Review : Powers of a matrix

Transcription

1 Review : Powers of a matrix Given a square matrix A and a positive integer k, we define A k = AA A } {{ } k times Note that the multiplications AA, AAA,... make sense. Example. Suppose A=. Then A 0 2 = = A 3 = (A 2 )A= = A 4 = (A 3 )A= = /22

2 Transpose of A Suppose A is an m n matrix. The transpose of A, which is denoted by A T, is an n m matrix whose rows are columns of A Example. Suppose A = 0 3. Then A T = Properties of transpose (A T ) T = A Suppose A and B are both m n matrices and c is a scalar. Then (A+B) T = A T + B T and (ca) T = ca T. Suppose A is m n and B is n p. Then (AB) T = B T A T 2/22

3 Remark. Note that (AB) T = A T B T is generally not true! For example, if A is 2 3 and B is 3 5, then (AB) T is defined and is a 5 2 matrix, whereas A T B T is not even defined! = } {{ } } {{ }} {{ } A B = }{{} A T } {{ } B T AB 3/22

4 Inverse of a matrix Suppose A is an n n matrix. We say that A is invertible if there exists an n n matrix C such that AC=CA=I n. The matrix C, if it exists, is called the inverse of A, and is usually denoted by A. If A is invertible, then it has a unique inverse Example. Suppose A =. Then if we take C= we will have AC= = and CA= = 2 0 This means that A is invertible and C is the inverse of A. 4/22

5 Question. When does an n nmatrix A have an inverse? Theorem (2 2 matrices) a b Suppose A=. Then : c d If ad bc=0 then A is not invertible. If ad bc 0 then A is invertible and we have A d b = ad bc c a The quantity ad bc is called the determinant of A. 3 Example. The matrix A= is not invertible because =6 6=0. 3 Example. The matrix A= is invertible because 2 3 2= Moreover, A = = /22

6 Example 4 3 Is A= invertible? If so, then compute A 3. Solution. Determinant of A is (4 ) (3 ( 3 ))=4 ( )=5 0. Therefore A is invertible and A = [ ] 5 = Then, and AA = [ [ A A= 5 ][ 5 3 ] 5 = ] 4 3 = /22

7 Basic properties of inverse Theorem Suppose A is an n n matrix. If A is invertible, then A is invertible, and (A ) = A Reason : A A=AA = I n, i.e., A is the inverse of A. 2 If A and B are n ninvertible matrices, then AB is also invertible and we have (AB) = B A Reason : (AB)(B A )=A(BB )A = AI n A = AA = I n. 3 If A is an n ninvertible matrix, then so is A T and the inverse of A T is the transpose of A. That is, (A T ) = (A ) T Reason : A T (A ) T = (A A) T = I T n= I n. 7/22

8 Calculating A in general Suppose you are given an n nmatrix A. Question. Is A invertible? If it is, how can we find its inverse? Notation Recall that I n is the n nidentity matrix, i.e., I n = e e 2 e 3 e n We denote the i-th column of I n by e i. Therefore I n = [ e e 2 e n ] The i-th entry of e i is, and the rest of its entries are 0. 8/22

9 Suppose A is invertible, i.e., we can find an n n matrix A such that AA = I n. Write A as A = [ x x 2 x n ] We have AA = Ax Ax 2... Ax n. But: AA = I n Ax Ax 2... Ax n = e e 2 e n It follows that Ax = e, Ax 2 = e 2,, Ax n = e n To determine if A is invertible, and to calculate A, we should solve the following n matrix equations : Ax = e, Ax 2 = e 2,, Ax n = e n 9/22

10 To determine if A is invertible, and to calculate A, we should solve n matrix equations : Ax = e, Ax 2 = e 2,, Ax n = e n This means we have to solve the linear systems with the following augmented matrices : [ A e ], [ A e2 ],, [ A en ] we can solve these linear systems simultaneously by finding the R.E.F. of [ A e e 2 e n ] = [ A In ] 0/22

11 Algorithm for finding A Suppose A is an n n matrix. Row reduce the super augmented matrix [ A In ] to obtain its R.E.F. If after row reduction we obtain a matrix of the form [ In B ] then A is invertible and B=A. Otherwise, A is not invertible. /22

12 Example 0 2 Is A= 0 3 invertible? If so, then find its inverse R R 2 R R 2 R 3 R 3 R R R +2R R 2 R 2 2R R.E.F Checking the answer: AA = = Similarly, we can check that A A=I 3. 2/22

13 Example Is A= 2 0 invertible? If it is, then find its inverse. 3 0 R R 2 2R 0 0 R R 3 3R R 3 R 3 R 2 R R R 2 R 2 +2R 3 0 R R R 3 R R 2 R R R It follows that A is not invertible. R.E.F /22

14 Example 0 2 Is A= 0 3 invertible? If it is, then find its inverse R Solution R R 3 R 3 4R R R 3 +3R R R 3 2 R R 2 2R R R 3R It follows that A is invertible and A = /22

15 Criteria for invertibility Theorem Let A be an n n matrix. Then the following statements are equivalent (i.e., they are either all true or all false). A is an invertible matrix. 2 The R.E.F. of A is I n. 3 A has n pivot positions. 4 The equation Ax=0 only has the trivial solution x=0. 5 The columns of A are linearly independent. 6 For every vector b inr n, the matrix equation Ax=bhas a unique solution. 7 The columns of A spanr n. 8 A T is invertible. 5/22

16 Example Suppose that A is an n n matrix which satisfies Is A invertible? Solution. One can write or A 5 2A 3 + A 2I=0. A ( A 4 2A 2 + I ) = 2I A ( 2 A4 A I) = ( 2 A4 A I) A=I It follows that A is invertible. In fact the above relation shows that : A = 2 A4 A I 6/22

17 Matrix Equations Example Solve the equation AXB=Cfor X where A= 0 3 B= and C= Solution. First of all, note that since A is 3 3 and B is 2 2, X should be 3 2. If A and B were invertible then we could write : A AXBB = A CB = I 3 XI 2 = A CB = X=A CB We now check to see if A and B indeed invertible. For A, we have Row Reduction Which means that A is invertible and A = /22

18 Example Solve the equation AXB=Cfor X where A= 0 3 B= and C= For B=, we can use the method of this lecture or the method 0 of the last lecture : determinant of B is 0, hence B = 2 2 = 0 0 Finally, we have X=A CB = = 5 9 = /22

19 Example 3 Let A= and B= 4 5 for X. 00. Solve the equation B(X T + A)B = 2I I= 0 0 Solution. Since A and B are 2 2, the matrix X is also 2 2. (We also have I=I 2.) B(X T + A)B = 2I B B(X T + A)B = 2B I=2B (X T + A)B = 2B (X T + A)B B=2B B=2I X T = 2I A X=(2I A) T = 2I A T = X T + A=2I = /22

20 Example Simplify A (BC) T (B ) T (C ) T A 2 Solution. We have A (BC) T (B ) T (C ) T A 2 = A C T B T (B ) T (C ) T A 2 = A C T( B T (B ) T) (C ) T A 2 = A C T( B T (B T ) ) (C ) T A 2 = A C T I (C ) T A 2 = A C T (C ) T A 2 = A C T (C T ) A 2 = A I A 2 = A A 2 = A 20/22

21 Example Suppose A= and B is a 2 3 matrix such that AB= What is the second column of B? Solution. If we write B as B=[ b b 2 b 3 ] then we have And therefore AB=[Ab Ab 2 Ab 3 ] Ab 2 = But determinant of A is 2 0, hence A is invertible and by the formula from the last lecture we have A = [ ] [ = 2 2 ] = b 2 2 = A = = /22

22 Today s key point : Algorithm for finding A Suppose A is an n n matrix. Row reduce the super augmented matrix [ A In ] to obtain its R.E.F. If after row reduction we obtain a matrix of the form [ In B ] then A is invertible and B=A. Otherwise, A is not invertible. 22/22