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

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.

• 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... – DonAntonio Feb 26 at 18:25
• I guess it should be $|\lambda_i(A)| < 1$. – SampleTime Feb 26 at 18:26
• @SampleTime You're correct. Fixed. – wyer33 Feb 26 at 18:31
• @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. – wyer33 Feb 26 at 18:33
• What's $\prec$ mean here? – 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.