# Spectral radius and positive definite matrix

Let $$A$$ be a $$m \times m$$ matrix. Show that there exists a $$m \times m$$ positive definite matrix $$B$$ such that $$B - A^H B A \succ 0$$ if the spectral radius $$\rho(A) < 1$$

Let $$B= B^{1/2}B^{1/2}$$, if we want to show $$B-A^HBA$$ is PD, it equivalents to show that

$$B^{-1/2}(B-A^HBA)B^{-1/2}= I-B^{-1/2}A^HB^{1/2}B^{1/2}AB^{-1/2}$$

is positive definite. Then, let $$C = B^{1/2}AB^{-1/2}$$, and $$A$$, $$C$$ are similar to each other, it equivalents to show that $$I-C^HC$$ is positive definite. Then, we only need to show that $$\lambda_i \left( C^H C \right) < 1$$ but I don't know how to prove the last inequality since $$C$$ is not Hermitian.

Could you please give me some ideas?

If $$\rho(A)<1,$$ then for any matrix norm $$\lim_{k\to \infty} \|A^k\| = 0.$$ In particular, this is true for the operator norm, and it is also easy to see (using the Jordan decomposition) that the convergence is eventually monotonic. So, pick $$B = \sum_{j=0}^k(A^j)^H A^j$$ for $$k$$ large enough.
• Your choice of $B$ will make $S:=B-A^HBA=(A^k)^H(I-A^HA)A^k$. It doesn't work when $A$ is singular, because $S$ is singular. It also doesn't work when $A$ is nonsingular, because $S$ is congruent to $I-A^HA$, which is not positive definite when $\|A\|_2\ge1$. Apr 26, 2021 at 14:29