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# jordan - MATH 513 JORDAN FORM Let A 1 A k be square...

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Unformatted text preview: MATH. 513. JORDAN FORM Let A 1 , . . . , A k be square matrices of size n 1 , . . . , n k , respectively with entries in a field F . We define the matrix A 1 ⊕ . . . ⊕ A k of size n = n 1 + . . . + n k as the block matrix A 1 . . . A 2 . . . . . . . . . . . . . . . . . . . . . . . . A k It is called the direct sum of the matrices A 1 , . . . , A k . A matrix of the form λ 1 . . . λ 1 . . . . . . . . . . . . . . . . . . . . . . . . λ 1 . . . . . . λ is called a Jordan block . If k is its size, it is denoted by J k ( λ ). A direct sum J = J k 1 ⊕ . . . ⊕ J k r ( λ r ) of Jordan blocks is called a Jordan matrix . Theorem. Let T : V → V be a linear operator in a finite-dimensional vector space over a field F . Assume that the characteristic polynomial of T is a product of linear polynimials. Then there exists a basis E in V such that [ T ] E is a Jordan matrix. Corollary. Let A ∈ M n ( F ) . Assume that its characteristic polynomial is a product of linear polynomials. Then there exists a Jordan matrix J and an invertible matrix C such that A = CJC- 1 . Notice that the Jordan matrix J (which is called a Jordan form of A ) is not defined uniquely. For example, we can permute its Jordan blocks. Otherwise the matrix J is defined uniquely (see Problem 7). On the other hand, there are many choices for C . We have seen this already in the diagonalization process. What is good about it? We have, as in the case when A is diagonalizable, A N = CJ N C- 1 . So, if we can compute J N , we can compute A N . It follows from the matrix multiplication that ( A 1 ⊕ . . . ⊕ A k ) N = A N 1 ⊕ . . . ⊕ A N k . Thus it is enough to learn how to raise a Jordan block in N th power. First consider the case when λ = 0. We have J k (0) 2 = 1 . . . 1 . . . . . . . . . . . . . . . . . . . . . . . . 1 . . . . . . 2 = 1 . . . 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 . . . . . . . . . . . . 1 We see that the ones go to the right until they disappear. Continuing in this way, we see that J k (0) k = 0 . Now we have J k ( λ ) N = ( λI n + J k (0)) N = λ N I n + N 1 λ N- 1 J k (0) + . . . + N i λ N- i J k (0) i + . . . + λ N N- 1 J k (0) N- 1 + J k (0) N . (1) This is proved in the same way as one proves the Newton formula: ( a + b ) N = N X i =0 N i a n- i b i . We look at the product of N factors ( a + b ) . . . ( a + b ). To get a monomial a n- i b i we choose i brackets from which we will take b . The number of choices is ( N i ) . Notice that in formula (1), the powers J k (0) i are equal to zero as soon as i ≥ k ....
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jordan - MATH 513 JORDAN FORM Let A 1 A k be square...

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