Eigenvalues of Positive semi-definite matrices after congruent transformations

Question

Suppose $B$ is a positive semi-definite matrix. Let $A$ be a matrix such that all eigenvalues of $A$ lie outside the unit circle. Clearly, eigenvalues of $A^n$ become unbounded (tend to infinity) as $n$ becomes unbounded (tends to infinity) $(n\in \mathbb{N})$. I want to show that the non zero eigenvalues of $A^nB(A^T)^n$ become unbounded as $n$ becomes unbounded.

Answer 1

It's equivalent to show that the eigenvalues of $A^{-n}C(A^T)^{n}$ go to zero as $n \to \infty$. To that effect, apply the spectral norm, noting that $\|A^{-1}\| < 1$. $$ \|A^{-n}C(A^T)^{n}\| \leq \|A^{-1}\|^n \|C\| \|A^{-1}\|^n $$ which goes to $0$ as $n \to \infty$.

Answer 2

Suppose $B$ has rank $k$. It suffices to show that the $k$-th largest singular value of $A^nB^{1/2}$ approaches infinity when $n\to\infty$. By Courant-Fischer minimax principle, \begin{align*} \sigma_k(A^nB^{1/2}) &=\max_{\dim V=k}\min_{x\in V,\ \|x\|=1}\|A^n B^{1/2}x\|\\ &\ge\max_{\dim V=k}\min_{x\in V,\ \|x\|=1}\sigma_\min(A^n)\|B^{1/2}x\|\\ &=\sigma_\min(A^n)\sigma_k(B^{1/2}). \end{align*} As $\sigma_k(B^{1/2})>0$, it in turn suffices to prove that the smallest singular value of $A^n$ goes to infinity when $n$ tends to infinity.

In general, even if all eigenvalues of $A$ have moduli greater than $1$, it may still occur that $\sigma_\min(A)<1$. Even worse, $\sigma_\min(A^n)$ could be smaller than $1$ for fairly large $n$. For example, suppose $A=\frac{65}{64}\pmatrix{1&1\\ 0&1}$. Its smallest singular value is about $0.63$. The value of $\sigma_\min(A^n)$ actually decreases at first when $n$ increases. It reaches its minimum when $n=64$, starts to increase afterwards and becomes greater than $1$ for the first time only when $n=190$.

However, when all eigenvalues of $A$ have moduli greater than $1$, $\sigma_\min(A^n)$ must eventually tend to infinity. This is because Gelfand's formula dictates that $\lim_{n\to\infty}\|A^{-n}\|^{1/n}=\rho(A^{-1})<1$. Hence $\|A^{-n}\|$ must approach zero when $n\to\infty$. So, if we we use the operator norm, we get $\|A^{-n}\|=\sigma_\max(A^{-n})=1/\sigma_\min(A^n)$ the assertion follows.