2.4 MATRIX PRODUCTS. The composite of two functions: and z = cos(y) is z = cos(sin(x)). y = A x, with A = z = B y, with B =

Transcription

1 24 MATRIX PRODUCTS The composite of two functions: and z = cos(y) is z = cos(sin(x)) y = sin(x) Consider two transformation systems: y = A x, with A = z = B y, with B = The composite of the two transformation systems is z = B(A x) Question: Is z = T ( x) linear? If so, what s the matrix? 1

2 (a) Find the matrix for the composite: z 1 = 6y 1 + 7y 2 z 2 = 8y 1 + 9y 2 and y 1 = x 1 + 2x 2 y 2 = 3x 1 + 5x 2 z 1 = 6(x 1 + 2x 2 ) + 7(3x 1 + 5x 2 ) = ( )x 1 + ( )x 2 = 27x x 2 z 2 = 8(x 1 + 2x 2 ) + 9(3x 1 + 5x 2 ) = ( )x 1 + ( )x 2 = 35x x 2 This shows the composite is linear with matrix =

3 (b)use Fact to show the transformation T ( x) = B(A x) is linear: T ( v + w) = B(A( v + w)) = B(A v + A w) = B(A v) + B(A w) = T ( v) + T ( w) T (k v) = B(A(k v)) = B(k(A v)) = k(b(a v)) = kt ( v) Once we know that T is linear, we can find its matrix by computing the vectors: T ( e 1 ) and T ( e 2 ): T ( e 1 ) = B(A( e 1 ))=B(first column of A) = = T ( e 2 ) = B(A( e 1 ))=B(second column of A) = = The matrix of T ( x) = B(A x) = BA( x): = T ( e 1 ) T ( e 2 ) =

4 Definition Matrix multiplication 1 Let B be an m n matrix and A a q p matrix The product BA is defined if (and only if) n = q 2 If B is an m n matrix and A an n p matrix, then the product BA is defined as the matrix of the linear transformation T ( x) = B(A x) This means that T ( x) = B(A x) = (BA) x, for all x in R p The product BA is an m p matrix 4

5 Let B be an m n matrix and A an n p matrix Let s think about the columns of the matrix BA: (ith columns of BA) = (BA) e i = B(A e i ) = B(ith column of A) If we denote the columns of A by v 1, v 2,, v p,we can write BA = B v 1 v 2 v p }{{} A B v 1 B v 2 B v p = 5

6 The matrix product, column by column Let B be an m n matrix and A an n p matrix with columns v 1, v 2,, v p Then, the product BA is BA = B v 1 v 2 v p = B v 1 B v 2 B v p To find BA, we can multiply B with the columns of A and combine the resulting vectors Fact Matrix multiplication is noncommutative: AB BA, in general However, at times it does happen that AB = BA ; then, we say that the matrices A and B commute 6

7 The matrix product, entry by entry Let B be an m n matrix and A an n p matrix The ijth entry of BA is the dot product of the ith row of B and the jth column of A b 11 b 12 b 1n b 21 b 22 b 2n b i1 b i2 b in b m1 b m2 b mn a 11 a 12 a 1j a 1p a 21 a 22 a 2j a 2p a n1 a n2 a nj a np is the m p matrix whose ijth entry is b i1 a 1j + b i2 a 2j + + b in a nj = n k=1 b ik a kj Example = = We have done these computations before(where?) 7

8 Matrix Algebra Fact (a) For an invertible n n matrix A AA 1 = I n and A 1 A = I n Fact (b) For an m n matrix A AI n = I m A = A Fact (c) Matrix multiplication is associative (AB)C = A(BC) We can write simply ABC for the product (AB)C = A(BC) 8

9 Proof (a) (AB)C = (AB) v 1 v 2 v q = (AB) v 1 (AB) v 2 (AB) v q and A(BC) = A B v 1 B v 2 B v q = A(B v 1 ) A(B v 2 ) A(B v q ) Since (AB) v i = A(B v i ), by definition of the matrix product, we find that (AB)C = A(BC) Proof (b) Consider two linear transformations and are identical because, T ( x) = ((AB)C) x L( x) = (A(BC)) x T ( x) = ((AB)C) x = (AB)(C x) = A(B(C x)) and L( x) = (A(BC)) x = A((BC) x) = A(B(C x)) 9

10 If A and B are invertible n n matrices, is BA invertible? y = BA x multiply both sides by B 1 B 1 y = B 1 BA x = I n A x = A x next, multiply both sides by A 1 A 1 B 1 y = A 1 A x = I n x = x This computation shows that the linear transformation is invertible since x = A 1 B 1 y Fact The inverse of a product of matrices If A and B are invertible n n matrices, then BA is invertible as well, and (BA) 1 = A 1 B 1 Pay attention to the order of the matrices Proof Verify it by yourself 10

11 Fact Let A and B be two n n matrices such that Then, BA = I n a A and B are both invertible b A 1 = B and B 1 = A, and c AB = I n Proof (a) To demonstrate A is invertible it suffices to show that the linear system A x = 0 has only the solution x = 0 (b) B = A 1 since and BA x = B 0 = 0 (BA)A 1 = (I n )A 1 = A 1 B 1 = (A 1 ) 1 = A (c) AB = AA 1 = I n 11

12 Example B= 1 a b ad bc d c b a is the inverse of A = c d it suffices to verify that BA = I 2 : BA= 1 ad bc = 1 ad bc Example d c b a a b c d ad bc bd bd ac ac ad bc = I 2 Suppose A, B and C are three n n matrices and ABC = I n Show that B is invertible, and express B 1 in term of A and C Solution Write ABC = (AB)C = I n We have C(AB) = I n Since matrix multiplication is associative, we can write (CA)B = I n We conclude that B is invertible, and B 1 = CA 12

13 Distributive property for matrices Fact If A, B are n n, and C, D are n p matrices, then A(C + D) = AC + AD and (A + B)C = AC + BC Fact If A is an m n matrix, B an n p matrix, and k a scalar, then (ka)b = A(kB) = k(ab) 13

14 Partitioned Matrices It is sometimes useful to break a large matrix down into smaller submatrices by slicing it up with horizontal or vertical lines that go all the way through the matrix For example, we can think of the 4 4 matrix A = as a 2 2 matrix whose entries are four 2 2 matrices: A = = A11 A 12 A 21 A 22, 14

15 with A 11 = , A 12 = , etc The submatrices in such a partition need not be of equal size; for example, we could have B = = = B11 B 12 B 21 B 22 A useful property of partitioned matrices is the following:

16 Multiplying partitioned matrices Partitioned matrices can be multiplied as though the submatrices were scalars: AB = A 11 A 12 A 1n A 21 A 22 A 2n A i1 A i2 A in B 11 B 12 B 1j B 1p B 21 B 22 B 2j B 2p B n1 B n2 B nj B np A m1 A m2 A mn is the partitioned matrix whose ijth entry is the matrix A i1 B 1j + A i2 B 2j + + A in B nj = n k=1 A ik B kj, provided that all the products A ik B kj are defined 15

17 Example = = Compute this product without using a partition, and see whether you find the same result 16

18 Example A = A11 A 12 0 A 22, where A 11 is an n n matrix, A 22 is an m m matrix, and A 12 is an n m matrix a For which choices of A 11, A 12, and A 22 is A invertible? b If A is invertible, what is A 1 (in terms of A 11, A 12, A 22 )? 17

19 Solution We are looking for an (n + m) (n + m) matrix B such that BA = I n+m = In 0 0 I m Let us partition B in the same way as A: B = B11 B 12 B 21 B 22, where B 11 is n n, B 22 is m m, etc The fact that B is the inverse of A means that B11 B 12 A11 A 12 In 0 B 21 B 22 0 B 22 = 0 I m, or using 18

20 B 11 A 11 = I n B 11 A 12 +B 12 A 22 = 0 B 21 A 11 = 0 B 21 A 12 +B 22 A 22 = I m We have to solve this system for the submatrices B ij 1 By Equation 1, A 11 must be invertible, and B 11 = A By Equation 3, B 21 = 0 (Multiply by A 1 11 form the right) 3 Equation 4 now simplifies to B 22 A 22 = I m Therefore, A 22 must be invertible, and B 22 = A Lastly, Solve for B 12 by Equation 2 A 1 11 A 12 + B 12 A 22 = 0

21 B 12 A 22 = A 1 11 A 12 B 12 = A 1 11 A 12A 1 22 So a A is invertible if (and only if) both A 11 and A 22 are invertible (no condition is imposed on A 12 ) b If A is invertible, then its inverse is A 1 = A 1 11 A 1 11 A 12A A 1 22

22 Verify this result for the following example: Example = Homework Exercise 24: 5, 13, 17, 23, 27, 35 19