Is this a correct realization? The eigenspaces corresponding to the eigenvalues of A are the same as the eigenspaces corresponding to the eigenvalues of A^-1, transpose of A, and A^k for any k > 1.
It took me some time to realize this but the v's, when you manipulate these equations, don't change. So I'm lead to believe that the eigenvectors are actually the same for all such variants of A.
matt grime
Jul15-08, 07:20 PM
If A is invertible, then clearly v is an eigenvector for A if and only if it is an eigenvector for A^-1.
Unless you define 'variants' of A then we can't answer your second question. I'll attempt a guess: no, the eigenvectors of A^2 are not the same as eigenvectors of A.
JinM
Jul15-08, 07:49 PM
Sorry for the ambiguity -- you knew what I meant anyway.
Why aren't the eigenvectors of A^k for k > 1 the same as the eigenvectors of A?
matt grime
Jul15-08, 08:14 PM
Why should they be?
Manchot
Jul15-08, 09:26 PM
Sorry for the ambiguity -- you knew what I meant anyway.
Why aren't the eigenvectors of A^k for k > 1 the same as the eigenvectors of A?
Assuming that you are referring to an integer k, it's true that eigenvectors of A are automatically eigenvectors of A^k. However, you don't have any guarantee that eigenvectors of A^k are eigenvectors of A.
JinM
Jul16-08, 12:58 AM
Wait a second.
If x is an eigenvalue of A, then x^k is an eigenvalue of A^k (k is an integer).
Are you guys saying that there could possibly be other eigenvalues of A^k that differ from all x^k's (the eigenvalues of A raised to the k)? That's why we can't guarantee matching eigenvectors -- is that it?
JinM
Jul16-08, 01:09 AM
But is that even possible? A^k is always a square matrix, whose order matches that of matrix A.
Hmm..
If we take a 3x3 matrix A with eigenvalues -1, 1, and 4. A^2 will have eigenvalues of 1 (with algebraic multiplicity 2), and 4. A will have three eigenspaces, while A^2 will have two eigenspaces. So, the eigenspace of A corresponding to the eigenvalue -1 will not "live" in the eigenspace of A^1 corresponding to the eigenvalue (-1)^2 = 1. But that contradicts what Manchot is saying.. hrmph
n_bourbaki
Jul16-08, 08:12 AM
Wait a second.
If x is an eigenvalue of A, then x^k is an eigenvalue of A^k (k is an integer).
Are you guys saying that there could possibly be other eigenvalues of A^k that differ from all x^k's (the eigenvalues of A raised to the k)?
No, we're definitely not saying that.
That's why we can't guarantee matching eigenvectors -- is that it?
No.
n_bourbaki
Jul16-08, 08:15 AM
But is that even possible? A^k is always a square matrix, whose order matches that of matrix A.
What do you mean by order?
Hmm..
If we take a 3x3 matrix A with eigenvalues -1, 1, and 4. A^2 will have eigenvalues of 1 (with algebraic multiplicity 2), and 4. A will have three eigenspaces, while A^2 will have two eigenspaces. So, the eigenspace of A corresponding to the eigenvalue -1 will not "live" in the eigenspace of A^1 corresponding to the eigenvalue (-1)^2 = 1. But that contradicts what Manchot is saying.. hrmph
You mean 16, not 4, for the e-value of A^2.
Manchot stated that you cannot assume an e-vector of A^k is an e-vector of A for all matrices. You're saying he's wrong just because you can do it for one (diagonalizable) matrix.
Certainly if A is diagonalizable, then e-vectors of A are e-vectors of A^k and vice versa. However, being diagonalizable is a very special property.