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.

  • 2
    $\begingroup$ What is your definition of "magnitude"? Because both because of the name and also because of your notation it would seem that magnitude must be a non-negative real number... $\endgroup$ – DonAntonio Feb 26 at 18:25
  • $\begingroup$ I guess it should be $|\lambda_i(A)| < 1$. $\endgroup$ – SampleTime Feb 26 at 18:26
  • $\begingroup$ @SampleTime You're correct. Fixed. $\endgroup$ – wyer33 Feb 26 at 18:31
  • $\begingroup$ @DonAntonio I made a mistake on the bound. Magnitude means the magnitude of the complex eigenvalue. This didn't really make any sense when the bound was 0, which was a mistake. Hopefully, it makes more sense now with the bound of 1. $\endgroup$ – wyer33 Feb 26 at 18:33
  • $\begingroup$ What's $\prec$ mean here? $\endgroup$ – chhro Feb 26 at 18:34

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.


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