Re: Eigenvalues and transformations
- From: Sensei <senseiwa@xxxxxxx>
- Date: Sat, 22 Apr 2006 17:04:29 +0200
On 2006-04-20 19:47:42 +0200, Sensei <senseiwa@xxxxxxx> said:
A = P^T X P
The matrix X as you pointed out can be obtained by the left inverse of P: QP=I, and X = Q^T A Q. X is a matrix over K, MxM. Let's have M > N.
Is there any connection between the eigenvalues/eigenvectors of X and A?
I read about the singular value decomposition, and since P is real (I'm not interested in somehting too advanced, it's too early!) and A is real too, it can be decomposed into three matrices, the same can be done on Q (QP=I) and I can say this, if I'm not mistaken:
Q = U S V^T
V is such that V^T V = V V^T = I, the same is for U (if I understand, since they are real matrices). So, I can write:
X = Q^T A Q = (U S V^T)^T A (U S V^T) = V S (U^T A U) S V^T
Another thing... If A is positive definite, can we say that X is positive definite? Probably not... In the last line, the matrix S ``preserves'' positive definiteness since singular values are non-negative, but U is just unitary (same for V)... but I'm new to algebra...
Well, I just don't know! Any hint, or link, or better, a book, is appreciated! :)
--
Sensei <senseiwa@xxxxxxx>
The optimist thinks this is the best of all possible worlds.
The pessimist fears it is true. [J. Robert Oppenheimer]
.
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