Matrix Arithmetic. j=1

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1 An m n matrix is an array A = Matrix Arithmetic a 11 a 12 a 1n a 21 a 22 a 2n a m1 a m2 a mn of real numbers a ij An m n matrix has m rows and n columns a ij is the entry in the i-th row and j-th column We often write A = a ij If A = a ij and B = b ij are two matrices of the same size they are both m n then they can be added The sum A + B is the m n matrix c ij where c ij = a ij + b ij for all rows i and columns j Suppose that c R The scalar product ca is the m n matrix d ij defined by d ij = ca ij 0 m,n is the m n zero matrix all entries are zero Suppose that A, B and C are m n matrices, and c, d R Then we have the following identities: A + B = B + A A + B + C = A + B + C cda = cda 0 m,n + A = A = A + 0 m,n The matrix 1A is an additive identity for A; that is, A + 1A = 1A + A = 0 m,n, so we write A = 1A Two matrices A = a ij and B = b jk can be multiplied to form a new matrix AB if A is m n and B is n l for some m, n, l The restriction is that the number of columns of A must be equal to the number of rows of B The product AB is defined to be the m l matrix e ik where n e ik = a ij b jk For example, = = j= I n is the n n identity matrix, I n = δ ij where { 1 if i = j δ ij = 0 if i j For instance, I 3 = 0 1

2 Suppose that A, B, C are matrices which can be multiplied where A is m n and c R Then we have the following identities: ABC = ABC AB + C = AB + AC A + BC = AC + BC cab = cab = AcB 0 l,m A = 0 l,n A0 n,l = 0 m,l I m A = A AI n = A Warning: Some of the familiar properties of multiplication in R are no longer true for matrices If A, B and C are n n with n 2, in general we have that AB BA It is possible for A and B to both be nonzero, but AB = 0 n,n Further, there are nonzero matrices A, B, C such that AB = AC but B C The transpose of an m n matrix A = a ij, written as A T, is the n n matrix b ij where b ij = a ji ; that is, A T is the matrix whose rows are the columns of A For example, T = The transpose satisfies the following formulas: A T T = A ca T = ca T A + B T = A T + B T AB T = B T A T Column vectors and row vectors R m is the set of all m 1 column vectors with real coefficients If v R m, then v 1 v 2 v = v m with v 1,, v m R R n is the set of all 1 n row vectors with real coefficients If w R n, then with w 1,, w n R We have a dot product on R m defined by w = w 1,, w n x y = x 1,, x m y 1,, y m = x 1 y 1 + x 2 y x m y m R for x = x 1,, x m, y = y 1,, y m R m Similarly, we have a dot product on R n defined by x y = x 1 y x n y n R 2

3 for x = x 1 x n, y = y 1 y n R n Some useful formulas Suppose that A is an m n matrix with coefficients in R, and x = x 1,, x n T R n Let v w = v T w be the dot product of the vectors v, w R n Writing A = A 1, A 2,, A n where A i R m are the columns of A, we obtain the formula 1 Ax = x 1 A x n A n Writing 2 Ax = A = A 1 A 2 A m where A j R n are the rows of A, we obtain the formula A 1 x A T 1 A 2 x x A T 2 x A m x = A T m x If A is an m n matrix and B = B 1,, B l is n l with each B j R n an n 1 column vector then the m l matrix 3 AB = AB 1, AB 2,, AB l where each AB i is an m 1 column matrix As a consequence, we obtain the following formula, which we will use later in this note 4 AB C = AB AC where A is m n, B is n s and C is n t, giving us an n s + t matrix B C, and after multiplication, an m s + t matrix AB AC Inverses of square n n matrices Definition 01 Suppose that A is an n n matrix A is invertible if there exists an n n matrix B such that AB = I n and BA = I n Theorem 02 Suppose that an n n matrix A is invertible matrix C such that AC = CA = I n Then there is a unique Proof Suppose that B, C are matrices such that AB = BA = I n and AC = CA = I n Then B = I n B = CAB = CAB = CI n = C Thus B = C is unique 3

4 Suppose that A is invertible We have seen that the matrix B satisfying AB = BA = I n is unique, so we may call this matrix the inverse of A, and write B = A 1 Example 03 Show, using the definition of invertibility, that 5 2 A = 7 3 is invertible with inverse A 1 = Solution: Let B = we cannot call this matrix A 1 until after we have shown that it really is the inverse to A We compute AB = = = I and BA = Thus A is invertible with inverse A 1 = B = 1 0 Example 04 Show, using the definition of invertibility, that A = 0 0 is not invertible = I 2 Solution: Suppose that A is invertible Let A 1 be the inverse of A We compute A 2 = 0 2,2 Then = A = I A = A 1 AA = A 1 A 2 = A 0 2,2 = 0 2,2 = 0 0 This is a contradiction to our assumption that A is invertible, so A is not invertible Theorem 05 Suppose that A, B are n n matrices which are invertible Then AB is invertible, with AB 1 = B 1 A 1 Proof We have that ABB 1 A 1 = AI n A 1 = AA 1 = I n and B 1 A 1 AB = B 1 I n B = B 1 B = I n, so that AB is invertible with inverse AB 1 = B 1 A 1 Corollary 06 Suppose that r is a positive integer and A 1, A 2,, A r are invertible n n matrices Then A 1 A 2 A r is invertible with A 1 A 2 A r 1 = A 1 r A 1 r 1 A 1 2 A 1 1 4

5 Definition 07 An n n elementary matrix is a matrix which is obtained by performing one elementary row operation on the n n identity matrix I n The 3 3 elementary matrix E 1 obtained by interchanging the first and second row of the identity matrix is E 1 = 0 The 3 3 elementary matrix E 2 obtained by multiplying the second row of the identity matrix by 2 is E 2 = The 3 3 elementary matrix E 3 obtained by adding 4 times the third row of the identity matrix to the first row is E 3 = Theorem 08 Suppose that E is an elementary matrix Then E is invertible and E 1 is an elementary matrix E 1 is obtained by performing the elementary row operation on I n which transforms E back to I n Considering the inverse of the above three examples of elementary matrices, we have that E1 1 is obtained by interchanging the first and second rows of the identity matrix so that E1 1 = 0 2 is obtained by multiplying the second row of the identity matrix by 1 2 so that E2 1 = E 1 E 1 3 is obtained by adding -4 times the third row of the identity matrix to the first row so that E3 1 = Theorem 09 Let A be an m n matrix, and suppose that B is obtained from A by performing a single elementary row operation on A Let E be the m m elementary matrix obtained by performing this row operation on I m Then EA = B An example illustrating the theorem is the case of the matrix A =

6 If we multiply A on the left by E 1, the elementary matrix obtained from interchanging the first and second rows of the 3 3 identity matrix, we obtain E 1 A = = which is the matrix obtained from A by interchanging the first and second rows We observe that: The RRE form of an n n matrix A is either the identity matrix I n, or the RRE form has a row of zeros for its last row Theorem 010 Suppose that A is an n n matrix Then A is invertible if and only if the RRE form of A is I n Proof First suppose that A is row equivalent to I n Since A is row equivalent to I n, we have by the previous theorem a product I n = E r E r 1 E 2 E 1 A where E 1, E 2,, E r are elementary matrices Since E 1,, E r are invertible, we have that A = E1 1 2 Er 1 is invertible, with inverse 5 A 1 = E r E r 1 E 2 E 1 Now suppose that A is invertible Let C be the RRE form of A We have a factorization C = E r E r 1 E 2 E 1 A where E 1, E 2,, E r are elementary matrices Multiplying both sides of this equation on the right by A 1, we have CA 1 = E r E 1 I n E r E 1 I n is computed by performing a sequence of elementary row operations on I n, so there can be no row consisting entirely of zeros in CA 1 However, if C is not I n, then the last row of C is the zero vector, and then the last row of CA 1 is also the zero vector, which is impossible Thus the RRE form of A is C = I n The proof of the above theorem is constructive and tells us how to a solve a problem like the following Example 011 Write as a product of elementary matrices A = Solution: First transform A into RRE form, keeping careful track of the elementary row operations and the precise order in which they are performed and making this clear from your work Multiply the second row by 1 10 Add -4 times the first row to the second row Add -3 times the second row to the first row 6 1 0

7 Let E 1 be the elementary matrix of the first operation, E 2 the elementary matrix of the second operation, E 3 be the elementary matrix of the third operation, so that E 1 =, E = 0 1, E 3 = 10 and E 3 E 2 E 1 A = I 2 We compute E = 4 1, E = 0 10, E = Thus we have an expression of A as a product of elementary matrices, A = E E 1 2 E 1 3 I 2 = Algorithm to compute the inverse of a matrix Suppose that A is an n n matrix Transform the n 2n matrix A I n into a reduced row echelon form C B by a sequence of elementary row operations A is invertible if and only if C = I n The RRE form of A is I n If A is invertible, then B = A 1 The algorithm is just a restatement of Theorem 010 Let E 1, E 2,, E r be the elementary matrices corresponding to the sequence of elementary row operations which transform A into its RRE form C Then using the formula 4, we have E r E r 1 E 2 E 1 A I n = E r E 1 A E r E 1 = C E r E 1 We have by Theorem 010 that A is invertible if and only if C = I n and if this holds, then A 1 = E r E 1 by 5 Example 012 Let A = Use the Algorithm to compute the inverse of a matrix to 1 Determine if A is invertible 2 If A is invertible, compute A 1 Solution: A I 3 = A is invertible since it is row equivalent to I

8 2 A 1 = Example 013 Let A = Use the Algorithm to compute the inverse of a matrix to 1 Determine if A is invertible 2 If A is invertible, compute A 1 Solution: A I 3 = A is not invertible since it is row equivalent to a matrix with a row of zeros and thus is not row equivalent to I 3 2 Not applicable the inverse does not exist Systems of Equations A system of m equations in n unknowns, can be written in matrix form as a 11 x 1 + a 12 x a 1n x n = b 1 a 21 x 1 + a 22 x a 2n x n = b 2 a m1 x 1 + a m2 x a mn x n = b m 6 A x = b where A is the m n matrix A = a ij, x = x 1 x n R n and b = b 1 b m R m If m = n and A is invertible, then 6 has a unique solution; it is x = A 1 b 8