Eigenvalues and transformations

Back again on matrices! I'll be probably be damned for my posts :)

I read a book on linear algebra, and I was thinking back on my post... so, let me formulate my question.

We have a matrix A over a field K, NxN, with KerA = { 0_K }, so I can say that A has full rank. We have a matrix P over K, MxN, with kerP = { 0_K }, so again P has full rank. Now, we can split A with something similar to a congruence with the difference that P is not square:

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

Between the parenthesis, U^T A U, is equal to U^-1 A U. Should this preserve the egienvalues of A? And what about the other matrices? Can I preserve the eigenvalues by restricting P (or X)?

I have an book on this, but it is only on square matrices... I cannot find anywhere something about non-square ones...

Thanks for any hint! :)

--
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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