Suppose $A$ is an n by n matrix whose eigenvalues have a magnitude larger than 1. $P$ is a covariance matrix. Is it true that $APA' \geq P$?

Question

Here, $A$ is an $n$ by $n$ matrix, whose eigenvalues all have a magnitude larger than 1. $P$ is a covariance matrix of dimension $n$. I guess it should be true that $APA' - P$ is positive semi-definite because $A$ is 'amplifying' the covariance $P$.

I guess to shows its correctness we may leverage on matrix decomposition such as $APA' - P = AQTQ'A' -QTQ'$. But I am not sure how to proceed especially how to incorporate the fact that the eigenvalues of $A$ all have magnititude larger than $1$.

Answer 1

This is true for all positive semidefinite $P$ if and only if $A=cI$ for some scalar $c$ with $|c|\ge1$.

Suppose the inequality is true for all PSD matrices $P$. By passing $P$ to the limit, the inequality is true for all rank-one PSD matrices $P=vv'$. Hence $(Av)(Av)'\succeq vv'$ and $Av$ is parallel to $v$. Since $v$ is arbitrary, $A$ must be a scalar matrix. Conversely, suppose $A=cI$ for some scalar $c$ of modulus $\ge1$. Then $APA'-P=(|c|^2-1)P\succeq0$ for all $P\succeq0$.

Answer 2

A counter example would be

$$ A = \begin{bmatrix} 2 & 0 \\ 0 & -2 \end{bmatrix}, \quad P = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}, \quad A\,P\,A'-P = \begin{bmatrix} 3 & -5 \\ -5 & 3 \end{bmatrix}, $$

which has eigenvalues $-2$ and $8$. If you prefer a positive definite covariance matrix you could also use

$$ P = \begin{bmatrix} 10 & 9 \\ 9 & 10 \end{bmatrix}, $$

which yields

$$ A\,P\,A'-P = \begin{bmatrix} 30 & -45 \\ -45 & 30 \end{bmatrix}, $$

and has eigenvalues $-15$ and $75$.