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04-30-03 - Lecture 32 Andrei Antonenko 1 Powers of...

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Lecture 32 Andrei Antonenko April 30, 2003 1 Powers of diagonalizable matrices In this section we will give 2 algorithms of computing the m -th power of a matrix. 1.1 Method 1 First method is based on diagonalization. Suppose A is a given matrix, and we want to find its m -th power, i.e. we want to get a formula for A m . We will suppose that the matrix A is diagonalizable. Let λ 1 , λ 2 , . . . , λ n be the eigenvalues of A , and e 1 , e 2 , . . . , e n be its linearly independent eigenvectors. Then we know, that there exists a matrix C , whose columns are eigenvectors, and a diagonal matrix D = λ 1 0 . . . 0 0 λ 2 . . . 0 . . . . . . . . . . . . . . . 0 0 . . . λ n such that D = C - 1 AC, or A = CDC - 1 . Now we can see that A m = ( CDC - 1 ) m = ( CDC - 1 )( CDC - 1 ) · · · ( CDC - 1 ) | {z } m = CD ( C - 1 C ) D ( C - 1 . . . C ) DC - 1 = CDID . . . IDC - 1 = CD m C - 1 . But D m = λ 1 0 . . . 0 0 λ 2 . . . 0 . . . . . . . . . . . . . . . 0 0 . . . λ n m = λ m 1 0 . . . 0 0 λ m 2 . . . 0 . . . . . . . . . . . . . . . . . 0 0 . . . λ m n 1
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So, A m = C λ m 1 0 . . . 0 0 λ m 2 . . . 0 . . . . . . . . . . . . . . . . . 0 0 . . . λ m n C - 1 . Example 1.1. Let’s find the formula for ˆ 3 - 2 1 0 ! m . The characteristic polynomial is p A ( λ ) = λ 2 - 3 λ + 2 . So, the eigenvalues are λ 1 = 1 , λ 2 = 2 . Let’s compute eigenvectors. λ 1 = 1 . After subtraction λ 1 = 1 from the diagonal, we have ˆ 2 - 2 1 - 1 !
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