Eigenvalues of a scaled matrix

[arunakkin]

Assume P is a symmetric positive-definite matrix,
and S to be a diagonal matrix with all its diagonal elements being greater than 1.

Let Q = SPS

then is Q-P symmetric positive-definite ?
i.e.
are the eigen-values of Q greater than P element-wise? or eig(Q)>= eig(P) in non-negative orthant

I tried a simulation to check if it is true. I did not come up with a case which disproves it.
I would be grateful if an analytical proof is provided.
Intuitively it makes sense. We scale any vector multiplied by Q (= SPS) before multiplying it with P. And as the eigen-values represent scaling, the resultant eigen-values of Q must be greater than P as long as the scaling by S increases vector in all dimensions (i.e. diag elements of S are >= 1)


Thanks so much,
arunakkin

 

[chiro]

Hey arunakkin and welcome to the forums.

For this particular problem I suggest you create a matrix P* where you absorb the S entries since they are diagonal. Absorbing these entries is simple and you can do this by multiplying each column by the appropriate diagonal value in that column. You do the same thing for RHS S matrix (expand this matrix out just for clarity and to check this yourself).

This means you will get P* where each column entry is multiplied by the square of the appropriate diagonal. You can then do an eigen-analysis on this new matrix P* and prove definitely whether your conjecture is correct.

 

[arunakkin]

Hi,
Thanks for the reply and interest.

I think what you are trying to say is that, in terms of above notation,

q(i,j) = p(i,j)*s(i,i)*s(j,j)

This is what I did in the simulation I mentioned in the post.
I randomly generated a symmetric positive definite matrix, and scaled (s(i,i)>=1 for all i) it as mentioned above and compared the eigenvalues of Q with that of P (after sorting them).
I wasn't able to find a result where eig(Q)<eig(P) (element-wise).


But this alone doesn't prove anything (as I might not have come across the case where this is disproved in my simulation).
Can you elaborate on eigen-analysis. Are you referring to diagonalization.
If so, I tried this but was not able to find any meaningful relationship which'll definitely prove that (Q-P) is positive definite (or eig(Q)> eig(P) ,element-wise)

Regards,
arunakkin.

 

[chiro]

Well, yes the idea is to show that the roots of the characteristic polynomial are greater (since you are trying to show eig(Q) >= eig(P)).

What I suggest is to relate the characteristic equation for Q to that of P taking into account that the columns have been scaled for the square of the diagonal entries.

You could if you wanted a general argument, use the properties of the determinant where |A^T| = |A| and the properties of where you factor out a column or a row of the matrix and how that affects the determinant.

I would start with the characteristic equation and compare the two in some way to show that the roots have the property you desire.

If you show that the roots are greater of char(Q) than char(P), that's a real proof and you're done.

 

[blue_raver22]

eni

I would like to first clarify that the statement "eig(Q) >= eig(P)" does not necessarily mean that all the eigenvalues of Q are greater than or equal to all the eigenvalues of P. It simply means that the eigenvalues of Q are greater than or equal to the corresponding eigenvalues of P in a non-negative orthant (i.e. all eigenvalues are positive or zero).

Now, to address the question at hand, let us first consider the definition of a symmetric positive-definite matrix. A matrix P is symmetric positive-definite if it satisfies the following conditions:

1. P is symmetric, i.e. P = P^T
2. All eigenvalues of P are positive, i.e. eig(P) > 0

Using these conditions, we can prove that Q-P is also symmetric and positive-definite.

1. Symmetry:
Since P is symmetric, we have Q-P = SPS - P = SPS - P^T = SPS - (SP)^T = SPS - S^T P^T = SP(S - S^T)P^T.
Since S is a diagonal matrix, S = S^T, and thus Q-P = SP(S - S)P^T = SP^2 (S - S) = SP^2 0 = 0. Therefore, Q-P is symmetric.

2. Positive-definite:
Since P is positive-definite, all its eigenvalues are positive. Additionally, since S is a diagonal matrix with all diagonal elements greater than 1, we can say that all eigenvalues of S are also greater than 1. This means that when we multiply P by S, the resulting eigenvalues will be larger than the original eigenvalues of P. And since we are squaring the diagonal elements of S, these eigenvalues will be even larger when we multiply by S again. This means that all eigenvalues of SP^2 will be greater than the eigenvalues of P (in a non-negative orthant). Therefore, all eigenvalues of Q-P = SP^2 0 will be greater than 0, making it a positive-definite matrix.

In conclusion, we can say that Q-P is indeed symmetric and positive-definite. And since the eigenvalues of a positive-definite matrix are always greater than 0, we can also say that the eigenvalues of Q-P will be greater than the eigen