Matrix trace minimization and zeros

[GoodSpirit]

Hello,
I would like to minimize and find the zeros of the function F(S,P)=trace(S-SP’(A+ PSP’)^-1PS) in respect to S and P.
S is symmetric square matrix.
P is a rectangular matrix
Could you help me?
Thank you very much
All the best
GoodSpirit

 

[GoodSpirit]

Hello everybody,

Perhaps I should explain a little bit.

The aim is to minimize an error metric and preferentially drive it to zero.
This should be done as function of S and P, as function of their rank and dimensions.
By the way, the matrix A is symmetric too and positive definite.

S' and P' are the transpose of S and P and ^-1 is the inversion operation.

By the way is there any of writing an equation and post here ( for example LateX).


Many thanks

GoodSpirit

 

[daon2]

I haven't the slightest idea how to approach this from a practical standpoint, but have you tried finding the eigenvalues of that mess? Here are some facts which may help:

You can assume that the resulting is not nilpotent.
You may diagonalize your symmetric matrices orthogonally.
Trace is linear and invariant under commutation.
Trace is invariant under conjugation, i.e., tr(XYX^(-1)) = tr(Y). - This combined with orthogonal diagonalization may help some, but not sure.
tr(AB)^2 <= tr(A)^2*tr(B)^2

 

[daon2]

^^ I must correct something I wrote above. You cannot assume the expression is not nilpotent.

But let me just show you what I was getting at with the other things. Let \(\displaystyle S=QDQ^{-1}\) be an orthogonal diagonalization of S, where the entries of the diagonal matrix D are the eigenvalues of S.

\(\displaystyle \text{tr}(S-SP’(A+ PSP’)^{-1}PS) = \text{tr}(S(I-P’(A+ PSP’)^{-1}PS))\\




= \text{tr}(QDQ^{-1}(I-P’(A+ PSP’)^{-1}PS))\\




= \text{tr}(QD(I-P’(A+ PSP’)^{-1}PS)Q^{-1})\\




= \text{tr}(D(I-P’(A+ PSP’)^{-1}PS))\)


So one of the S's has been replaced by D. You might be able to simplify this a lot more by attempting the same for A and the other S's. Anyway, good luck

 

[GoodSpirit]

Hi daon2,

I really thank you for answering!

Your suggestions I really interesting. For instance, tr(AB)^2 <= tr(A)^2*tr(B)^2 may help since if tr(A)^2*tr(B)^2 to zero so tr(AB)^2, and it much easier to analyse .
The second suggestions is also interesting but, isn't the matrices commutation circular that is : Tr(ABC)=tr(BCA)=tr(CAB) ?
Well at least I didn't understand the step from the second to third equation ...

Ive been trying using diagonalization and the representing the equation using a sum of matrices entries.
I didn´t get any satisfatory expression ... yet ...

All the best

GoodSpirit