Eigen2 - 1 Eigenvectors continued The eigenvector train continues on and stops for no one OK as most of you figured out the eigenvalues of a matrix

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: 1 Eigenvectors continued The eigenvector train continues on and stops for no one. OK, as most of you figured out, the eigenvalues of a matrix are the solutions of the equation det( A- λI ) = 0 and the eigenspace E λ i is the space N ( A- λ i I ). Let A be an n × n matrix. The characteristic polynomial is det( A- λI ). We usually call this polynomial the Here are some relevent/useful facts: · A is diagonalizable if and only if: 1) p ( λ ) splits into linear factors. 2) The sum of the dimensions of all of the eigenspaces equals n . · Suppose that the characteristic polynomial is equal to p ( λ ) = ± ( λ- α 1 ) n 1 ··· ( λ- α k ) n k . Don’t be afraid by this expression. This says the characteristic poly- nomail splits and its eigenvalues are α 1 ,...,α k . Then the dimension of each eigenspace E α i is bounded: 1 ≤ dim E α i ≤ n i . As ∑ n i = n , we see that condition 2) is equivalent to saying that E α i = n i for each i . · If there are n distinct eigenvalues, then A is has an eigenbasis. (This follows from the previous two statements) · The product of two diagonalizable matrices is NOT necessarily diagonalizable. example A = 1 1 2- 1 ,B = 1 2 2 1 are both diagonalizable (both have 2 distinct eigenvalues, so we can use the previous fact to conclude this). However their product AB = C C = 3 3 3 which is not diagonlizable. Its char. polynomial is ( λ- 3) 2 , so its only eigenvalue is 3. But C- 3 I = 3 and the dimension of its nullspace is only 1. 1 · Not all matrices have characteristic polynomials that split (this is a sign they don’t have very many eigenvalues). An extreme example of this is the rotation matrix 1- 1 ....
View Full Document

This note was uploaded on 01/11/2012 for the course MATH 100 taught by Professor Loveys during the Fall '11 term at McGill.

Page1 / 6

Eigen2 - 1 Eigenvectors continued The eigenvector train continues on and stops for no one OK as most of you figured out the eigenvalues of a matrix

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