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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?

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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.

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    $\begingroup$ 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 $\|A\|_2\ge1$. $\endgroup$
    – user1551
    Apr 26, 2021 at 14:29

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