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# Linear Algebra MT 210

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This 101 page Class Notes was uploaded by Mr. Halie Wilkinson on Saturday October 3, 2015. The Class Notes belongs to MT 210 at Boston College taught by Jenny Baglivo in Fall. Since its upload, it has received 10 views. For similar materials see /class/218063/mt-210-boston-college in Mathematics (M) at Boston College.

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Date Created: 10/03/15

MT210 Notebook 2 prepared by Professor Jenny Baglivo Copyright 2009 by Jenny A Baglivor All Rights Reserved 2 MT210 Notebook 2 2 1 Matrices and Their Operations r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r 2 11 Rectangular Arrays r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r 212 Square Matrices7 Types of Square Matrices7 Identity Matrix r r r r r r r r r r r 2 13 Matrix Sum and Scalar Product Linear Combination r r r r r r r r r r r r r r 2 14 Matrix Transpose r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r 2 15 Dot Product and Matrix Product Composition of Transformations r r r r r r r 2 16 Properties of the Basic Operations r r r r r r r r r r r r r r r r r r r r r r r r r 2L7 Powers of Square Matrices r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r 2 18 lnverses of Square Matrices Special Cases Properties of lnverses r r r r r r r r 2L9 lnverses and Solving Systems Finding lnverses r r r r r r r r r r r r r r r r r r r 2110 Elementary Matrices Relationship to lnverses r r r r r r r r r r r r r r r r r r r Factoring and Blocking r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r 2 21 Factorization LU Factorization r r r r r r r r r r r r r r r r r r r r r r r r r r r 2 22 Permutation Matrices PALU Form r r r r r r r r r r r r r r r r r r r r r r r 2 23 Partitioned Matrix7 Block Matrix lnverses in Special Cases r r r r r r r r r r r Determinants r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r 2 31 Determinants and Areas r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r Recursive De nition of Determinants Volumes r r r r r r r r r r r r r r r r r r Properties of Determinants Factorizations r r r r r r r r r r r r r r r r r r r r r Elementary Row Operations and Determinants r r r r r r r r r r r r r r r r r r lnvertibility and Determinants Cramer s Rule7 Cofactors and lnverses r r r r r r r r r r r r r r r r r r r r r r r r 26 27 29 30 32 2 MT210 Notebook 2 This notebook is concerned with matrix methods in linear algebra The notes correspond to material in Chapters 2 and 3 of the Lay textbook 21 211 Matrices and Their Operations Rectangular Arrays An in x n matrix A is a rectangular array with in rows and n columns7 1 2 all 03912 aln 03921 03922 39 39 39 a2n A i i i a1 12 an laijl am am amn m x n is the size of the matrix A there are in rows and n columns aij the number in the row i7 column j position is the ij entry of A The term after the rst equal sign shows the entries explicitly In the term after the second equal sign7 A is written as an array of n columns where each column is a vector in R The last term is used as shorthand when the size of the matrix is understood 212 Square Matrices Types of Square Matrices Identity Matrix 1 2 03 F U A square matrix is a matrix with m n An n x n square matrix is often called a square matrix of order ii If A is a square matrix of order n then the diagonal elements of A are the elements an 01227 7 am If A is a square matrix7 then The identity matrix of order n denoted by In is the n x n diagonal matrix all of whose diagonal elements equal 1 We sometimes just write I for the identity matrix when the order is clear from the problem For example7 let a b c a b c a 0 0 a 0 0 A1 I d 6 7 A2 0 d 6 7 A3 I C 0 and A4 0 b 0 7 c e f 0 0 f d e f 0 0 c where a through 1 are any numbers Then A1 is a symmetric matrix7 A2 is an upper triangular matrix7 A3 is a lower triangular matrix and A4 is a diagonal matrix As a diagonal matrix7 A4 is also upper triangular7 lower triangular and symmetric Note Coe icient matrices and row operations If A is the coefficient matrix for an n by n system7 then after the forward pass7 A is transformed to an upper triangular matrix If the system has a unique solution7 then the reduced echelon form of A is the identity matrix After the forward and backward passes7 A is transformed to the identity matrix Note Identity matriw and standard basis vectors The columns of the identity matrix are the standard basis vectors we discussed in the last section That is we can write In as follows 1 0 0 0 1 0 177 el 82 e747 0 0 1 where ej is the vector with 1 in position 37 and 0 everywhere else7 for j 17 27 7n 213 Matrix Sum and Scalar Product Linear Combination Let A aw and B blj be m x n matrices7 and let c7d E R be scalars Then matrix sum and scalar product are de ned as follows 1 The matrix sum of A and B is the m x n matrix AB a1b1 a2b2 anbn Equivalently7 A B is the matrix with i7j entry A B aij bij 2 The scalar product of A by c is the m x n matrix cA cal caz can Equivalently7 cA is the matrix whose i7j entry is 31477 caij Further7 the linear combination CA 1 dB is the m x n matrix cA dB cal dbl caz db2 m can dbn Equivalently7 CA 1 dB is the matrix whose i7 j entry is CA 1 dB7j cal7 dblj 2 3 For example if A 7 72 0 AB iii 5A 777 and5A72B 777 214 Matrix Transpose The transpose of the m x 71 matrix A denoted by AT A transpose is the n x m matrix Whose tj entry is aji ATLJ aji For example if 1 712 3 T A7 4 5 6 thenA 7 g OJC gtgt 1 Note that if A is a symmetric matrix then AT A For convenience let 1i 6 R for t 1 2 m be the columns of AT Then A is the matrix Whose rows are the aiT s That is if ATa1 a2 amthenA 215 Dot Product and Matrix Product Composition of Transformations 1 If 39v and 39w are vectors in R then their dot product is the scalar 39U39Lu 39UT39w Ulwl 12ng Unwn 2 Let A be an m x 71 matrix and let B b1 b2 bp be an n x p matrix Then the matrix product AB is the m x p matrix Whose jth column is the matrix vector product of A with the jth column of B ABAb1 Ab2 Abp Equivalently AB is the m x p matrix Whose t j entry is the dot product of the ith row of A with the jth column of B ABM ai bj Otinj ailblj 11392ij 39 39 39 ambnj where a is the ith row of A for t 1 m Dot products are de ned for vectors of the same length For example7 if 2 2 2 39v 3 and 39w 0 7 then 39U39Lu 2 3 6 0 22 30 64 28 6 4 Matrix products are de ned for compatible matrices that is7 when the number of columns of the matrix on the left equals the number of rows of the matrix on the right 2 71 For example7 if A g f and B 2 73 l 0 1 4 7 then 4 77 0 but BA is unde ned since they are not compatible 3 7 2 2 73 Problem 1 Let A j g f l and B 0 1 Find AB and BA ls AB BA 4 77 Matrix product and composition The matrix product has many applications For ex ample the matrix product can be interpreted as the standard matrix of the composition of two linear transformations Speci cally if TB RP a R has matrix B TA R a R has matrix A and T TA 0 TB is the composite function then T RP a Rm de ned by Tw TATBJB has standard matrix AB Problem 2 Let TB R2 a R2 and TA R2 a R2 be linear transformations with rules a 2 3 g TBw Ban Z 3 an and 71433 Aw E 73 an 5 5 13 13 Find the standard matrix of the linear transformation T TA 0 TB Footnote Geometric Interpretation In the problem above transformation TB corre sponds to rotation about the origin through the positive angle 0 arctan g and the trans formation TA corresponds to re ection in the line y Under the composite function T TA 0 T3 the unit square for example is rst rotated by 0 and then re ected in the line as illustrated below TATBR 216 Properties of the Basic Operations Some properties of the basic operations are given below In each case7 the sizes of the matrices are assumed to be compatible H Commutative A B B A to Associative A B C A B C and ABC ABC 03 Distributive AB C AB AC7 B CA BA CA F Scalar Multiples Given 0 E R CAB cAB ACB7 CA B CA CB and CAT CAT 0 Identity ImA A AI when A is an m x 71 matrix a Transpose of Transpose ATT A 1 Transpose of Sum A BT AT BT 00 Transpose of Product ABT BTAT Note 1 Commutativity Matrix sum is commutative but matrix product is not In fact7 as we have already seen7 it is possible that AB is de ned but that BA is unde ned Note 2 Associativity The associative law for products says that products like ABC are well de ned It is up to you whether you would like to compute AB rst and then multiply on the right by C7 or to compute BC rst and then multiply on the left by A 217 Powers of Square Matrices If A is an n x n matrix7 then we can use a shorthand for the powers of A Namely7 A2 AA7 A3 AAA7 A4 AAAA7 and so forth Problem 3 Let A l 1 j Find A2 A3 A4 and A100 a 0 Problem 4 Let A 0 b J a Find A27 A37 A4 and AP where p is a positive integer b What properties are suggested by your solution to part a Be as complete as possible a1 0 0 b1 0 0 Problem 5 Let A 0 a2 0 and B 0 b2 0 a Find AB and BA 0 0 a3 0 0 b3 b What properties are suggested by your solution to part a Be as complete as possible 218 Inverses of Square Matrices Special Cases Properties of Inverses The n x n matrix A is said to be invertible if there exists a square matrix C satisfying AC CA In where In is the identity matrix of order n Otherwise A is said to be not invertible Square matrices that are not invertible are also called singular matrices invertible square matrices are also called nonsingvlar matrices Problem 6 Use the properties of the basic matrix operations given in Section 216 page 8 to demonstrate that inverses are unique Notation for the matrix inverse If A is invertible then A 1 read A inverse is used to denote the matrix inverse of A Special cases Finding the inverse of a square matrix or determining that an inverse does not exist can often involve many computations Here are a few special cases H Powers If A is a square matrix and p is the smallest positive integer satisfying AP I then A is invertible with inverse A 1 Ap l ln symbols I AP A A174 A174 A gt A 1 AM to 2 by 2 Matrices Let A Z be a 2 x 2 matrix Then A 1 d 2 whenadibcy O adibc it If ad 7 be 0 then A does not have an inverse Note ad 7 be is the determinant of A 3 Diagonal Matrices Let A be the diagonal matrix with nonzero diagonal elements that is with an 0 for all Then A is invertible with inverse 1 a11 0 0 A71 0 1a22 0 l 0 0 1 amj That is A 1 is the diagonal matrix Whose diagonal elements are the reciprocals of the diagonal elements of A If an 0 for some i then A is not invertible 10 For example 1 If A 1 then from the work we did in Problem 3 we know that A3 I 7 then the determinant is and A 1 J Thus the inverse of A is A 1 A2 77 2 lfA 2 3 HA 3 0 0 0 0 1 0 then A 1 0 1 4 An example of a square matrix A 7 I satisfying A 1 A is Properties of inverses Let A and B be invertible square matrices of order n and let 0 be a nonzero constant Then 1 Inverse of Identity Igl In for each n 2 Inverse of Inverse A gil A 3 Inverse of Product AB71 B 1A 1 4 Inverse of Scalar Multiple 61471 1CA 1 5 Inverse of Transpose A7371 A71 Demonstration of Property 5 It is instructive to demonstrate the third property Assume that A and B are invertible square matrices of size n and let C B lA l We need to show 1 ABC In and 2 CAB In NOW complete the proof 219 Inverses and Solving Systems Finding Inverses The following theorem tells us that if the coef cient matrix of an n by n system of linear equations is invertible then the unique solution to the system can be found using inverses Speci cally Theorem Uniqueness Theorem Let A be an invertible n x 71 matrix Then the matrix equation A21 b has the unique solution 1 A lb for any b E R To see this note that A11 b A 1A1A 1b A lA1A 1b InmA 1b 11A 1b Problem 7 Use inverses to solve the 2 by 2 system 77ml 8x2 73 and 2x1 7 2x2 5 Finding the inverse and solving 11 linear systems An algorithm for nding the inverse can be developed using the following observations Let A be an invertible matrix of order n and In 81 82 en be the identity matrix of order n For convenience let C A l Then AC In can be written as follows AC A01 A02 Acn 81 82 en Thus the jth column of the inverse is the solution to the system ACj ej To turn the observations which assume that the inverse exists into an algorithm for nding the inverse we need the following facts Fact 1 The n x 71 matrix A is invertible if and only if A is row equivalent to In If A is row equivalent to In then any sequence of elementary row operations that reduces A to In also transforms In to A l Fact 2 Suppose that A and D are square matrices of order n Then if AD In then A is an invertible matrix Algorithm for nding an inverse Let A be an n x 71 matrix Form the n x 271 augmented matrix Whose last 71 columns are the columns of the identity matrix If A is row equivalent to the identity matrix7 then the n x 271 augmented matrix will be transformed as follows Allnl1nlA 1l That is7 the last 71 columns will be the inverse of A Otherwise7 A does not have an inverse 2 0 0 2 0 0 l 0 0 ForexampleletA 73 01 ThenAl13 73 01010 0 4 0 0 4 0 0 0 l 2 0 0 1 0 0 1 0 0 0 0 0 04000101000 A 100 001310 001g10 10 6 l 2 Problem 8 Find the inverse ofA 3 0 1 or state that the inverse does not exist 0 l 0 l a 12 Problem 9 Find the inverse ofA 0 1 c or state that the inverse does not exist 0 0 l 2110 Elementary Matrices Relationship to Inverses An elementary matrix E7 is one obtained by performing a single elementary row operation on an identity matrix Since each elementary row operation is reversible7 each elementary matrix is invertible For example7 let n 2 1 If the operation is R2 9 R2 2R1 with reverse operation R2 9 R2 7 2Rl7 then 710 17 10 E21andE 721 2 If the operation is R2 9 3R2 with reverse operation R2 E R2 then E and E 1 3 If the operation is R2 lt gt R1 with reverse operation R2 lt gt Rl7 then Left multiplication performs operation An elementary row operation performed on a matrix A corresponds to left multiplication by the corresponding elementary matrix For example7 if A 7 1 72 J and E is the rst matrix above7 then the product 7 1 O 3 1 O 7 3 1 O EA7217607470274 has the effect of replacing R2 by R2 2R1 Relationship to inverse of A Let A be an n x 71 matrix Suppose that A can be trans formed to the identity matrix using a sequence of p elementary row operations with elementary matrices E17 E27 Ep Then EpEp1E2E1A EpEp1E2E1A In a A 1 EPEZFL E2E1 Thus7 A 1 is the product of the elementary matrices used to transform A to In Further7 A is the product of the inverses of the elementary matrices in the opposite order A 1411 1 EpEp1E2E11 EflEglE1 Problem 10 Let A 7 The work on page 13 implies that A can be transformed DOOM HgtOO 0 1 0 to the identity matrix using the following four elementary row operations 3 1 1 1 RZ H 32 5317 2 Rz lt gt R37 3 32 H 11327 431 H 5R1 Find the corresponding elementary matrices E17 E27 E3 and E47 and verify that the inverse of A equals the product E4E3E2E1 22 Factoring and Blocking 221 Factorization LU Factorization A factorization of matrix A is an equation that expresses A as the product of two or more matrices A useful rst example is the LU factorization of a coef cient matrix The in x n matrix A has an LU faotorization if it is row equivalent to an echelon form matrix using row replacements only In this case7 A LU where 1 L is an m x in lower triangular matrix with 1 s on the diagonal and 2 U is an m x n echelon form matrix Note that a lower triangular matrix with 1 s on the diagonal is called a unit lower triangular matrix If the coef cient matrix A is square7 then U will be an upper triangular matrix 3 l 76 0 74 only one elementary row operation7 R2 9 R2 2R1 Let E be the elementary matrix for the operation Then For example7 if A 7 then A is row equivalent to an echelon form matrix using 7 10 31 O 7 31 O EA7217607470274 7 1 31 0 7 10 31 0 7 A E 0274 721H0274 LU 31 0 Problem 1 Find an LU factorization for A 76 0 74 0 6 710 Problem 2 Find an LU factorization for A Ozugtm on to roam l l H Problem 5 What patterns7 if any7 do you see in the solutions to the problems above Solving linear systems ef ciently Suppose that A LU where L is a unit lower triangu lar matrix and U is an echelon form matrix Then the matrix equation A11 b can be solved efficiently as follows Consider b A21 LU1 LU1 Ly7 where y U113 The efficient method is 1 Solve Ly b for y 2 Solve U11 y of 13 For example7 let A 7 7 J and b 73 J and assume that A LU 7 1 J 2 7i J is the LU factorization of A To solve A11 b using the method above7 10 2 10 2 2 1 Lib72172N012y2 2 73 2 2 0 8 10 4 4 2 Uiy0 12 012N012x2 is the solution to A11 b Problem 1 continued Let A Use the LU factorization 3 1 72 76 0 74 and b 2 0 6 2 of A we computed earlier to 1 solve Ly b for y and 2 solve U11 y for 13 Ef ciency of LU factorization The method above solves the matrix equation A11 b by 1 factoring A 2 using the equivalent of forward substitution to nd y and 3 using the equivalent of backward substitution to nd 13 For n by n systems for example 1 solving A11 b directly takes roughly 713 operations while 2 solving Ly b and U11 y together take roughly 712 operations So the method is useful when n is very large and the same coefficient matrix will be used with many different b s Large scale heat ow and largescale air ow problems are examples of situations where the method above is useful In addition since scalings are held off as long as possible the method is useful in situations where scalings could cause numerical roundoff problems 222 Permutation Matrices PALU Form A permutation matn39r P is elementary matrix corresponding to a row interchange or a product of elementary matrices corresponding to multiple row interchanges For example if n 4 o The permutation matrix corresponding to R1 lt gt R3 is o The permutation matrix corresponding to 1 R1 lt gt R3 followed by 2 R3 lt gt R4 is Use of permutation matrices in factoring lf row interchanges are needed to transform A to echelon form7 then an LU factorization is not possible But7 we can write PA LU7 where P is a permutation matrix 00 20 731001 For example let A 7 1 0 0 0 Since 00781 00 20 1000 1000 1000 A73100N3100N0100N0100 710 00 00 20 0020 00207 00781 00781 00781 0001 using operations 1 R1 lt gt R3 2 R2 9 R2 7 3Rl7 and 3 R4 9 R4 4ng7 we can construct the PA LU form as follows 223 Partitioned Matrix Block Matrix Inverses in Special Cases A partitioned matrix or block matrix is one Whose entries have been grouped into smaller matrices using vertical and horizontal cuts through the matrix For example 120 3110 A 3 4 0 A11 12andB 74 3 01 006 22 2 00 B11 312 A21 74 321 322 are examples of 2 x 2 block matrices Note that 1 Matrices of the same size that is with the same numbers of rows and columns and blocked in the same way that is so that corresponding blocks have the same numbers of rows and columns can be added by adding blocks 2 Matrices compatible for multiplication and blocked in compatible ways can be multiplied in blocks For example for compatible 2 x 2 block matrices A11 A12B11 312 A21 A22 321 322 1411311 1412321 1411312 1412322 1421311 1422321 1421312 1422322 as long as all multiplications and additions are valid For example the 2 x 2 block matrices given above cannot be added but they can be multiplied Note that ll7 where O is a zero matrix of an appropriate size and I is an identity matrix of an appropriate size using block structure simpli es the process AB A11 0 Bu 1 1411311 A11 0 A22 B21 0 A22B21 O Problem 4 Let A Compute the product AB Take advantage of the block structure as much as possible 2 x 2 block diagonal matrices and their inverses Let A be a square matrix of order n Then A is a 2 X 2 block diagonal matrix if A A51 A0 J 7 Where All and A22 are square matrices of orders p and q7 22 respectively7 and each 0 is a zero matrix of an appropriate size For example7 the following matrix is a 2 x 2 block diagonal matrix 7 A11 0 L AZZTT Where All is a square matrix of order 2 and A22 is a square matrix of order 3 The following theorem tells us that inverses of block diagonal matrices and block diagonal Speci cally7 Theorem Inverses Let A be a 2 x 2 block diagonal matrix as de ned above Then A is invertible if and only if both A11 and A22 are invertible 71 Further7 if A is invertible7 then A 1 A11 31 0 A22 It is instructive to write out the proof of the theorem on inverses 71 lt Su ose that All and A22 are invertible and let C A11 0 Then a pp 0 A 2 71 2 direct calculation shows that AC CA p O In Thus7 C A l 0 q a Suppose that A is invertible and let C Cu 012 be the inverse of A C C 2 Then AC In implies 2 A11 0 011 012 1411011 1411012 Ip 0 0 A22 021 022 1422021 1422022 0 q 39 NOW7 complete the proof Problem 5 In each case use block methods to nd the inverse of A 530 1A640 002 OOgtgtCA3 00000 5 0 7 6 0 21470 4 0 3 2 x 2 block upper triangular matrices and their inverses Let A be a square matrix of order n Then A is a 2 x 2 block upper triangular matrix if A A i A 51 A12 Where All and A22 are square matrices of orders p and q 22 respectively and O is a zero matrix of an appropriate size For example the following matrix is a 2 x 2 block upper triangular matrix 7 A11 A12 7 Where All is a square matrix of order 2 and A22 is a square matrix of order 3 The following theorem gives us the form of the inverse of a block upper triangular matrix when it exists Speci cally Theorem Inverses Let A be a 2 x 2 block upper triangular matrix as above Then A is invertible if and only if both A11 and A22 are invertible Ail AfilAIZAgzl Further if A is invertible then A 1 0 A231 Problem 6 a In each case7 use block methods to nd the inverse of A 37001 12010 37001 12010 24100 01000 10000 12 Suggest special cases of the inverse formula for 2 X 2 block upper triangular matrices based on the results of the last two inverses in part a Be as complete as possible 23 Determinants 231 Determinants and Areas If A Z is a 2 X 2 matrix7 then the determinant of A is the number detA 2 adi be Let A a1 a2 Analytic geometry can be used to show that the area of the parallelogram with corners 07 a1 a2 and a1 a2 is the absolute value of the determinant Area of Parallelogram ldetAl lad 7 bcl For example if 1 6 A a1 a2 5 2 then 1 detA and 2 the area of the parallelogram shown in the plot is 232 Recursive De nition of Determinants Volumes If A is an n X 71 matrix then the determinant of A can be de ned recursively as an alternating sum of multiples of determinants of smaller matrices Using expansion in the rst row of A detA w Zeuwal detA1 j1 Where A1 is the submatrix obtained by eliminating the 1St row and j column of Al The form for a 3 X 3 matrix is an 112 113 122 123 132 133 121 a ail 03922 03923 all 03931 03932 03933 ail 03922 03931 03932 i 112 113 23 131 133 The computation is completed once you ll in the determinants of each 2 X 2 submatrix and simplify the results For example if A 2 2 12 l 10 4 1 then the determinant of A is 8 Determinants and volumes Let A a1 a2 a3 be a 3 X 3 matrix Analytic geometry can be used to show that the volume of the solid With corners 0 a1 a2 a3 a1 a2 a1 a3 a2 a3 and a1 a2 a3 is the absolute value of the determinant Volume of Parallelopiped ldetAli For example if N H N Aa1 a2 a3 1 NC then the volume of the parallelopiped shown 0 in the plot is 2 Theorem Equivalent Recursive De nitions The determinant of the square matrix A can be obtained recursively by expanding in any row or in any column That is7 7L detA 1A1 271ija1j d6tltAijgt for any xed row i and j1 n detA 1A1 271 7aij detAj for any xed column j i 1 where AH is the submatrix obtained by removing the ith row and jth column of A 2 12 1 For example7 let A 10 4 1 7 as above If we expand in the 2nd row7 we get 2 2 8 12 1 2 1 2 12 detA710 2 8 4 2 8 71 2 2 Problem 1 Find each determinant using expansion 150 12 471 0720 1004 0030 50120 0011 aOOO 0b00 0050 000d aOOO 12500 5def0 yhij 233 Properties of Determinants Factorizations Let A B and E be square matrices of order n Then 1 det AT dam 2 detAB detAdetB 3 If A is a triangular matrix then detA 01110122quot an 4 If E is an elementary matrix for a row replacement then detE 1 5 If E is an elementary matrix for a row interchange then detE 71 6 If E is an elementary matrix for scaling a row by c then detE 0 Problem 2 Let A be an invertible square matrix Use the properties of determinants to explain why the determinant of A must be nonzero LU factorization and determinants Suppose that the square matrix A has an LU factorization where L is a unit lower triangular matrix and U is an echelon form matrix Then detA Permutation matrices and determinants Recall that a product of elementary matrices for row interchanges is called a permutation matrix If P is a permutation matrix then detP PALU form and determinants Suppose that the square matrix A does not have an LU factorization but that you can write PA LU for some permutation matrix P unit lower triangular matrix L and echelon form matrix U Then detA 234 Elementary Row Operations and Determinants The following theorem relates elementary row operations to determinants Theorem Row Operations and Determinants Let A be a square matrix Then 1 If B is obtained from A by a single row replacement7 then detB detA 2 If B is obtained from A by a single row interchange7 then detB idetA 3 If B is obtained from A by multiplying a row by c then detB c detA This theorem and the properties of determinants can be used to develop an alternative method for nding a determinant based on row operations For example7 l 74 2 l 74 2 l 74 2 71 7 0 0 3 2 0 3 2 72 8 79 72 8 79 0 0 75 In step 17 the row replacement R2 9 R2 R1 does not change the value of the determinant Likewise7 in step 27 the row replacement R3 9 R3 2R1 does not change the value of the determinant Since the third matrix is in echelon form7 its determinant is easy to compute Problem 5 Find each determinant using the method based on row operations 2784 1717 0 72879 366 2123 144 00722 1012 671121 02730 Methods for nding deteminants include expansion in rowscolumns using elementary row operations to reduce the matrix to echelon form or some combination of these approaches Properties of determinants can also be used in conjunction with these methods 0 A22 Problem 4 Let A All 0 J 7 A11 0 p O i i 7 0 q 0 A22 be a 2 x 2 block diagonal matrix Use the factorization above and the properties of determinants to demonstrate that the deter minant of A can be computed as follows detA detA11detA22 235 Invertibility and Determinants Theorem Invertible Matrices and Determinants Let A be a square matrix Then A is invertible if and only if the determinant of A is nonzero d2 1 1 1 1 1 1 1 1 1 detA1 c c2 0 071 6271Ci1d710 1 51 1d d2 0 d71d271 01d1 1 1 1 671d710 1 61 0 0 d C we know that A is invertible iff It is instructive to write out the proof of the theorem relating invertibility to determinants a Assume that A is invertible Since A is row equivalent to the identity matrix A can be written as a product of elementary matrices say A E1E2 Ek NOW complete this part 6 Assume that detA 7 0 and that A can be transformed to echelon form using k elementary operations EkEk1E1A U where U is an echelon form matrix NOW complete this part 236 Cramer7s Rule Cofactors and Inverses Suppose that A is an invertible matrix of order n and b E R Then we know that A21 b has the unique solution 1 A Cramer s rule tells us that each component of the unique solution is a ratio of determinants Speci cally d5tAib detA where Aib is the matrix obtained from A by replacing column i with b for each i For example ifn 2 A a1 a2 all an and b le7 then 2 0421 0422 b2 0422 0421 b2 1411 b1 an and 1421 all b1 Further A113 b gt 951 and x2 ugtto Problem 5 Use Cramer s rule to solve the matrix equation 7 Problem 6 Let A 1 1 1 b 1 6 c2 and b 0 assume that A is invertible7 and consider the 1 d d2 0 matrix equation A11 b Use Cramer s rule7 and the work we did on page 317 to nd a simpli ed formula for the second component of the solution7 m2 ijCofact0rs lf AM is the submatrix of A obtained by eliminating the ith row and jth column then the ij cofactor of A is the number CM 1ijd6tAij Given the i j cofactor de nition the recursive de nition of the determinant becomes detA 21 JCi if we expand in row i and detA ELI JCi if we expand in column j For this reason the recursive de nition is often referred to as the cofactor de nition of the determinant Cofactors and Inverses Let A be an invertible matrix Cramer s rule can be used to nd an explicit formula for each element of the inverse of A Speci cally since the jth column of the inverse can be obtained by solving A21 ej Cramer s rule tells us that the ij entry of the inverse of A is i detAej 1 A H 7 W for each z and 3 Interestingly detAiej C the ji cofactor and you get A4 2 all 0412 0413 an agg agg and consider nding the numerators of the entries 0431 0432 0433 For example let n 3 A in the rst column of A 1 the numerators of the 11 21 and 3 1 entries 1 a12 dis detA1el 0 a22 agg 0 a32 ass all 1 dis d6tA281 an 0 agg an 0 ass an an 1 detA3el agl agg 0 an a32 0 Explicit Formula for A l Based on Cramer s rule7 an explicit formula for the inverse of A is 011 012 39 39 39 Cm T A71 21 22 C27 detA E E Cnl Cng Cm That is7 A 1 is the reciprocal of detA times the transpose of the cofactor matrix 1 fl 0 3 0 2 l 71 Problem 7 Let A 7 0 0 72 1 2 72 0 9 Use determinants based on Cramer s rule to nd the 21 and 34 entries of A l MT210 Notebook 3 prepared by Professor Jenny Baglivo Copyright 2009 by Jenny A Baglivor All Rights Reserved 3 MT210 Notebook 3 31 Vector Spaces and Subspaces r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r Silrl Vector Spaces Over the Reals r r r r r r r r r r r r r r r r r r r r r r r r r r r r 3 12 Vector Subspaces and Spans r r r r r r r r r r r r r r r r r r r r r r r r r r r r r 3 2 Subspaces Related to Matrices and Linear Transformations r r r r r r r r r r r r r r r 3 21 Null Space and Column Space r r r r r r r r r r r r r r r r r r r r r r r r r r r r 322 Linear Transformations Range and Kernel r r r r r r r r r r r r r r r r r r r r 33 Linear Independence and Bases r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r 3 31 Linearly Independent Sets Linearly Dependent Sets r r r r r r r r r r r r r r r 3 32 Basis of a Vector Space r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r 3 33 Bases for Null and Column Spaces r r r r r r r r r r r r r r r r r r r r r r r r r 34 Unique Representations and Coordinate Systems r r r r r r r r r r r r r r r r r r r r r 341 Unique Representation Theorem Coordinate Vectors r r r r r r r r r r r r r r r r 342 Change of Coordinates in kSpace r r r r r r r r r r r r r r r r r r r r r r r r r r 343 Coordinate Mapping Functions lsomorphic Vector Spaces r r r r r r r r r r r r 3 5 Dimension and Rank r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r 3 51 Dimension of a Vector Space 352 Dimensions of Row and Column Spaces of a Matrix 353 Rank and Nullity of a Matrix Row Space r r r r r r r r r r r r r r r r r r r r r r 3 MT210 Notebook 3 This notebook is concerned with vector space concepts in linear algebra The notes correspond to material in Chapter 4 of the Lay textbook 31 Vector Spaces and Subspaces 311 Vector Spaces Over the Reals A vector space V is a nonempty set of objects called vectors on which are de ned the oper ations of addition and scalar multiplication satisfying the following rules 1 Closed Under Vector Addition For each 17 y 6 V7 1 y E V 2 Closed Under Scalar Multiplication For each 1 E V and c E R7 01 6 V 3 Properties of Addition Vector addition is commutative and associative 4 Zero Vector There is a zero vector satisfying 1 0 0 1 13 5 Additive Inverse For each 1 E V there is a y E V satisfying 1 y y 1 0 6 Properties of Scalar Multiplication Scalar multiplication satis es Cm y 0111 Cy 0 111 0111 dill am Cdm 111 11 for all cd E R7 my 6 V Ewample R For each n7 R with the usual de nitions of vector sum and scalar product is a vector space In fact7 R is our model vector space Ewample men Let men be the collection of all m x n matrices with the usual de ni tions of matrix sum and scalar product Then men satis es the vector space axioms7 where the zero vector of the matrix space is the zero matrix 0 of size in x n A particular example we will use often in illustrations is the vector space of 2 x 2 matrices szzz 3a75707d6R with the usual de nitions of matrix sum and scalar product In many ways7 the vector space MZXZ is just like the vector space R47 where R4 abcd R R DTQ with the usual de nitions of vector sum and scalar product More generally the vector spaces men and Rm have a lot in common Ewample Pm Let Pm be the collection of polynomials of degree at most n Pn pt a0a1ta2t2antn a0a1an E R with addition and scalar multiplication de ned in the following way H Addition lf p1 and 112 are polynomials in Pn then 111 112 is the polynomial whose value at each 25 is the sum of the values p1t and p2t 121 p2t p1t p2t for each t e R to Scalar Multiplication If p is a polynomial in Pm and k E R then kp is the polynomial whose value at each 25 is k times the value pt kpt k pt for each 25 E R You add polynomials by adding their values at each 25 you scale a polynomial by scaling its value at each 25 Then Pm satis es the vector space axioms where the zero vector77 of the polynomial space is the constant function all of whose values are zero 0t 0 for all t E R In many ways the vector space Pm is just like77 the vector space Rn where do Rn 11 a0a1an R a with the usual de nitions of vector sum and scalar product Each column vector corresponds to the coef cients of a polynomial of degree at most Note that Pm consists of polynomials whose degrees are at most n not exactly n Thus P0CP1 CCPn1 CPn For example if n 2 then P0pta aER the vector space of constant functions is a subspace of P1 pt abt ab E R the vector space of constant and linear functions which is in turn a subspace of P2 pt abtct2 abc R the vector space of constant linear and quadratic functions Ewample Ca b Let 1 b C R be an interval and Ca b f 1 b a R f is continuous be the collection of real valued continuous functions with domain 1 b with addition and scalar multiplication de ned in the following way H Addition lf f1 and f2 are functions in Ca b then f1 f2 is the function whose value at each x E 1 b is the sum of the values f1 and f2m f1 f2 f1z f2m for each x E 1b to Scalar Multiplication If f is a function in Ca b and k E R then kf is the function whose value at each x is k times the value fm for each x E 1 b You add functions by adding their values at each x in the domain you scale a function by scaling its value at each Then Ca b satis es the vector space axioms where the zero vector77 of the continuous function space is the function all of whose values are zero 0m 0 for all z E 1 b Some important notes 1 The vector space of continuous functions on 1 b is Lot like R for any n 2 Polynomial functions are continuous for all reals Thus if we let Pm 1 b be the collection of polynomial functions with domains restricted to the interval 1 b then P0ab C P1ab C C Pnab C C Cab Each Pna b satis es the vector space axioms 3 Differentiable functions are continuous Thus if we let Da b be the collection of differentiable real valued functions with domain 1 b then Dab C Cab Da b satis es the vector space axioms 312 Vector Subspaces and Spans Let H be a subset of the vector space V7 H Q V H is said to be a subspace of V if the following three conditions are satis ed 1 Contains Zero Vector 0 E H 2 Closed under Addition lf 17 y 6 H7 then 1 y E H 3 Closed Under Scalar Multiplication lf 1 E H and c E R7 then 61 6 H For example7 the collection of vectors corresponding to the m axis in 2 space7 aeRcR2 satis es the three subspace conditions Thus7 H is a subspace of R2 Note that if H is a subspace of V7 then H is a vector space in its own right That is7 if the three conditions above hold7 then all conditions needed for a vector space will hold Linear combinations and spans A linear combination of vectors vhvz gvk E V is a vector of the form then 013901 cg39vz ck39vk where 0102 ck E R are scalars The span of the set 39017 3902 mk is the collection of all linear combinations of the 3905s Spanv139vz39vkc139v1 0202 ckvk cl 6 R for all Q V For example7 the collection of vectors corresponding to the m axis in 2 space is the span of the rst standard basis vector7 Hspaniaiaiaiisi The following theorem tells us that spans are subspaces7 and thus vector spaces Theorem If H Q V is the span of a nite set of vectors7 then H is a subspace of V Not all subspaces can be written as the span of a nite set of vectors Thus7 to demonstrate that a subset is a subspace7 you need to either 0 demonstrate that all three subspace conditions hold7 or 0 demonstrate that the subset is the span of a nite set of vectors Further7 to demonstrate that a subset is not a subspace7 you need to demonstrate that one of the three conditions fails Problem 1 Which7 if any7 of the following subsets are subspaces Why y mac C R27 where m is a xed constant y mac b C R27 where m and b 7 0 are xed constants c D x2y2 1CR2 a a7dER CMZXz a b C d 2dabc cMZXZ IadibcOCM2gtlt2 gHptat aERCP3 h Zptabtct2dt3 p t0f0r allt CP3 E f 11 a R f is continuous and at fb C Cab 32 Subspaces Related to Matrices and Linear Transformations 321 Null Space and Column Space Let A be an m x 71 matrix There are two important subspaces related to A 1 Null Space The null space of A is the set of solutions to the matrix equation Am O NullA m Am 0 Q R NullA is a subspace since it can be written as the span of a nite set of vectors 2 Column Space The column space of A is the set of all b for which Am b is consistent ColA b Am b is consistent Q R ColA is a subspace since it is the span of the columns of A l 2 For example7 let A a1 a2 a3 73 1 76 1 0 72mg 72 0 a m 0 m3 0 where m3 is free7 0 1 13 nbozm COO 10 SinceA 0 7317 21 0 10 0 N 01 0 00 72 1 NullA Span 0 C R3 and l 2vowelmamspanmmcw since 13 2111 COM 1 0 2 N 0 1 0 0 1 0 we know that 7 ii if g Write NullA and ColA as spans of nite sets of vectors Simplify your answers as much as possible Problem 1 Let A 322 Linear Transformations Range and Kernel Let V and W be vector spaces7 and T V a W be a function which assigns to each 1 E V a value or image b T1 E W T is said to be a linear transformation if the following two conditions are satis ed 1 T Respects Addition T1 y T1 Ty for every my 6 V 2 T Respects Scalar Multiplication Tc1 cT1 for every 1 E V and c E R Note If T is a linear transformation then TO 0 To demonstrate this fact7 let 0 2 in the second condition Then T0 T20 2T0 0 2T0 7 T0 T0 Problem 2 Which7 if any7 of the following are linear transformations Why a T M2X2 a R with rule Ta adi bc 1 T M2gtlt2 a MZXZ with rule TA MA7 where M is a xed 2 x 2 matrix c T P2 a P27 where the image of the polynomial pt a bt 0252 is the polynomial Tpt a 1 b 1tc 1t2 d T MW e M2X2 with rule TA A AT Subspaces related to T There are two subspaces related to a linear transformation T 1 Range of T The range of T is the set of all images RangeT b b T1 for some 13 Q W 2 Kemel of T The kemel of T is the set of all pre images of O KernelT 113 T1 0 Q V 1 0 2 For example7 if T R3 a R3 with rule T1 A11 73 1 76 J23 then the work shown 2 1 4 on page 10 implies that RangeT ColA Span 7 1 C R3 2 1 72 and KernelT NullA Span 0 C R3 1 13 Problem 5 The following three functions are linear transformations In each case7 give explicit descriptions of the kernel and range of the transformation a T P3 a P37 where the image of p is the function Tpt p39t for each 25 b T R2 a MZXZ is the function with rule ajbb I c T P3 a R2 is the function with rule Tp 33 Linear Independence and Bases 331 Linearly Independent Sets Linearly Dependent Sets Let V be a vector space The subset 017027 7016 Q V is said to be linearly independent when the linear combination 01m cg39vz ck39vk equals the zero vector only when all scalars ci are zero That is7 when 610162 2Ckvk0 gt 6162Ck0 Otherwise7 the set 017027 7016 is said to be linearly dependent Note that 1 If vi 0 for some i7 then the set 017027 7 we is linearly dependent 2 If each vi 7 07 then the set 017027 7016 is linearly dependent iff one vector can be written as a linear combination of the others Problem 1 In each case7 determine if the set is linearly independent or linearly dependent 1 2 3 4 5 6 1 01702703 1 7 0 71 Q R4 7 8 9 ltbgt M17M27M37 1112211 11ng C P1t7172t7p3t Q Pz7 Where p1t 17t7p2t17t2and p3t1 2t 7 3252 for all t E R 332 Basis of a Vector Space A basis for the vector space V is a linearly independent set that spans V Specifically7 the set 3901 3902 7016 is said to be a basis for V if 1 01702 7016 is linearly independent and 2 V Span39v139uz7 7 Uk Problem 1 continued Use the work you did in Problem 1 to nd bases for a V Spanivl71 27v3 E R4 1 V SpanM17M27M3 Q M2gtlt2 C V SPaHP1t7P2t7P3t Q P2 Problem 2 In each case7 nd a basis for V ltagt vspanr 3H8 5H3 www b V A Each column of A has sum 0 Q MZXg 333 Bases for Null and Column Spaces Let A be an m x 71 matrix Then 1 The method we use to write NullA as the span of a nite set of vectors automatically produces a linearly independent set7 hence a basis for NullA 2 The pivot columns of A form a basis for Col 1 2 O 4 Problem 5 Let A 2 4 71 3 Use the fact that 3 6 2 22 1 2 O 4 O 1 2 O 4 O 1 2 O 4 0 A10 2 4 71 3 O N O O 71 75 O N O O 1 5 O 3 6 2 22 O O O 2 1O 0 O O O O O to nd bases for NullA and ColA 34 Unique Representations and Coordinate Systems 341 Unique Representation Theorem Coordinate Vectors Theorem Unique Representations Let B b17 b2 7bk be a basis for the vector space V Then for each 390 6 V7 390 c1b1 cgbz ckbk for unique constants 0102 ck Proof The proof of the unique representation theorem is straightforward Suppose that 1 C1b162526kbk d1b1 dzbzdkbk for scalars ci di 6 R Then7 by subtracting the second representation from the rst7 we get 0 Cl d1b1 t 02 d2bz ck 7 dkbk NOW complete the proof7 Coordinates and coordinate vectors The ci s are called the coordinates of 39v in basis 3 Listed as a vector in Rk7 c MB is called the coordinate vector of 39v in basis 3 For example7 consider the standard basis in R27 1 0 L M v H l and let 390 be a vector in R2 Since 390 81 yez the coordinate vector of 39v in the standard basis is the vector itself7 lvlg 1 Problem 1 Consider the basis Bb17b2 i 7i of R27 and let 1 be a vector Find the coordinate vector of v in basis B 148 The plot shows a grid of linear combinations of the form 6121 dbg7 Where either 6 or d is an integer 1 1 Problem 2 B b17172 0 0 J is a 0 2 basis for a subspace V of Rgi Let 1 5 0 6 Vi Find 148i 4 V corresponds to the zz plane in 3spacei The plot shows a grid of linear combinations of the form 6171 d122 Where either 6 or d is an integer Problem 5 B 17t21t2 is a basis for a subspace V of P2 a bt ct2 Let 1125 2 7252 Find plg za7b706R 342 Change of Coordinates in kSpace Let B b17 b2 bk be a basis for Rk and P5 be the matrix Whose columns are the his Pg b1 b2 bk Since the columns of P5 are linearly independent7 we know that P5 is invertible and P5 01 gt cPg139v or vlg Pgl39v For this reason7 P5 is called the change of coordmates matrix Problem 1 continued As above7 let 3 bhbz i J ii and 39v Find P57 and verify that vlg Pglv is the same as the vector obtained earlier 343 Coordinate Mapping Functions Isomorphic Vector Spaces Let B b17 b27 bk be a basis for the vector space V The transformation T V a Rk with rule T39v MB is called the coordinate mapping function For example7 B 172529 is the standard basis for P2 a bt cit2 a 125 6 R With respect to this basis7 the coordinate mapping function is the transformation a T P2 a R3 which maps 1125 a bt 0252 to Tp b C Theorem Coordinate Mapping The coordinate mapping function T V a Rk described above is a linear transformation that is both one to one and onto Isomorphisms Isomorphic vector spaces An isomomhism is an invertible linear trans formation Under the conditions of the coordinate mapping theorem7 T is an isomorphism and the vector spaces V and Rk are said to be isomomhic have the same structure A problem involving vectors in V can be translated to Rk via the coordinate mapping func tion7 solved in Rk and translated back Example continued Consider the set 111t7 112t7 p3t 1 22527 4 1 25 52527 3 1 225 To determine if 121t7 112t7 p3t is a linearly independent set in P27 for example7 we could equivalently determine if 1 4 3 0171273903 O 1 2 is a linearly independent set in R3 0 SincelvlL v2 valO 1 4 3 0 1 4 3 0 1 4 3 0 1 0 75 0 0 1 2 0 N 0 1 2 0 N 0 1 2 0 N 0 1 2 0 7 2 5 0 0 0 73 76 0 0 0 0 0 0 0 0 0 we know that both 0 the set of vectors in R3 and o the set of pre images in P2 are linearly dependent Further7 we know that p3t 75p1t 2p2t for all t E R 1 is the standard basis for MZXZ With respect 1 O O 1 Problem 4 B 0 O O O to this loasis7 the coordinate mapping function is the transformation a TM2X2HR4withruleTlt 1 Using the standard basis and coordinate mapping function7 a Determine if 2 73 1 71 71 3 lawman 2H3 3H 1 Si is a linearly independent set in MZXZ If the set is linearly dependent7 express one of the matrices as a linear combination of the others 1 Demonstrate that A1A27A37A4 HHHH is a linearly independent set in MZXZ c Express 3 1 0 2 as a linear combination of the 4 matrices given in part 35 Dimension and Rank 351 Dimension of a Vector Space The following theorem tells us that if a vector space V has a basis with k elements then all bases of V must have exactly k elements Theorem Finite Bases If B b1b2 bk is a basis for the vector space V then 1 Any subset of V with more than k vectors must be linearly dependent 2 Every basis of V must contain exactly k vectors 3 Every linearly independent set with exactly k vectors is a basis of V Note that each part of this theorem can be proven using the coordinate mapping theorem kdimensional vector spaces If V has a basis with k elements then V is said to be k dimensional and we write dimV k the dimension of V equals k For example given positive integers m and n dimR n dimPn n 1 and dimMmgtltn mn Trivial vector space By convention the dimension of the trivial vector space is zero dim0 0 The trivial vector space does not have a basis since 0 is linearly dependent oodimensional vector spaces If V is not trivial and is not spanned by a nite set of vectors then V is said to be in nite dimensional Examples of in nite dimensional vector spaces include 1 The vector space of polynomials of all orders PDo pt pt is a polynomial function of any degree with the usual de nitions of addition and scalar multiplication 2 The vector space of continuous real valued functions on the interval 11 Cab f 11 a R l f is continuous with the usual de nitions of addition and scalar multiplication Theorem Subspaces If H Q V is a subspace and dimV k then dimH dimV k Problem 1 In each case7 nd the dimension of the subspace H ab25 2a2b4cd bcd 3a30d a H za7b707d6R ERA b H Spantt271t71 425 3252 g P2 352 Dimensions of Row and Column Spaces of a Matrix Let A be an m x 71 matrix The dimension of the null space corresponds to the number of free variables when solving A11 0 The dimension of the column space corresponds to the number of pivot columns of A Problem 2 In each case7 nd the dimensions of the null and column spaces of A 176 0 0 720 0 12 74 1A 1 3 2 6 aA 0 0 0 5 5 bA 2 6 75 712 0 0 0 0 0 1 3 3 6 353 Rank and Nullity of a Matrix Row Space Let A be an m x 71 matrix Then the rank of A is the dimension of its column space and the nullity of A is the dimension of its null space rankA dimColA and nullityA dimNullA Since each pivot column of A corresponds to a basic variable when solving A11 0 and the number of basic variables plus the number of free values must equal the total number of variables we have the following theorem Rank Theomm If A is an m x 71 matrix then rankA nullityA 71 Row space of A In many applications we need to work with both a matrix and its trans pose Let A be an m x 71 matrix The row space of A is the column space of AT Row A Col AT g R Row Space Theorem If A and B are row equivalent then the row space of A and the row space of B are equal RowA RowB Further if B is in echelon form then the nonzero rows of B form a basis for the row space of B and for the row space of A 1 76 O O 720 O 1 2 74 1 i i i i For example let A 0 0 0 5 5 Since A is already in echelon form a basis for O O O O O the row space of A is 177670707 2070717277471707070757539 lt is common practice to write these vectors in row form 71 73 2 6 Problem 21 continued As above let A 2 6 75 712 1 3 73 76 Find rankA and a basis for RowA Problem 3 Let A 1 Find rankA and bases for ColA7 NullA and RowA Rank and nullity continued Let A be an m x 71 matrix If A is row equivalent to B and B is in echelon form7 then the number of nonzero rows of B is the same as the number of pivots of A Thus7 dimRowA dimColA Further7 1 rankAT rankA 2 rankA nullityA n and 3 rankAT nullityAT m Problem 4 Let In be the n x 71 identity matrix and Oan be the m x n zero matrix Fill in the following table Size of A rankA m X n rankAT nullityA nullityAT A 3 03x2 7 3 3 02x3 02x3 A 03x2 Problem 5 Let A be an m x 71 matrix In each case7 determine if the statment is TRUE or FALSE7 and give a reason for each choice a ColA Collt gt b NullA Nullq 0 Col AT ColQ T Col AT ATD d Null AT NullQ Tgt Null AT ATD MT210 Notebook 2 prepared by Professor Jenny Baglivo Copyright 2009 by Jenny A Baglivor All Rights Reserved 2 MT210 Notebook 2 2 1 Matrices and Their Operations r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r 2 11 Rectangular Arrays r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r 212 Square Matrices7 Types of Square Matrices7 Identity Matrix r r r r r r r r r r r 2 13 Matrix Sum and Scalar Product Linear Combination r r r r r r r r r r r r r r 2 14 Matrix Transpose r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r 2 15 Dot Product and Matrix Product Composition of Transformations r r r r r r r 2 16 Properties of the Basic Operations r r r r r r r r r r r r r r r r r r r r r r r r r 2L7 Powers of Square Matrices r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r 2 18 lnverses of Square Matrices Special Cases Properties of lnverses r r r r r r r r 2L9 lnverses and Solving Systems Finding lnverses r r r r r r r r r r r r r r r r r r r 2110 Elementary Matrices Relationship to lnverses r r r r r r r r r r r r r r r r r r r Factoring and Blocking r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r 2 21 Factorization LU Factorization r r r r r r r r r r r r r r r r r r r r r r r r r r r 2 22 Permutation Matrices PALU Form r r r r r r r r r r r r r r r r r r r r r r r 2 23 Partitioned Matrix7 Block Matrix lnverses in Special Cases r r r r r r r r r r r Determinants r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r 2 31 Determinants and Areas r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r Recursive De nition of Determinants Volumes r r r r r r r r r r r r r r r r r r Properties of Determinants Factorizations r r r r r r r r r r r r r r r r r r r r r Elementary Row Operations and Determinants r r r r r r r r r r r r r r r r r r lnvertibility and Determinants Cramer s Rule7 Cofactors and lnverses r r r r r r r r r r r r r r r r r r r r r r r r 26 27 29 30 32 2 MT210 Notebook 2 This notebook is concerned with matrix methods in linear algebra The notes correspond to material in Chapters 2 and 3 of the Lay textbook 21 211 Matrices and Their Operations Rectangular Arrays An in x n matrix A is a rectangular array with in rows and n columns7 1 2 all 03912 aln 03921 03922 39 39 39 a2n A i i i a1 12 an laijl am am amn m x n is the size of the matrix A there are in rows and n columns aij the number in the row i7 column j position is the ij entry of A The term after the rst equal sign shows the entries explicitly In the term after the second equal sign7 A is written as an array of n columns where each column is a vector in R The last term is used as shorthand when the size of the matrix is understood 212 Square Matrices Types of Square Matrices Identity Matrix 1 2 03 F U A square matrix is a matrix with m n An n x n square matrix is often called a square matrix of order ii If A is a square matrix of order n then the diagonal elements of A are the elements an 01227 7 am If A is a square matrix7 then The identity matrix of order n denoted by In is the n x n diagonal matrix all of whose diagonal elements equal 1 We sometimes just write I for the identity matrix when the order is clear from the problem For example7 let a b c a b c a 0 0 a 0 0 A1 I d 6 7 A2 0 d 6 7 A3 I C 0 and A4 0 b 0 7 c e f 0 0 f d e f 0 0 c where a through 1 are any numbers Then A1 is a symmetric matrix7 A2 is an upper triangular matrix7 A3 is a lower triangular matrix and A4 is a diagonal matrix As a diagonal matrix7 A4 is also upper triangular7 lower triangular and symmetric Note Coe icient matrices and row operations If A is the coefficient matrix for an n by n system7 then after the forward pass7 A is transformed to an upper triangular matrix If the system has a unique solution7 then the reduced echelon form of A is the identity matrix After the forward and backward passes7 A is transformed to the identity matrix Note Identity matriw and standard basis vectors The columns of the identity matrix are the standard basis vectors we discussed in the last section That is we can write In as follows 1 0 0 0 1 0 177 el 82 e747 0 0 1 where ej is the vector with 1 in position 37 and 0 everywhere else7 for j 17 27 7n 213 Matrix Sum and Scalar Product Linear Combination Let A aw and B blj be m x n matrices7 and let c7d E R be scalars Then matrix sum and scalar product are de ned as follows 1 The matrix sum of A and B is the m x n matrix AB a1b1 a2b2 anbn Equivalently7 A B is the matrix with i7j entry A B aij bij 2 The scalar product of A by c is the m x n matrix cA cal caz can Equivalently7 cA is the matrix whose i7j entry is 31477 caij Further7 the linear combination CA 1 dB is the m x n matrix cA dB cal dbl caz db2 m can dbn Equivalently7 CA 1 dB is the matrix whose i7 j entry is CA 1 dB7j cal7 dblj 2 3 For example if A 7 72 0 AB iii 5A 777 and5A72B 777 214 Matrix Transpose The transpose of the m x 71 matrix A denoted by AT A transpose is the n x m matrix Whose tj entry is aji ATLJ aji For example if 1 712 3 T A7 4 5 6 thenA 7 g OJC gtgt 1 Note that if A is a symmetric matrix then AT A For convenience let 1i 6 R for t 1 2 m be the columns of AT Then A is the matrix Whose rows are the aiT s That is if ATa1 a2 amthenA 215 Dot Product and Matrix Product Composition of Transformations 1 If 39v and 39w are vectors in R then their dot product is the scalar 39U39Lu 39UT39w Ulwl 12ng Unwn 2 Let A be an m x 71 matrix and let B b1 b2 bp be an n x p matrix Then the matrix product AB is the m x p matrix Whose jth column is the matrix vector product of A with the jth column of B ABAb1 Ab2 Abp Equivalently AB is the m x p matrix Whose t j entry is the dot product of the ith row of A with the jth column of B ABM ai bj Otinj ailblj 11392ij 39 39 39 ambnj where a is the ith row of A for t 1 m Dot products are de ned for vectors of the same length For example7 if 2 2 2 39v 3 and 39w 0 7 then 39U39Lu 2 3 6 0 22 30 64 28 6 4 Matrix products are de ned for compatible matrices that is7 when the number of columns of the matrix on the left equals the number of rows of the matrix on the right 2 71 For example7 if A g f and B 2 73 l 0 1 4 7 then 4 77 0 but BA is unde ned since they are not compatible 3 7 2 2 73 Problem 1 Let A j g f l and B 0 1 Find AB and BA ls AB BA 4 77 Matrix product and composition The matrix product has many applications For ex ample the matrix product can be interpreted as the standard matrix of the composition of two linear transformations Speci cally if TB RP a R has matrix B TA R a R has matrix A and T TA 0 TB is the composite function then T RP a Rm de ned by Tw TATBJB has standard matrix AB Problem 2 Let TB R2 a R2 and TA R2 a R2 be linear transformations with rules a 2 3 g TBw Ban Z 3 an and 71433 Aw E 73 an 5 5 13 13 Find the standard matrix of the linear transformation T TA 0 TB Footnote Geometric Interpretation In the problem above transformation TB corre sponds to rotation about the origin through the positive angle 0 arctan g and the trans formation TA corresponds to re ection in the line y Under the composite function T TA 0 T3 the unit square for example is rst rotated by 0 and then re ected in the line as illustrated below TATBR 216 Properties of the Basic Operations Some properties of the basic operations are given below In each case7 the sizes of the matrices are assumed to be compatible H Commutative A B B A to Associative A B C A B C and ABC ABC 03 Distributive AB C AB AC7 B CA BA CA F Scalar Multiples Given 0 E R CAB cAB ACB7 CA B CA CB and CAT CAT 0 Identity ImA A AI when A is an m x 71 matrix a Transpose of Transpose ATT A 1 Transpose of Sum A BT AT BT 00 Transpose of Product ABT BTAT Note 1 Commutativity Matrix sum is commutative but matrix product is not In fact7 as we have already seen7 it is possible that AB is de ned but that BA is unde ned Note 2 Associativity The associative law for products says that products like ABC are well de ned It is up to you whether you would like to compute AB rst and then multiply on the right by C7 or to compute BC rst and then multiply on the left by A 217 Powers of Square Matrices If A is an n x n matrix7 then we can use a shorthand for the powers of A Namely7 A2 AA7 A3 AAA7 A4 AAAA7 and so forth Problem 3 Let A l 1 j Find A2 A3 A4 and A100 a 0 Problem 4 Let A 0 b J a Find A27 A37 A4 and AP where p is a positive integer b What properties are suggested by your solution to part a Be as complete as possible a1 0 0 b1 0 0 Problem 5 Let A 0 a2 0 and B 0 b2 0 a Find AB and BA 0 0 a3 0 0 b3 b What properties are suggested by your solution to part a Be as complete as possible 218 Inverses of Square Matrices Special Cases Properties of Inverses The n x n matrix A is said to be invertible if there exists a square matrix C satisfying AC CA In where In is the identity matrix of order n Otherwise A is said to be not invertible Square matrices that are not invertible are also called singular matrices invertible square matrices are also called nonsingvlar matrices Problem 6 Use the properties of the basic matrix operations given in Section 216 page 8 to demonstrate that inverses are unique Notation for the matrix inverse If A is invertible then A 1 read A inverse is used to denote the matrix inverse of A Special cases Finding the inverse of a square matrix or determining that an inverse does not exist can often involve many computations Here are a few special cases H Powers If A is a square matrix and p is the smallest positive integer satisfying AP I then A is invertible with inverse A 1 Ap l ln symbols I AP A A174 A174 A gt A 1 AM to 2 by 2 Matrices Let A Z be a 2 x 2 matrix Then A 1 d 2 whenadibcy O adibc it If ad 7 be 0 then A does not have an inverse Note ad 7 be is the determinant of A 3 Diagonal Matrices Let A be the diagonal matrix with nonzero diagonal elements that is with an 0 for all Then A is invertible with inverse 1 a11 0 0 A71 0 1a22 0 l 0 0 1 amj That is A 1 is the diagonal matrix Whose diagonal elements are the reciprocals of the diagonal elements of A If an 0 for some i then A is not invertible 10 For example 1 If A 1 then from the work we did in Problem 3 we know that A3 I 7 then the determinant is and A 1 J Thus the inverse of A is A 1 A2 77 2 lfA 2 3 HA 3 0 0 0 0 1 0 then A 1 0 1 4 An example of a square matrix A 7 I satisfying A 1 A is Properties of inverses Let A and B be invertible square matrices of order n and let 0 be a nonzero constant Then 1 Inverse of Identity Igl In for each n 2 Inverse of Inverse A gil A 3 Inverse of Product AB71 B 1A 1 4 Inverse of Scalar Multiple 61471 1CA 1 5 Inverse of Transpose A7371 A71 Demonstration of Property 5 It is instructive to demonstrate the third property Assume that A and B are invertible square matrices of size n and let C B lA l We need to show 1 ABC In and 2 CAB In NOW complete the proof 219 Inverses and Solving Systems Finding Inverses The following theorem tells us that if the coef cient matrix of an n by n system of linear equations is invertible then the unique solution to the system can be found using inverses Speci cally Theorem Uniqueness Theorem Let A be an invertible n x 71 matrix Then the matrix equation A21 b has the unique solution 1 A lb for any b E R To see this note that A11 b A 1A1A 1b A lA1A 1b InmA 1b 11A 1b Problem 7 Use inverses to solve the 2 by 2 system 77ml 8x2 73 and 2x1 7 2x2 5 Finding the inverse and solving 11 linear systems An algorithm for nding the inverse can be developed using the following observations Let A be an invertible matrix of order n and In 81 82 en be the identity matrix of order n For convenience let C A l Then AC In can be written as follows AC A01 A02 Acn 81 82 en Thus the jth column of the inverse is the solution to the system ACj ej To turn the observations which assume that the inverse exists into an algorithm for nding the inverse we need the following facts Fact 1 The n x 71 matrix A is invertible if and only if A is row equivalent to In If A is row equivalent to In then any sequence of elementary row operations that reduces A to In also transforms In to A l Fact 2 Suppose that A and D are square matrices of order n Then if AD In then A is an invertible matrix Algorithm for nding an inverse Let A be an n x 71 matrix Form the n x 271 augmented matrix Whose last 71 columns are the columns of the identity matrix If A is row equivalent to the identity matrix7 then the n x 271 augmented matrix will be transformed as follows Allnl1nlA 1l That is7 the last 71 columns will be the inverse of A Otherwise7 A does not have an inverse 2 0 0 2 0 0 l 0 0 ForexampleletA 73 01 ThenAl13 73 01010 0 4 0 0 4 0 0 0 l 2 0 0 1 0 0 1 0 0 0 0 0 04000101000 A 100 001310 001g10 10 6 l 2 Problem 8 Find the inverse ofA 3 0 1 or state that the inverse does not exist 0 l 0 l a 12 Problem 9 Find the inverse ofA 0 1 c or state that the inverse does not exist 0 0 l 2110 Elementary Matrices Relationship to Inverses An elementary matrix E7 is one obtained by performing a single elementary row operation on an identity matrix Since each elementary row operation is reversible7 each elementary matrix is invertible For example7 let n 2 1 If the operation is R2 9 R2 2R1 with reverse operation R2 9 R2 7 2Rl7 then 710 17 10 E21andE 721 2 If the operation is R2 9 3R2 with reverse operation R2 E R2 then E and E 1 3 If the operation is R2 lt gt R1 with reverse operation R2 lt gt Rl7 then Left multiplication performs operation An elementary row operation performed on a matrix A corresponds to left multiplication by the corresponding elementary matrix For example7 if A 7 1 72 J and E is the rst matrix above7 then the product 7 1 O 3 1 O 7 3 1 O EA7217607470274 has the effect of replacing R2 by R2 2R1 Relationship to inverse of A Let A be an n x 71 matrix Suppose that A can be trans formed to the identity matrix using a sequence of p elementary row operations with elementary matrices E17 E27 Ep Then EpEp1E2E1A EpEp1E2E1A In a A 1 EPEZFL E2E1 Thus7 A 1 is the product of the elementary matrices used to transform A to In Further7 A is the product of the inverses of the elementary matrices in the opposite order A 1411 1 EpEp1E2E11 EflEglE1 Problem 10 Let A 7 The work on page 13 implies that A can be transformed DOOM HgtOO 0 1 0 to the identity matrix using the following four elementary row operations 3 1 1 1 RZ H 32 5317 2 Rz lt gt R37 3 32 H 11327 431 H 5R1 Find the corresponding elementary matrices E17 E27 E3 and E47 and verify that the inverse of A equals the product E4E3E2E1 22 Factoring and Blocking 221 Factorization LU Factorization A factorization of matrix A is an equation that expresses A as the product of two or more matrices A useful rst example is the LU factorization of a coef cient matrix The in x n matrix A has an LU faotorization if it is row equivalent to an echelon form matrix using row replacements only In this case7 A LU where 1 L is an m x in lower triangular matrix with 1 s on the diagonal and 2 U is an m x n echelon form matrix Note that a lower triangular matrix with 1 s on the diagonal is called a unit lower triangular matrix If the coef cient matrix A is square7 then U will be an upper triangular matrix 3 l 76 0 74 only one elementary row operation7 R2 9 R2 2R1 Let E be the elementary matrix for the operation Then For example7 if A 7 then A is row equivalent to an echelon form matrix using 7 10 31 O 7 31 O EA7217607470274 7 1 31 0 7 10 31 0 7 A E 0274 721H0274 LU 31 0 Problem 1 Find an LU factorization for A 76 0 74 0 6 710 Problem 2 Find an LU factorization for A Ozugtm on to roam l l H Problem 5 What patterns7 if any7 do you see in the solutions to the problems above Solving linear systems ef ciently Suppose that A LU where L is a unit lower triangu lar matrix and U is an echelon form matrix Then the matrix equation A11 b can be solved efficiently as follows Consider b A21 LU1 LU1 Ly7 where y U113 The efficient method is 1 Solve Ly b for y 2 Solve U11 y of 13 For example7 let A 7 7 J and b 73 J and assume that A LU 7 1 J 2 7i J is the LU factorization of A To solve A11 b using the method above7 10 2 10 2 2 1 Lib72172N012y2 2 73 2 2 0 8 10 4 4 2 Uiy0 12 012N012x2 is the solution to A11 b Problem 1 continued Let A Use the LU factorization 3 1 72 76 0 74 and b 2 0 6 2 of A we computed earlier to 1 solve Ly b for y and 2 solve U11 y for 13 Ef ciency of LU factorization The method above solves the matrix equation A11 b by 1 factoring A 2 using the equivalent of forward substitution to nd y and 3 using the equivalent of backward substitution to nd 13 For n by n systems for example 1 solving A11 b directly takes roughly 713 operations while 2 solving Ly b and U11 y together take roughly 712 operations So the method is useful when n is very large and the same coefficient matrix will be used with many different b s Large scale heat ow and largescale air ow problems are examples of situations where the method above is useful In addition since scalings are held off as long as possible the method is useful in situations where scalings could cause numerical roundoff problems 222 Permutation Matrices PALU Form A permutation matn39r P is elementary matrix corresponding to a row interchange or a product of elementary matrices corresponding to multiple row interchanges For example if n 4 o The permutation matrix corresponding to R1 lt gt R3 is o The permutation matrix corresponding to 1 R1 lt gt R3 followed by 2 R3 lt gt R4 is Use of permutation matrices in factoring lf row interchanges are needed to transform A to echelon form7 then an LU factorization is not possible But7 we can write PA LU7 where P is a permutation matrix 00 20 731001 For example let A 7 1 0 0 0 Since 00781 00 20 1000 1000 1000 A73100N3100N0100N0100 710 00 00 20 0020 00207 00781 00781 00781 0001 using operations 1 R1 lt gt R3 2 R2 9 R2 7 3Rl7 and 3 R4 9 R4 4ng7 we can construct the PA LU form as follows 223 Partitioned Matrix Block Matrix Inverses in Special Cases A partitioned matrix or block matrix is one Whose entries have been grouped into smaller matrices using vertical and horizontal cuts through the matrix For example 120 3110 A 3 4 0 A11 12andB 74 3 01 006 22 2 00 B11 312 A21 74 321 322 are examples of 2 x 2 block matrices Note that 1 Matrices of the same size that is with the same numbers of rows and columns and blocked in the same way that is so that corresponding blocks have the same numbers of rows and columns can be added by adding blocks 2 Matrices compatible for multiplication and blocked in compatible ways can be multiplied in blocks For example for compatible 2 x 2 block matrices A11 A12B11 312 A21 A22 321 322 1411311 1412321 1411312 1412322 1421311 1422321 1421312 1422322 as long as all multiplications and additions are valid For example the 2 x 2 block matrices given above cannot be added but they can be multiplied Note that ll7 where O is a zero matrix of an appropriate size and I is an identity matrix of an appropriate size using block structure simpli es the process AB A11 0 Bu 1 1411311 A11 0 A22 B21 0 A22B21 O Problem 4 Let A Compute the product AB Take advantage of the block structure as much as possible 2 x 2 block diagonal matrices and their inverses Let A be a square matrix of order n Then A is a 2 X 2 block diagonal matrix if A A51 A0 J 7 Where All and A22 are square matrices of orders p and q7 22 respectively7 and each 0 is a zero matrix of an appropriate size For example7 the following matrix is a 2 x 2 block diagonal matrix 7 A11 0 L AZZTT Where All is a square matrix of order 2 and A22 is a square matrix of order 3 The following theorem tells us that inverses of block diagonal matrices and block diagonal Speci cally7 Theorem Inverses Let A be a 2 x 2 block diagonal matrix as de ned above Then A is invertible if and only if both A11 and A22 are invertible 71 Further7 if A is invertible7 then A 1 A11 31 0 A22 It is instructive to write out the proof of the theorem on inverses 71 lt Su ose that All and A22 are invertible and let C A11 0 Then a pp 0 A 2 71 2 direct calculation shows that AC CA p O In Thus7 C A l 0 q a Suppose that A is invertible and let C Cu 012 be the inverse of A C C 2 Then AC In implies 2 A11 0 011 012 1411011 1411012 Ip 0 0 A22 021 022 1422021 1422022 0 q 39 NOW7 complete the proof Problem 5 In each case use block methods to nd the inverse of A 530 1A640 002 OOgtgtCA3 00000 5 0 7 6 0 21470 4 0 3 2 x 2 block upper triangular matrices and their inverses Let A be a square matrix of order n Then A is a 2 x 2 block upper triangular matrix if A A i A 51 A12 Where All and A22 are square matrices of orders p and q 22 respectively and O is a zero matrix of an appropriate size For example the following matrix is a 2 x 2 block upper triangular matrix 7 A11 A12 7 Where All is a square matrix of order 2 and A22 is a square matrix of order 3 The following theorem gives us the form of the inverse of a block upper triangular matrix when it exists Speci cally Theorem Inverses Let A be a 2 x 2 block upper triangular matrix as above Then A is invertible if and only if both A11 and A22 are invertible Ail AfilAIZAgzl Further if A is invertible then A 1 0 A231 Problem 6 a In each case7 use block methods to nd the inverse of A 37001 12010 37001 12010 24100 01000 10000 12 Suggest special cases of the inverse formula for 2 X 2 block upper triangular matrices based on the results of the last two inverses in part a Be as complete as possible 23 Determinants 231 Determinants and Areas If A Z is a 2 X 2 matrix7 then the determinant of A is the number detA 2 adi be Let A a1 a2 Analytic geometry can be used to show that the area of the parallelogram with corners 07 a1 a2 and a1 a2 is the absolute value of the determinant Area of Parallelogram ldetAl lad 7 bcl For example if 1 6 A a1 a2 5 2 then 1 detA and 2 the area of the parallelogram shown in the plot is 232 Recursive De nition of Determinants Volumes If A is an n X 71 matrix then the determinant of A can be de ned recursively as an alternating sum of multiples of determinants of smaller matrices Using expansion in the rst row of A detA w Zeuwal detA1 j1 Where A1 is the submatrix obtained by eliminating the 1St row and j column of Al The form for a 3 X 3 matrix is an 112 113 122 123 132 133 121 a ail 03922 03923 all 03931 03932 03933 ail 03922 03931 03932 i 112 113 23 131 133 The computation is completed once you ll in the determinants of each 2 X 2 submatrix and simplify the results For example if A 2 2 12 l 10 4 1 then the determinant of A is 8 Determinants and volumes Let A a1 a2 a3 be a 3 X 3 matrix Analytic geometry can be used to show that the volume of the solid With corners 0 a1 a2 a3 a1 a2 a1 a3 a2 a3 and a1 a2 a3 is the absolute value of the determinant Volume of Parallelopiped ldetAli For example if N H N Aa1 a2 a3 1 NC then the volume of the parallelopiped shown 0 in the plot is 2 Theorem Equivalent Recursive De nitions The determinant of the square matrix A can be obtained recursively by expanding in any row or in any column That is7 7L detA 1A1 271ija1j d6tltAijgt for any xed row i and j1 n detA 1A1 271 7aij detAj for any xed column j i 1 where AH is the submatrix obtained by removing the ith row and jth column of A 2 12 1 For example7 let A 10 4 1 7 as above If we expand in the 2nd row7 we get 2 2 8 12 1 2 1 2 12 detA710 2 8 4 2 8 71 2 2 Problem 1 Find each determinant using expansion 150 12 471 0720 1004 0030 50120 0011 aOOO 0b00 0050 000d aOOO 12500 5def0 yhij 233 Properties of Determinants Factorizations Let A B and E be square matrices of order n Then 1 det AT dam 2 detAB detAdetB 3 If A is a triangular matrix then detA 01110122quot an 4 If E is an elementary matrix for a row replacement then detE 1 5 If E is an elementary matrix for a row interchange then detE 71 6 If E is an elementary matrix for scaling a row by c then detE 0 Problem 2 Let A be an invertible square matrix Use the properties of determinants to explain why the determinant of A must be nonzero LU factorization and determinants Suppose that the square matrix A has an LU factorization where L is a unit lower triangular matrix and U is an echelon form matrix Then detA Permutation matrices and determinants Recall that a product of elementary matrices for row interchanges is called a permutation matrix If P is a permutation matrix then detP PALU form and determinants Suppose that the square matrix A does not have an LU factorization but that you can write PA LU for some permutation matrix P unit lower triangular matrix L and echelon form matrix U Then detA 234 Elementary Row Operations and Determinants The following theorem relates elementary row operations to determinants Theorem Row Operations and Determinants Let A be a square matrix Then 1 If B is obtained from A by a single row replacement7 then detB detA 2 If B is obtained from A by a single row interchange7 then detB idetA 3 If B is obtained from A by multiplying a row by c then detB c detA This theorem and the properties of determinants can be used to develop an alternative method for nding a determinant based on row operations For example7 l 74 2 l 74 2 l 74 2 71 7 0 0 3 2 0 3 2 72 8 79 72 8 79 0 0 75 In step 17 the row replacement R2 9 R2 R1 does not change the value of the determinant Likewise7 in step 27 the row replacement R3 9 R3 2R1 does not change the value of the determinant Since the third matrix is in echelon form7 its determinant is easy to compute Problem 5 Find each determinant using the method based on row operations 2784 1717 0 72879 366 2123 144 00722 1012 671121 02730 Methods for nding deteminants include expansion in rowscolumns using elementary row operations to reduce the matrix to echelon form or some combination of these approaches Properties of determinants can also be used in conjunction with these methods 0 A22 Problem 4 Let A All 0 J 7 A11 0 p O i i 7 0 q 0 A22 be a 2 x 2 block diagonal matrix Use the factorization above and the properties of determinants to demonstrate that the deter minant of A can be computed as follows detA detA11detA22 235 Invertibility and Determinants Theorem Invertible Matrices and Determinants Let A be a square matrix Then A is invertible if and only if the determinant of A is nonzero d2 1 1 1 1 1 1 1 1 1 detA1 c c2 0 071 6271Ci1d710 1 51 1d d2 0 d71d271 01d1 1 1 1 671d710 1 61 0 0 d C we know that A is invertible iff It is instructive to write out the proof of the theorem relating invertibility to determinants a Assume that A is invertible Since A is row equivalent to the identity matrix A can be written as a product of elementary matrices say A E1E2 Ek NOW complete this part 6 Assume that detA 7 0 and that A can be transformed to echelon form using k elementary operations EkEk1E1A U where U is an echelon form matrix NOW complete this part 236 Cramer7s Rule Cofactors and Inverses Suppose that A is an invertible matrix of order n and b E R Then we know that A21 b has the unique solution 1 A Cramer s rule tells us that each component of the unique solution is a ratio of determinants Speci cally d5tAib detA where Aib is the matrix obtained from A by replacing column i with b for each i For example ifn 2 A a1 a2 all an and b le7 then 2 0421 0422 b2 0422 0421 b2 1411 b1 an and 1421 all b1 Further A113 b gt 951 and x2 ugtto Problem 5 Use Cramer s rule to solve the matrix equation 7 Problem 6 Let A 1 1 1 b 1 6 c2 and b 0 assume that A is invertible7 and consider the 1 d d2 0 matrix equation A11 b Use Cramer s rule7 and the work we did on page 317 to nd a simpli ed formula for the second component of the solution7 m2 ijCofact0rs lf AM is the submatrix of A obtained by eliminating the ith row and jth column then the ij cofactor of A is the number CM 1ijd6tAij Given the i j cofactor de nition the recursive de nition of the determinant becomes detA 21 JCi if we expand in row i and detA ELI JCi if we expand in column j For this reason the recursive de nition is often referred to as the cofactor de nition of the determinant Cofactors and Inverses Let A be an invertible matrix Cramer s rule can be used to nd an explicit formula for each element of the inverse of A Speci cally since the jth column of the inverse can be obtained by solving A21 ej Cramer s rule tells us that the ij entry of the inverse of A is i detAej 1 A H 7 W for each z and 3 Interestingly detAiej C the ji cofactor and you get A4 2 all 0412 0413 an agg agg and consider nding the numerators of the entries 0431 0432 0433 For example let n 3 A in the rst column of A 1 the numerators of the 11 21 and 3 1 entries 1 a12 dis detA1el 0 a22 agg 0 a32 ass all 1 dis d6tA281 an 0 agg an 0 ass an an 1 detA3el agl agg 0 an a32 0 Explicit Formula for A l Based on Cramer s rule7 an explicit formula for the inverse of A is 011 012 39 39 39 Cm T A71 21 22 C27 detA E E Cnl Cng Cm That is7 A 1 is the reciprocal of detA times the transpose of the cofactor matrix 1 fl 0 3 0 2 l 71 Problem 7 Let A 7 0 0 72 1 2 72 0 9 Use determinants based on Cramer s rule to nd the 21 and 34 entries of A l

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