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

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 1 . . . 1 . . . . . . . . . . . . . . . . . . . . . . . . 1 . . . . . . is called a Jordan block . If k is its size, it is denoted by J k ( ). A direct sum J = J k 1 . . . J k r ( 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 = 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 ( ) N = ( I n + J k (0)) N = N I n + N 1 N- 1 J k (0) + . . . + N i N- i J k (0) i + . . . + 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 ....
View Full Document

This note was uploaded on 02/24/2012 for the course MATH 513 taught by Professor Igordolgachev during the Winter '09 term at University of Michigan-Dearborn.

Page1 / 9

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online