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## Main Question or Discussion Point

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

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