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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 ....
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