2005 DR GRIGORE CALUGAREANU Department of Mathematics and Computer Sciences The Algebra Group Kuwait University 2006 Contents Preface List of symbols H Matrices 3 Determinants 4 n Vectors 5 Lines and Planes 6 Vector Spaces 8 Diagonalization References iii 11 29 37 41 45 55 59 CONTENTS Preface Back in 1997 somebody asked in the Mathematics Department 77Why are the results in 111 Course Linear Algebra so bad77 The solution was to cancel some sections of the 6 chapters selected for this one semester course The solutions of some of the so called theoretical Exercises were to be covered in the lectures But this takes time and less time remains for covering the long material in these 6 chapters Our collection of solutions is intended to help the students in 111 course and provides to the lecturers a precious additional time in order to cover carefully all the large number of notions results examples and procedures to be taught in the lectures Moreover this collection yields all the solutions of the Chapter Tests and as a Bonus some special Exercises to be solved by the students in their Home Work Because often these Exercises are required in Midterms and Final Exam the students are warmly encouraged to prepare carefully these solutions and if some of them are not understood to use the Of ce Hours of their teachers for supplementary explanations The author Vi PREFACE List of symbols Symbol FUDNZ Description the set of all positive integer numbers the set of all integer numbers the set of all rational numbers the set of all real numbers for R any of the above numerical sets the set R7 removing zero the set of all n vectors With entries in R the set of all m X n matrices With entries in R the set of all square n X n matrices the set of all permutations of n elements the set of all subsets of M the set of all polynomials of indeterminate X With coe icients in R Viii LIST OF SYMBOLS Second Edition updated to eighth Edition All rights reserved to the Department of Mathematics and Computer Sciences7 Faculty of Science7 Kuwait University 10 Chapter 1 Matrices Page 20 T5 A SQUARE matrix A 127 is called upper triangular if 127 0 for 239 gt j It is called lower triangular if 127 0 for 239 lt j 111 112 lln 0 122 12 0 0 133 13 l 0 0 0 0 aml Upper triangular matrix The entries below the main diagonal are zero 111 0 0 0 121 122 0 0 131 132 133 0 0 0 am 1n2 1n3 Inn 12 CHAPTER 1 MATRICES Lower triangular matrix The entries above the main diagonal are zero a Show that the sum and di erence of two upper triangular matrices is upper triangular b Show that the sum and di erence of two lower triangular matrices is lower triangular c Show that if a matrix is upper and lower triangular then it is a diagonal matrix Solution a As A above let B by be also an upper triangular matrix S AB 327 be the sum and D A7B dij be the di erence of these matrices Then for every 239 gtj we have 32 127 bij 0 0 0 respectively 127 127 7 bij 0 7 0 0 Thus the sum and di erence of two upper triangular matrices is upper triangular 2 3 3 2 1 4 4 4 Example 0 1 2 0 3 2 0 4 4 sum of two 0 0 1 0 0 3 0 0 4 upper triangular matrices which is also upper triangular 0 1 0 0 0 0 0 0 7 1 1 0 1 0 0 di erence of two lower trian 1 1 1 1 2 1 0 atrices which is also lower triangular b Similar c If a matrix is upper and lower triangular then the entries above the main diagonal and the entries below the main diagonal are zero Hence all the entries o the main diagonal are zero and the matrix is diagonal T6 a Show that if A is an upper triangular matrix then AT is lower triangular b Show that if A is an lower triangular matrix then AT is upper triangular Solution a By the de nition of the transpose if AT a5 and A is an upper triangular matrix 127 0 for every 239 gt j and so a 127 0 Hence AT is lower triangular b Similar Page 3738 TA Show that the product of two diagonal matrices is a diagonal matrix Solution Just verify that 111 0 0 bn 0 0 0 122 0 0 bgg 0 i 0 0 am 0 0 b7m 111511 0 0 0 122b22 0 0 0 ambm T5 Show that the product of two scalar matrices is a scalar matrix Solution Just verify that a 0 0 b 0 0 ab 0 0 0 a 0 0 b 0 i 0 ab 0 0 0 a 0 0 b 0 0 ab Short solution Notice that any scalar matrix has the form 11 a 0 a 0 Then7obV1ously alnbln abln shows that prod 0 0 a ucts of scalar matrices are scalar matrices T6 a Show that the product of two upper triangular matrices is an upper triangular matrix 14 CHAPTER 1 MATRICES b Show that the product of two lower triangular matrices is a lower triangular matrix Sketched solution a A direct computation shows that the product of two upper triangular matrices 111 112 am 511 512 bln 0 122 lgn 0 bgg bgn 0 0 am 0 0 bm is upper triangular too Indeed this is 111 511 111511 a12522 aiibin 112527 ainbnn 122522 a22b2n 123537 aznbnn 0 0 ambm an upper triangular matrix Complete solution Let P p27 AB be the product of two upper triangular matrices lf 239 gt j then 1 221 aikbkj aikbkj n 2162 likbkj ailblj li iilbiil j 04239in linbnj SlHCe matrices A and B are upper triangular in the rst sum the as are zero and in the second sum the b 7s are zero Hence p27 0 and P is upper triangular too b Analogous T9 a Show that the j th column of the matrix product AB is equal to the matrix product AcoljB b Show that the i th row of the matrix product AB is equal to the matrix product r0w AB Solution a With usual notations consider A 127 an m X 72 matrix B bij an n X p matrix and C AB 027 the corresponding matrix product an m X p matrix 15 As already seen in the Lectures7 an arbitrary 239j entry 027 in the TL product is given by the dot product rowiA o cobB Zaikbkj or 191 blj bgj anannam o Hence the j th column of the product the matrix bnj C is the following 013 r0w1A o colj B c row A ocol B 2 2lt A AcoljB cm rownA o colj B using the product just do the computation of the initial m X 72 matrix A blj i 52j and the n X 1 matrix cobB bnj b Analogous Page 51 T9 Find a 2 X 2 matrix B a 02 and B a 2 such that AB BA7 1 Where A 0 1 Solution Obviously B A satis es the required conditions indeed7 AA AA7 A a 02 and A 12 Solution for a 7better statement nd all the matrices B With this prop erty We search for B as an unknown matrix Z S J 16 CHAPTER 1 MATRICES Thus 1 i i S i S 5 iandso7 usingthe de nition of matrix equality7 a 20 a b 2d 2a b c c 39 d 20 1 These equalities are equivalent to c 0 and a d Therefore every matrix B of the form 5 S with arbitrary real numbers a b veri es AB BA Example g 5 T13 Show that 71A 7A Solution In the proof of Theorem 11 Properties ofthe matrix addition we took D 71A the scalar multiplication and we have veri ed that AD DA 07 that is 7A71A T123 Let A and B be symmetric matrices a Show that A B is symmetric b Show that AB is symmetric if and only if AB BA Solution a This follows at once from A BT Th391394bgt AT BT A B7 A and B being symmetric above some of the equalities7 their justi cation is given b First observe that AB symmetric matrices A7 B Now7 if AB is symmetric7 ABT AB and thus using the previous equality AB BA Conversely7 if AB BA then ABT BA AB7 that is7 AB is symmetric T26 If A is an n X n matrix7 show that AAT and ATA are symmetric Solution For instance AATT 1138 ATTAT Th391394a AAT7 so that AAT is symmetric Likewise ATA is symmetric T Th148 BTAT BA holds for arbitrary 17 We recall here a DEFINITION given in Exercise T24 a matrix A 127 is called skewsymmetric if AT 7A T27 If A is an n X n matrix7 a Show that A AT is symmetric b Show that A 7 AT is skew symmetric Solution a We have only to observe that A ATT Th391394bgt AT ATT Th14a AT Th11a b Analogously7 A 7 ATT 1103 AT 7 AT AT 7 A Th11a T28 Show that if A is an n X n matrix7 then A can be written uniquely as A S K7 Where S is symmetric and K is skew symmetric Solution Suppose such a decomposition exists Then AT ST KT S7Kso that AAT 2S andA7AT 2K NOW take S AAT and K A7AT One veri es A SK7 S ST and KT 7K similarly to the previous Exercise 26 T Th14a T32 Show that if AX b is a linear system that has more than one solution7 then it has in nitely many solutions Solution Suppose uL a 112 are two di erent solutions of the given linear system For an arbitrary real number 7 7 such that 0 lt r lt 17 consider WT T111 1 7 ru2 This is also a solution of the given system AW Aru1 17 ru2 TA11117 TA112 Tb 17 rb b First observe that WT 111112 lndeed7 WT 111 implies 111 T111 177 1127 17ru17u2 0 and hence 111 1127 acontradiction Similarly7 WT Next7 observe that for 0 lt r73 lt 17 r 3 the corresponding solutions are di erent indeed7 WT W5 implies T111 1 7 ru2 3111 1 7 3u2 and so 7 7 3u1 7 112 07 a contradiction Page 89 T11 Let u and V be solutions to the homogeneous linear system Ax 0 18 CHAPTER 1 MATRICES a Show that u V is a solution b For any scalar 7 show that T11 is a solution c Show that u 7 V is a solution d For any scalars r and 37 show that T11 8V is a solution Remark We have interchanged b and c from the book7 on purpose Solution We use the properties of matrix operations a Au V Th39l ub Au AV 0 0 0 b Am 1 mm r0 0 c We can use our preVious b and a by b7 for r 71 we have 71V 7V is a solution by a u7V u 7 V is a solution d Using twice our b7 T11 and 8V are solutions and7 by a7 T11 8V is a solution T112 Show that if u and V are solutions to the linear system Ax b7 then u 7 V is a solution to the associated homogeneous system Ax 0 Solution Our hypothesis assures that Au b and AV b Hence Au 7 V Th391392bgt Au 7 AV b 7 b 0 and so u 7 V is a solution to the associated homogeneous system Ax 0 Page 86 Bonus Find all values of a for which the resulting linear system has a no solution7 b a unique solution7 and c in nitely many solutions m y 7 2 23 v 2y 2 3 ya2752 a Solution We use Gauss Jordan method The augmented matrix is 1 1 71 1 2 1 3 The rst elementary operations give 1 1 1275 a 1 1 71 2 7R R 1 1 71 2 1 2 1 3 1N 0 1 2 1 11 1275 a 0 0 1274 172 CASE 1 a2 7 4 07 that is a 6 i2 i a 2 Then the last matrix gives nal Step after Step 8 in what follows we refer to the steps in the Procedure p 65 68 the following reduced row echelon form 1 1 1 2 1 0 73 1 0 1 2 1 R Rl 0 1 2 1 0 0 0 0 0 0 0 0 i i It 1 32 i i i The corresponding system is 1 i 22 7 Which has 2 is an arbi trary real number7 in nitely many solutions 1 1 71 2 ii a 72 The last matrix is 0 1 2 1 and so our last 0 0 0 74 equation is 0 X x 0 X y 0 X 2 74 Hence this is an inconsistent system it has no solutions CASE 2 a2 7 4 a 0 ie7 1 i2 In the 1 X 4 submatrix which remains neglecting the rst and second rows7 we must only use Step 4 multiply by m and after this7 the nal Step from REF to RREF 1112 1102 01210101iaT2 1 001a2 001 m 3 1001 0101 001 m Finally the solution depending on a in this case is m 1 7 y 2 1 1imandzm xy2 2 24 23y22 3 23ya2712 11 20 CHAPTER 1 MATRICES 1 1 1 2 Solution The augmented matrix is 2 3 2 5 The rst 2 3 a2 7 1 a 1 elementary operations give 1 1 1 72R1R2 3 1 1 2 2 3 2 5 N 1 0 1 7 2 3 1271 11 0 1 1273 173 and 1 1 1 2 1 0 1 1 0 1 0 1 N 0 1 0 1 0 0 a2 3 a 7 4 0 0 a2 7 3 a 7 4 CASE 1 a2 7 3 a 07 that is 1 Then a2 7 3 is our third pivot and we continue by Step 4 multiply the third row by 2173 1 0 1 1 1 0 0 1752143 0 1 0 1 N 0 1 0 1 0 0 1 543 0 0 1 5243 This is a consistent system With the unique solution m 17 12743 y 17 74 3273 CASE 2 a2 7 3 07 that is a E Hence a 7 4 a 0 and the last equation is 0 X m 0 X y 0 X 2 a 7 4 a 07 an inconsistent system no solutions 2 Pages 105106 Bonus 16 Find all the values of a for Which the inverse of 1 1 0 A 1 0 0 1 2 1 exists What is A 1 7 21 Solution We use the practical procedure for computing the inverse see p 957 textbook 10100RR110100 00010 1i2 307107110 52 20001 0107101 110100 10100 010 1710 R2R3 0101710 R Rl 0107101 0007211 100010 010 1710 OED 0007211 CASE 1 If a 0 then C has a zero row 0 a 3 Hence A is singular that is7 has no inverse CASE 2 If a 0 we use Step 4 multiply the third row by 100010 010 1710 CED 001ill so that A is nonsingular and D A l lNHO 077 gt7 Ell 00 22 If A and B are nonsingular7 are A B7 A 7 B and 7A nonsingular 7 Explain Solution If A7 B are nonsingular generally it is not true that A B or A 7 B are nonsingular lt su ices to give suitable counterexamples for A In and B 7 we have A B 0n7 Which is not invertible see the de nition p 917 and7 for A B7 the di erence A 7 B 0n7 again a non invertible rnatrix 22 CHAPTER 1 MATRICES Finally7 if A is nonsingular we easily Check that 7A is also invertible and 7A 1 7A4 Remark More can be proven see Exercise 207 b for every 0 a 07 if A is riorisirigular7 CA is also nonsirigular and CA 1 Aquotlindeed7 one veri es CAM1 1 Th3 cAA 1 13 cAA 1 1AA 1 In and similarly A 1CA In Chapter Test 1 1 Find all solutions to the linear system 1237243 21233245 72 3 64 Solution Gauss Jordan method is used 111723 1172 2 13 2 5 41 0 11 6 i1 52 071163 07116 111723 1723 1 1 0 171761 Ewes 0171761 0 71 1 6 3 0 0 0 0 4 The corresponding equivalent system has the third equation 0 gtlt10 gtlt20 gtlt30 gtlt447 Which has no solution 2 Find all values of a for Which the resulting linear system has a no solution7 b a unique solution7 and C in nitely many solutions xz4 2xy325 7373ya275a2a78 Solution Gauss Jordan method is used 1 0 1 4 2 1 3 5 72R1E3R1R3 73 73 a2 7 5a a 7 8 1 0 1 4 0 1 1 73 WES 0 73 a2 7 5a 3 a 4 24 CHAPTER 1 MATRICES 1 0 1 4 0 1 1 73 0 0 a275a6 175 As usually we distinguish two cases notice that a275a6a72a730 gta6 273 CASE 1 a 2 or a 3 In both cases a 1 a 0 and so the third equation of the corresponding system is 0 X 0 X y 0 X 2 a 17 With no solution CASE 2 If 1 27 3 then a2 7 5a 6 a 0 and the procedure continues with Step 4 1 0 1 WR3 0 1 1 73 7R3tR2A 75 0 0 1 a275a6 175 1 0 0 4 aw 0 1 0 737aLEM 0 0 1 127511443 With the corresponding equivalent system unique solution 7 7 175 7 7 7 175 7 175 7 4 a275a67 y 7 3 12751114372 7 a275a6 39 3 If possible7 nd the inverse of the following matrix 1 2 71 0 1 1 1 0 71 Solution We use the Practical Procedure see p 957 textbook 1271100 1271100 011010 RSR30110102R R3 1071001 07207101 12711001R1271100 01 1 010 23 01 1 010 7R3R R3R1 00 27121 00171 1 2 0 1 1 0 0 7 1 g 010 1071 2gt112R1 010 071 1 1 1 1 0 0 1 75 1 5 0 0 1 75 1 5 CED Since 0 has no zero rows7 A 1L exists that is7 A is nonsingular and 1 E 1417 i o i 1 1 1 2 2 4 If A 73 J 7 nd all values of A for Which the homogeneous system A12 7 AX 0 has a nontrivial solution Solution The homogeneous system has a nontrivial solution if and only i A 1 if A12 7 A 7 2 A i 2 the practical procedure for nding the inverse CASE 1 A 717 1 a 0 then this is the rst pivot in is singular see Theorem 1137 p 99 Use 1 2 1 0 17mm 1 A2 fl 0 72123122 2 A72 0 1 2 A72 0 1 1 L L 0 1 L L 0 I A11 A31 A2311 A 013 2 A Z m ATll 2 M16 ALH 1 Novv7ifA27A76 A2A730lt AE 7273 thenChasa zero row and A12 7 A is singular as required CASE 2 If A 1 0 then the initial matrix is g 73 1 1 so that using Step 3 we obtain 3 1 3 1 27301N17 05N17505 0 2 1 0 0 2 1 0 0 1 0 26 CHAPTER 1 MATRICES N10 010 Hence the coef cient matrix of the system is a nonsingular matrix with in 3 oh Apia verse 3 and therefore the homogeneous system has a unique trivial 5 solution 1 3 0 2 1 1 5 a If A 1 0 1 1 and B 1 0 0 2 7 compute 1 71 4 1 1 71 AB 1 1 0 2 b Solve Ax b for x if A 1 2 1 3 and b 1 4 2 5 3 3 6 8 Solution a AB 1 Th39l lmb B lA 1L 72 2 78 0 5 3 b If A 1 exists that is7 A is nonsingular then see p98 x A lb 14 25 7 Answer each of the following as true or false a If A and B are n X n matrices7 then ABAB A22ABB2 b H 111 and 112 are solutions to the linear system AX b7 then W iul iug is also a solution to AX b c If A is a nonsingular matrix7 then the homogeneous system AX 0 has a nontrivial solution 1 A homogeneous system of three equations in four unknowns has a nontrivial solution 27 e If A7 B and C are n X n nonsingular matrices7 then ABC 1 CilAilBil Solution a False7 since generally AB BA fails Indeed7 take A 1 andB ThenAB andBA As a matter of fact A FB2 ABAB ABA A Th12c b True7 indeed if AuL b and 14112 b then AW 146111 Th12b Th13d iAul 314112 ib 3b b C False see Theorem 113 p 99 1 True special case of Theorem 18 p77 for m 3 lt 4 n e False a special case of Corollary 12 p 94 gives ABC39 1 C lB lA l 1t su ices to give an example for B lA l a A lB l ThLZ b 112 28 CHAPTER 1 MATRICES Chapter 3 Determinants Page 194195 T3 Show that if c is a real number and A is an n X 72 matrix then detcA c detA Solution You only have to repeatedly use 72 times Theorem 35 p 187 the scalar multiplication of a matrix by 0 consists of the multiplication of the rst row7 of the second row7 7 and the multiplication ofthe n th row7 by the same real number 0 Hence detcA cccA c detA T5 Show that if detAB 0 then detA 0 or detB 0 Solution lndeed7 by Theorem 38 p 191 detAB detA detB 0 and so detA 0 or detB 0 as a zero product of two real numbers T6 ls detAB detBA 7 Justify your answer Solution Yes Generally AB a BA but because of Theorem 38 detAB detA detB detB detA detBA determinants are real numbers T8 Show that if AB In then detA a 0 and detB a 0 Solution Using Theorem 387 1 detln detAB detA detB and so detA a 0 and detB a 0 as a nonzero product of two real numbers 29 30 CHAPTER 3 DETERMINANTS T9 a Show that if A A l then detA i1 b Show that if AT A l then detA i1 Solution a We use Corollary 32 p 191 from A A lL if the inverse exists detA a 0 by Exercise T8 above we derive detA detA 1 deg and so detA2 1 Hence detA i1 b By Theorem 31 p 185 detAT detA One uses now a T10 Show that if A is a nonsingular matrix such that A2 A then detA 1 Solution Using again Theorem 38 detA detA2 detAA detA detA and so detAdetA 7 1 0 If A is nonsingular by Exercise T8 above detA y 0 and so detA71 0 and nally detA 1 T16 Show that if A is n X n with A skew symmetric AT 7A see Section 14 Exercise T24 and n is odd then detA 0 Solution By Theorem 31 detAT detA By the Exercise T3 above det7A det71A 71 detA idetA because 72 is odd Hence detA 7 detA and so 2detA 0 and detA 0 Page 210 T7 Show that if A is singular then ade is singular Solution If A is singular then detA 0 Since Aade detAIn this is Theorem 311 Aade 0 First of all if A 0 then ade O by the de nition of the adjoint matrix all cofactors are zero In the remaining case if A a 0 then ade cannot be nonsingular because otherwise multiplying to the right with ade 1 the equality Aade O we obtain A 0 So 1de is singular T8 Show that if A is an n X 72 matrix then detade detA 1 Solution Use the equality given in Theorem 311 Aade detAln Taking the determinants of both members of the equality one obtains detA detade detA notice that detAln is a scalar matrix hav ing 72 copies of the real number detA on the diagonal Theorem 37 p 188 is used 31 If detA 07 according to the previous Exercise7 together with A7 ade is also singular7 and so detade 0 and the formula holds lf detA a 0 we can divide both members in the last equality by detA and thus detade detA detA detA 1 T10 Let AB AC Show that if detA a 07 then B 0 Solution Using Theorem 312 p 2037 A is nonsingular and so7 A l exists Multiplying AB AC to the left with A 1 we obtain A 1AB A 1AC and nally B C T12 Show that if A is nonsingular7 then ade is nonsingular and 1 UAW detA A adjA 1 Solution First use Exercise T8 previously solved if A is nonsingular then detA a 0 and so detade detA 1L a 0 Thus ade is also nonsingular Eurther7 the formulas in Theorem 3117 Aade adeA detAIn7 give ademA Aade In and so ade 1 mA by de nition Finally7 one can write the equalities given in Theorem 311 for A4 A 1adjA 1 detA 1In detlEAgtIn by Corollary 327 p 191 Hence7 by left multiplication with A7 one nds adjA 1 detlwA detlw Supplementary Exercises Bonus 1 THE PROOF OF THEOREM 111 SECTION 17 Suppose that A and B are n X n matrices a If AB In then BA In b If BA In then AB In PROOF a If AB In7 taking determinants and using suitable prop erties7 detA detB detAB detIn 1 shows that detA a 0 and detB a 0 and so A and B are nonsingular By left multiplication with 32 CHAPTER 3 DETERMINANTS A 1 which exists7 A being nonsingular7 we obtain A 1AB A lln and B A l Hence BA A lA In b Similarly 2 EXERCISE T3 Show that if A is symmetric7 then ade is also symmetric Solution Take an arbitrary cofactor A27 71 7 detMij where Mi is the submatrix of A obtained by deleting the i th row and j th column Observe that the following procedure gives also 11127 1 consider the transpose AT 2 in AT delete the j th row and i th column 3 transpose the resulting submatrix Further7 ifA is symmetric ie7 A AT the above procedure shows that M7 Therefore A fly2 1ij 71 7 detMj A27 and so ade is symmetric Chapter Test 3 1 Evaluate 1 1 2 71 0 1 0 3 71 2 i3 4 39 5 0 72 J Solution Use the expansion of detA along the second row fourth row7 rst column and third column are also good choices7 having also two zero entries detA0 gtltAzi1gtltA220 XA233 gtltA24A223A24 71 1 1 2 71 73 4 3 712 73 21517 0 0 72 0 5 0 2 Let A be 3 X 3 and suppose that lAl 2 Compute a l3Al b l3A 1l C K319711 33 Solution Notice that 1A1 is here and above only an alternative nota tion for detA Thus a 1314 331141 54 b 1314 331M 013A ll 7 3 For what value of a is 2 1 0 0 a 1 0 71 3 1 3a 0 14 7 0 1 a 72 a 2 Solution Evaluating the sum of the two determinants7 we obtain 27a 7 3 a 1 6a 7 2a 147 a simple equation with the solution a 4 Find all values of a for which the matrix a2 0 3 5 a 2 3 0 1 is singular a2 0 3 Solution The determinant 5 a 2 13 79a aa 7 3a 3 0 3 0 1 if and only if a E 737073 By Theorem 312 p 203 these are the required values 5 Not requested Cramer7s rule 6 Answer each of the following as true or false a detAAT detA2 b det7A 7 detA C If AT A l7 then detA 1 d If detA 0 then A 0 e If detA 7 then Ax 0 has only the trivial solution 34 CHAPTER 3 DETERMINANTS f The sign of the term a15a23a31a42a54 in the expansion of the deter minant of a 5 X 5 matrix is g If detA 0 then detade 0 h If B PAP l and P is nonsingular then detB detA i If A4 In then detA 1 j If A2 A and A y In then detA 0 Solution a True detAAT This detA detAT detA detA detA2 b False detiA det71A 235 71 detA idetA only if n is odd For instance if A 2 then det7A det 071 71 17 71 idetA c False indeed AT A 1L detAT detA 1 detA m Which implies detA 6 i1 and not necessarily detA 1 For instance for A 1 91 AT A 1L holds but detA 71 TL31 d False obviously A 1 g a 0 but detA 0 e True indeed 7 a 0 and one applies Corollary 34 p 203 f True indeed the permutation 53124 has 4 2 6 inversions and so is even g True use Exercise T8 p 210 Which tells us that detade detA 1 h True using Theorems 38 and Corollary 32 notice that P is non singular we have detB detPAP 1 detP detA detP 1 detPdet1Pgt detA detA i False for instance ifn 2 and A 5 holds and detA 71 j True by the way of contradiction lf detA a 0 then A is non singular see Theorem 312 p 203 and thus the inverse A l exists By 31 J the equality A4 2 35 left multiplication of A2 A With this inverse A 1 we obtain at once A 1A2 A lA and A In 36 CHAPTER 3 DETERMINANTS Chapter 4 n Vectors Page 246 In the following Exercises we denote the n vectors considered by u 11171127 H711 7 V 1mm7 7vn and W 1017102 H710 T7 Show that uoVW uovuoW Solution Easy computation u 0 V W urv1 w1 u2v2 wz um w mm mm mm mm mm unwn uivi U202 unvn u1w1 uzwz unwn u 0 V u o W TS Show that if u 0 V u o W for all 117 then V W Solution First take u el 17070 Then eloV 10W gives 01 LUl Secondly take u 82 017070 Then 2V 2W implies v2 1027 and so on Finally7 take u en 07071 Then enoV enoW gives vn 10 Hence V v17v277vn W 101102710 T9 Show that if c is a scalar7 then chH lcl Hull Where lcl is the absolute value of c 38 CHAPTER 4 N VECTORS Solution Indeed by the norm length de nition chH cu12 cu22 cum2 czu12 021122 czunz 02u12 1122 1172 lcl M Hull T10 Pythagorean Theorem in R Show that Hu vH2 HuH2 Hle if and only if 11 0 V 0 Solution First notice that for arbitrary n vectors we have Hu vH2 11 v o 11 V HuH2 2u o v Hle using Theorem 43 p 236 and the previous Exercise T7 Finally if the equality in the statement holds then 2ro 0 and 11 o 31 0 Conversely 11 0V 0 implies 2n 0 V 0 and Hu vH2 Hqu M Chapter Test 4 1 Find the cosine of the angle between the vectors 12 714 and 37241 1 Solution Use the formula of the angle p 237 cos A HuHHVl 374744 7 1 WW 7 mm 2 Find the unit vector in the direction of 2 71 1 3 Solution Use the remark after the de nition p 239 if x is a nonzero vector then 11 H x is a unit vector in the direction of x Therefore H2 71 1 v4 11 V15 and the required vector is g 71 13 3 ls the vector 123 a linear combination of the vectors 132 2271 and 370 7 Solution Yes searching for a linear combination 1 3 2y2 2 71 237 770 127 37 yields a linear system x2y32 1 3x2y72 2 2m 7 y 3 1 2 3 The coef cient matrix 3 2 7 is nonsingular because compute 2 i1 0 its determinant is 14 a 0 Hence the system has a unique solution 4 Not requested linear transformations 5 Not requested linear transformations 6 Answer each of the following as true or false a lnR 7 ifro07 thenu00rV0 b In R 7 iquVuOW7theHVW c lan7ifcu0thenc00ru0 dgt1n R chH 0M M R HuvH M M f Not requested linear transformations g The vectors 17071 and 717170 are orthogonal h In R 7 if 07 then u 0 i In R 7 if u is orthogonal to V and W7 then u is orthogonal to 2v3w j Not requested linear transformations lt Solution a False for instance u e1 1707 0 a 0 and V e2 017070 0 but u 0 V 182 0 b False for 117V as above take W 0 then 81082 e100 but e2 0 c True indeed7 ifcu 0 and c a 0 then 0111701127 7cun 0707 0 or cul cug cum 07 together With c a 07 imply ul ug un 07 that is7 u 0 d False compare With the correct formula on p 2467 Theoretical Exercise T9 for instance7 if c 717 and u a 0 then HiuH a 7 40 CHAPTER 4 N VECTORS e False compare With the correct Theorem 45 p238 if u a 0 and V 11 then H11Vll HOH 0 i 2 Hull Hull ll ull Hull HVH g False the dot product u 0 V 71 0 0 a 07 so that the vectors are not orthogonal see De nition p238 h True indeed7 Mug u5 ufl 0 implies u 11 ui 0 and for real numbers ul ug un 07 that is7 u 0 i True indeed7 compute once again by the de nition p 238 uo2v 3w 2246 2uov3uow 2 gtlt 03 gtlt 0 0 Chapter 5 Lines and Planes Page 263 T2 Show that 11 X V oW 110 V X W Solution Indeed7 11 X V o W ugvg 7 u3v27u3v1 7 u1v37u1v2 7 ugvl 0 10171027103 U203 u3v2wl U301 711030102 U102 uzvl vg 71102103 03102 71203101 01103 u3viw2 02101 71177127713 39 12103 031027 03101 01103701102 02101 u 0 V X W T4 Show that U1 U2 U3 ugtltVoW v1 v2 v3 101 U12 103 Solution In the previous Exercise we already obtained 11 X V o W u1v2w3 7 Ug LUg u2v3w1 7 Ul LUg u3v1w2 7 Ug LUl But this computation can be continued as follows 71102103 03102 71203101 1103 u3viw2 02101 U1 U2 U3 02 03 vi 03 vi 02 U1 U2 U3 vi 02 03 7 U12 U13 U11 U13 U11 102 101 U12 U13 41 42 CHAPTER 5 LINES AND PLANES using the expansion of the determinant along the rst row T5 Show that u and V are parallel if and only if u gtlt V 0 Solution In Section 42 p 238 the following de nition is given two nonzero vectors u and V are parallel if luovl that is cos 6 i1 or equivalently sin 0 6 denoting the angle of u and V Notice that a 0 a for nonzero vectors and u gtlt V 0 gt Hu gtlt vll 0 Using the length formula Hu gtlt vll sin we obtain sin 0 if and only if u gtlt V 0 the required result Page 271 T5 Show that an equation of the plane through the noncollinear points P117y17217131532712722 and 131533713723 is y 2 1 1 11 21 1 0 2 12 22 1 39 3 13 23 1 Solution Any three noncollinear points P11y121 P12y222 and P13 y323 determine a plane whose equation has the form abyczd0 where a b c and d are real numbers and a b c are not all zero Since P11y121 P12y222 and P13y323 lie on the plane their coordi nates satisfy the previous Equation a1by1czld0 a2by2czgd0 a3by3czgd0 43 We can write these relations as a homogeneous linear system in the unknowns 1 b7 0 and d abyc2d0 azlby1c21d0 a2by2C22d0 azgby3c23d0 Which must have a nontrivial solution This happens if and only if see Corollary 34 p 203 the determinant of the coe icient matrix is zero7 that is7 if and only if 2 y 2 1 1 11 21 1 0 2 12 22 1 39 3 13 23 1 Chapter Test 5 3 Find parametric equations of the line through the point 57271 that is parallel to the vector u 37 27 5 Solution 2 53t7 y 22t7 2 15t7oo lt t lt 00 see p266 4 Find an equation of the plane passing through the points 17 27 17 3747 57 07 17 x y 2 1 1 2 1 1 i i Solution 3 4 5 1 0 see the preVious T ExerC1se 57 p 271 0 1 1 1 Thus 2 1 1 1 1 2 1 1 2 1 4 1y35123413450 1 1 1 1 1 0 1 1 0 1 1 ory10 44 CHAPTER 5 LINES AND PLANES 5 Answer eaCh of the following as true or false b lfugtltv0andugtltW0thenugtltVW0 C lfV 73u7 then u gtlt V 0 d The point 27 37 4 lies in the plane 22 7 3y 2 5 e The planes 22 7 3y 32 2 and 22 y 7 2 4 are perpendicular Solution b True7 using Theorem 51 p260 properties C and a uxvwugtltvugtltw000 C True7 using TheoretiCal ExerCise T5 p2637 or direCtly if u u17u27u3 then for V 73ujL7 731127 73u3 the Cross produCt i j k U1 uz U3 0 73111 73112 73113 beCause factoring out 73 it has two equal rows d False Veri Cation 2 X 2 7 3 X 3 1gtlt 4 71 a 0 e False The planes are perpendiCular if and only if the Corresponding norrnal veCtors are perpendiCular7 or7 if and only if these veCtors have the dot produCt zero 27 737 3 o 27 17 71 4 7 3 7 3 72 a 0 Chapter 6 Vector Spaces Page 302 T2 Let 81 and 82 be nite subsets of R and let 81 be a subset of 82 Show that a If 81 is linearly dependent so is 82 b If 82 is linearly independent so is 31 Solution Since 81 is a subset of 32 denote the vectors in the nite sets as follows 31 V1V2 Vk and 82 V1V2VkVk1Vm a If 81 is linearly dependent there are scalars cl Cg ck not all zero such that clvl 02V2 ckvk 0 Hence clvl 02V2 ckvk 0vk1 0Vm 0 Where not all the scalars are zero and so 82 is linearly dependent too b If SI is not linearly independent it is by De nition linearly de pendent By a 82 is also linearly dependent a contradiction Hence 81 is linearly independent T4 Suppose that S V1V2V3 is a linearly independent set of vectors in R Show that T W1W2W3 is also linearly independent Where W1 V1 V2 V3 W2 V2 V3 and W3 V3 Solution Take clwl CgWg 03W3 0 that is 01V1 V2 V3 45 46 CHAPTER 6 VECTOR SPACES 02V2 V3 Cng 0 Hence clvl 01 2V2 C1 02 03V3 0 and by linearly independence of S7 01 cl 02 cl 02 Cg 0 But this homogeneous system has obviously only the zero solution Thus T is also linearly independent T6 Suppose that S V17V27V3 is a linearly independent set of vectors in R ls T W17W27W37 where W1 V17 W1 V1 V2 and W1 V1 V2 V37 linearly dependent or linearly independent 7 Justify your answer Solution T is linearly independent We prove this in a similar way to the previous Exercise T4 Page 316 28 Bonus Find all values of a for which 1270717 071727 1707 1 is a basis for R3 Solution 1 Use the procedure given in Section 637 p 2947 in order to determine the values of a for which the vectors are linearly independent this amounts to nd the reduced row echelon form for the matrix we have reversed the order of the vectors to simplify computation of RREF 101 2 10 1 0a2quot R20a 2 1201 0017a2 If a 0 or a 6 i1 the reduced row echelon form has a zero row For 1 717071 the three vectors are linearly independent7 and so form a basis in R3 by Theorem 69 a7 p312 Solution 2 Use the Corollary 64 from Section 667 p335 it is su icient to nd the values of a for which the determinant 02 1 0 2 aa2717 0 1 1 090 These are a E R7 7101 T9 Show that if V1V2Vn is a basis for a vector space V and c a 0 then CV1V2 Vn is also a basis for V Solution By de nition if k1cvl k2V2 knvn 0 then the system V1V2 Vn being linearly independent klc k2 kn 0 and c a 0 k1 k2 kn 0 The rest is covered by Theorem 69 a T10 Let S V1V2V3 be a basis for vector space V Then show that T w1wzw3 Where W1 V1V2V3 W2 V2V3 and W3 V37 is also a basis for V Solution Using Theorem 69 a it is su icient to show that T W1W2W3 is linearly independent If01W1 csz 63W3 0 then 01V1 V2 V3 62V2 V3 3V3 clvl cL 2V2 cl 02 03V3 0 and V1V2 V3 being linearly independent cl 01 02 01 02 Cg 0 Hence by elementary computations 01 02 Cg 0 T12 Suppose that V1V2 Vn is a basis for R Show that if A is an n X n nonsingular matrix then AV1AV2AVn is also a basis for 71 Solution First we give a solution for Exercise T10 Section 63 Suppose that V1V2Vn is a linearly independent set of vectors in R Show that if A is an n X n nonsingular matrix then AV1AV2 Avn is linearly independent SOLUTION USING SECTION 66 Let M v1v2vn be the matrix Which has the given vectors as columns cobM Vi 1 S 239 S Thus the product AM AVlAngAVn that is cohAM AV 1 S 239 S n The given vectors being linearly independent detM a 0 The matrix A being nonsingular detA a 0 Hence detAM detA detM a 0 so that AV1AV2AVn is a linearly independent set of vectors by Corollary 64 in Section 66 48 CHAPTER 6 VECTOR SPACES Finally7 if AV17AV27 7Avn is linearly independent and has n vectors notice that dimR n7 it only remains to use Theorem 69 a7 p312 T13 Suppose that V17V277Vn is a linearly independent set of vectors in R and let A be a singular matrix Prove or disprove that AV17AV27 7 Avn is linearly independent Solution Disprove see also the previous Exercise lndeed7 AV17AV27 Avn can be linearly dependent take for instance A 0 a set of vectors Which contains a zero vector is not linearly inde pendent Page 337 338 6 Bonus Exercises 1 2 3 3 LetSv17V27V5WhereV1 i 7V2 2 7V3 3 7 3 5 3 3 4 V4 3 7 V5 5 Find a basis for the subspace V spanS of R 3 3 Solution We use the procedure given after the proof of Theorem 667 p 308 1 2 3 3 5 1 2 3 3 5 A i 2 1 2 3 3 0 i3 i4 i3 i7 7 1 2 3 3 5 0 0 0 0 0 2 1 2 3 3 0 i3 i4 i3 i7 1 2 3 3 5 0 1 1 Z 3 3 0 0 0 0 0 and so V1 and V2 is a basis 0 0 0 0 0 49 11 Compute the row and column ranks of A verifying Theorem 6117 1 2 3 A 3 1 75 i2 1 7 8 71 2 5 Solution We transform A into row echelon form 1 2 3 2 1 1 2 3 2 1 3 1 i5 72 1 N 0 i5 714 78 72 RSN 7 8 1 2 5 0 6 722 712 i2 1 2 3 2 1 1 2 3 2 1 0 1 8 4 0 N 0 1 8 4 0 0 i6 722 712 i2 0 0 26 12 i2 1 2 2 1 N 0 1 8 4 0 and so the column rank is 3 0 0 1 i i Further7 for the row rank7 we transform in row echelon form the trans pose AT7 1 3 7 1 3 7 1 3 7 1 8 0 75 76 0 1 1 i5 71 N 0 714 722 R2213 0 414 46 N 72 2 0 is 712 0 is 412 1 51 0 72 72 l0 45 461 OOOHOJ 3 1 0 1 and so the row rank is also 3 0 0 lg 0 lo 01 18 If A is a 3 X 4 matrix7 What is the largest possible value for rankA Solution The matrix A has three rows so the row rank is S 3 and four 50 CHAPTER 6 VECTOR SPACES columns and so the column rank is g 4 By Theorem 611 the rank of A is at most 3 20 If A is a 5 X 3 matrix show that the rows are linearly dependent Solution The matrix haVing only 3 columns the column rank is g 3 Hence also the row rank must be S 3 and 5 rows are necessarily dependent otherwise the row rank would be 2 5 1 1 4 i1 1 2 3 2 25 Determine whether the matrix A 71 3 2 1 is singular i2 6 12 i or nonsingular using Theorem 613 Solution We transform the matrix into row echelon form in order to compute its rank 1 1 4 71 1 1 4 71 1 1 4 i1 1 2 3 2 0 1 71 3 0 1 i1 3 i1 3 2 1 0 4 6 0 0 0 2 712 i2 6 12 i4 0 8 20 i6 0 0 28 738 1 1 4 i1 1 1 4 i1 0 1 i1 3 0 1 71 3 N 0 0 1 76 N 0 0 1 i6 Hence the rank the number 0 0 28 730 0 0 0 1 of nonzero rows is 4 and the matrix is nonsingular 2 29 ls S 1 5 71 a linearly independent set of 75 vectors 7 Solution The matrix whose columns are the vectors in S 422 1571 2753 has a zero determinant verifyl Hence by Corollary 64 the set S of vectors is linearly dependent T7 Let A be an m X 72 matrix Show that the linear system Ax b has a solution for every m X 1 matrix b if and only if rankA m Solution Case 1 m S 72 If the linear system Ax b has a solution for every m X 1 matrix b then every m X 1 matrix b belongs to the column space of A Hence this column space must be all Rm and has dimension m Therefore rankA column rankA dimcolumn spaceA m Conversely if rankA m the equality rankA rankAlb follows at once because generally rankA rankAlb and rankAlb S m because Alb has only m rows Finally we use Theorem 614 Case 2 m gt n We observe that in this case rankA m is impossible because A has only 72 lt m rows and so rankA 72 But if m gt n the linear system Ax b has no solution for every m X 1 matrix b indeed if we have more equations than unknowns and for a given b the system has a solution it su ices to modify a coe icient in b and the corresponding system is no more veri ed by the same previous solution Chapter Test 6 1 Consider the set W of all vectors in R3 of the form a bc where abc 0 ls W asubspace ofR3 7 Solution Yes 000 6 W so that W a V For abca b c E W also aa b b cc E W because aa bb cc abca b c 000 Finally for every k E R and abc E W also kabc E W because kakbkckabc0 52 CHAPTER 6 VECTOR SPACES 2 Find a basis for the solution space of the homogeneous system 1 32 33 i4 2 1 22 23 7 24 25 123i3425 COO Solution We transform the augmented matrix A in row echelon form 1 3 3 71 2 0 1 2 2 i2 2 0 N 1 1 1 i3 2 0 1 0 0 i4 2 0 N 0 1 1 1 0 0 0 0 0 0 0 0 1 2 Hence X 3 3 4 5 4 if7 5 u7 so that basis 1 0 0 1 2 0 71 1 0 0 3 1 2 3 71 72 71 2 0 71 0 0 2 0 0 The corresponding system is now 44 7 25 i3 4 39 4 72 71 0 75 0 u 0 taking 3 37 1 0 0 1 0 4 72 1 71 0 1 7 0 7 0 is the required 0 1 0 0 0 1 3 Does the set of vectors 17 71717 17 7371 1272 form a basis for R3 7 53 1 1 1 Solution If A 71 73 2 is the matrix consisting7 as columns7 1 1 2 of the given vectors7 detA 76 71 2 3 7 2 2 72 a 0 so that the vectors are linearly independent Using Theorem 697 a7 this is a basis in 3 4 For What values of A is the set of vectors x2 7 517027723727373 linearly dependent 7 A2 7 5 2 2 Solution If A 1 72 3 is the matrix consisting7 as 0 3 73 columns7 of the given vectors7 then detA 62 7 5 6 7 903 7 5 6 73A2 27 Thus detA 0 if and only if A 6 i3 5 Not required 6 Answer each of the following as true or false a All vectors of the form 1707 7a form a subspace of R3 ltbgt1n R chH 7 01le c Every set of vectors in R3 containing two vectors is linearly inde pendent d The solution space of the homogeneous system AX 0 is spanned by the columns e If the columns of an n X 72 matrix form a basis for R 7 so do the rows f If A is an 8 X 8 matrix such that the homogeneous system Ax 0 has only the trivial solution then rankA lt 8 g Not required h Every linearly independent set of vectors in R3 contains three vec tors i If A is an n X n symmetric matrix7 then rankA n j Every set of vectors spanning R3 contains at least three vectors Solution a True lndeed7 these vectors form exactly 54 CHAPTER 6 VECTOR SPACES span10 71 Which is a subspace by Theorem 63 p 285 b False Just take 0 71 and X a 0 You contradict the fact that the length norm of any vector is Z 0 c False x 111 and y 222 are linearly dependent in R3 because 2x 7 y 0 d False The solution space of the homogeneous system AX 0 is spanned by the columns which correspond to the columns of the reduced row echelon form Which do not contain the leading ones e True In this case the column rank of A is 72 But then also the row rank is n and so the rows form a basis f False Just look to Corollary 65 p 335 for n 8 h False For instance each nonzero vector alone in R3 forms a linearly independent set of vectors i False For example the zero n X 72 matrix is symmetric but has not the rank n it has zero determinant j True The dimension of the subspace of R3 spanned by one or two vectors is S 2 but dim R3 3 Chapter 8 Diagonalization Page 421 T1 Let A be a particular eigenvalue of A Show that the set W of all the eigenvectors of A associated with M as well as the zero vector7 is a subspace of R called the eigenspace associated with A7 Solution First7 0 E W and so W a V Secondly7 if x7 y E W then Ax ij Ay Ajy and consequently Ax y AxAy ij My Ajxy Hence xy E W Finally7 if X E W then Ax ij and so Acx CAx Cij Ajcx Hence ex 6 W T3 Show that if A is an upper lower triangular matrix7 then the eigenvalues of A are the elements on the main diagonal of A Solution The corresponding matrix AI 7 A is also upper lower tri angular and by Theorem 37 Section 317 p 1887 the characteristic poly nomial f is given by A 7 111 70412 7041 0 A7092 7042 z z z A7a117122717m 0 0 A7 am 55 56 CHAPTER 8 DIAGONALIZATION expanding successively along the rst column The corresponding char acteristic equation has the solutions 1117122 quot41 T4 Show that A and AT have the same eigenvalues Solution lndeed7 these two matrices have the same characteristic poly nomials detAIn 7 AT detAInT 7 AT detAIn i AT Th31 detAIn i A T7 Let A be an n X 72 matrix a Show that detA is the product of all the roots of the characteristic polynomial of A b Show that A is singular if and only if 0 is an eigenvalue of A Solution The characteristic polynomial A i 111 70412 7041 70421 A 7 122 7042 7am 7am A 7 am is a degree n polynomial in A which has the form A clAn l CgAn 2 cn1A on one uses the de nition of the determinant in order to check that the leading coe icient actually is 1 If in the equality fA detAIn 7 A A clAn 1L 02A 2 cn1A on we let A 0 we obtain an det7A 71 det A By the other way7 if fA A 7 A1A 7 A2A 7 An is the decompo sition using the real eigenvalues of A7 then in the same way7 A 0 we obtain f0 on 71 A1A2An Hence A1A2An detA b By Theorem 312 p 2037 we know that a matrix A is singular if and only if detA 0 Hence7 using a7 A is singular if and only if A1A2An 07 or7 if and only if A has 0 as an eigenvalue Chapter Test 8 1 If possible7 nd a nonsingular matrix P and a diagonal matrix D so that A is similar to D where 100 A520 432 Solution The matrix A is just tested for diagonalizability Since the matrix A13 7A is lower triangular using Theorem 377 p 1887 the charac teristic polynomial is det13 7 A A 7 1 7 227 so we have a simple eigenvalue A1 1 and a double eigenvalue A2 2 For A1 1 the eigenvector is given by the system 3 7 AX 0 solve 1 it x1 75 11 For the double eigenvalue A2 2 only one eigenvector is given by the system 213 7 AX 0 solve it x2 0 1 Therefore A is not diagonalizable 2 and 3 not required 4 Answer each of the following as true or false a Not required b If A is diagonalizable7 then each of its eigenvalues has multiplicity one c If none of the eigenvalues of A are zero7 then detA a 0 d If A and B are similar7 then detA detB e If x and y are eigenvectors of A associated with the distinct eigen values Al and A2 respectively7 then x y is an eigenvector of A associated with the eigenvalue A1 A2 58 CHAPTER 8 DIAGONALIZATION Solution b False see the Example 67 p 427 c True according to T Exercise 77 p 421 section 817 detA is the product of all roots of the characteristic polynomial But these are the eigenvalues of A see Theorem 827 p 413 If none of these are zero7 neither is their product d True indeed7 if B P lAP then detB detP 1AP det P l det Adet P det P l det P det A det A7 since all these are real numbers e False hypothesis imply Ax MK and Ay Agy By addition of columns AX y AX Ay Alx My is not generally equal to A1 A2x y An easy example any 2 X 2 matrix having two di erent nonzero eigenvalues The sum is a real number di erent from both a 2 X 2 matrix cannot have 3 eigenvalues speci c example7 see Example 37 p 424

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