Linear algebra based proof that if there exists $P\succ 0$ and $P\succ A^TPA$ then $|\lambda_i(A)|<1$

Question

Is there a proof based on linear algebra that shows the following?

If there exist $P \succ 0$ and $P \succ A^TPA$, then $| \lambda_i (A) | < 1$ for all $i$.

Here, $|\lambda_i(A)|$ denotes the magnitude of the $i$th eigenvalue, which may be complex.

Unless I've made a mistake, this is simply a Lyapunov stability condition for the discrete-time linear time-invariant system $x_{k+1} = Ax_k$. That said, the original statement above contains no statement about stability and is simply a statement about the eigenvalues of a matrix $A$. As such, it seems like there should be a direct proof, but I've not seen one and am not sure how to derive it.

Answer 1

As another user has pointed out in a comment, the correct statement should be: if $P\succ0$ and $P\succ A^TPA$, then $|\lambda_i(A)|<1$ for each $i$, i.e. $\rho(A)<1$.

The proof is simple. As $P$ is positive definite, it has a (unique) positive definite square root $P^{1/2}$. Multiply by $P^{-1/2}$ on both sides of $P\succ A^TPA$, we get $I\succ B^TB$, where $B=P^{1/2}AP^{-1/2}$. It follows that $\rho(A)^2=\rho(B)^2\le\|B\|^2=\rho(B^TB)<1$ where $\|\cdot\|$ denotes the operator norm.