Let $A$ be a square matrix with elements in $\mathbb{R}$ or $\mathbb{C}$, $\rho\left(A\right)$ stands for the spectral radius of $A$, i.e., the maximum absolute eigenvalue of $A$; $A^{*}$ is the conjugate transpose of $A$; an induced matrix norm $\left\Vert *\right\Vert $ besides the usual vector norm properties, has the sub-multiplicative property.

Question: Given a matrix $A$, there exists a matrix norm $\left\Vert *\right\Vert $ such that

$\rho\left(A\right)<1\Rightarrow\left\Vert A\right\Vert <1$ and $\left\Vert A^{*}\right\Vert <1$?

Motivation: In [Horn and Johnson 1985, Matrix Analysis, Lemma 5.6.10] it is shown how to construct a matrix norm such that $\left\Vert A\right\Vert \leq\rho\left(A\right)+\epsilon$ for any given scalar $\epsilon>0$. With this norm it is sufficient to choose $\epsilon$ sufficiently small to guarantee $\left\Vert A\right\Vert <1$. However, it is possible to find examples where any of the above choice of $\epsilon$ leads to $\left\Vert A^{*}\right\Vert >1$. The problem is to find a matrix norm with sub-multiplicative property that guarantees both, $\left\Vert A\right\Vert <1$ and $\left\Vert A^{*}\right\Vert <1.$

Someone know how to prove or a reference with a proof for that?


Consider the matrix $A = \pmatrix{0 & 1\cr 0 & 0\cr}$ which has spectral radius $0$. But $A A^*$ has spectral radius $1$, so for any sub-multiplicative norm $$1 \le \|A A^*\| \le \|A\| \|A^*\| \le \max(\|A\|,\|A^*\|)^2$$


Here's another argument (when $n > 1$): all norms on the algebra of $n \times n$ matrices are equivalent since it is a finite dimensional Banach space ( this is true of all linear space norms, whether submultiplicative or not). Hence, given any algebra ( ie submultiplicative as well as subadditive) norm on the algebra of $n \times n$ matrices, we have $\rho(A) = \lim_{k \to \infty} \|A^{k}\|^{\frac{1}{k}} \leq \|A \|$ for any matrix $A$.

If there were an algebra norm of the form you ask, then if $\rho(A) <1$ we would have $\|A \| < 1$ and $\|A^{\ast}\| < 1$, so $\rho(AA^{\ast}) \leq \| AA^{\ast} \| \leq \|A\| \|A^{\ast}\| < 1$. But consider an upper triangular matrix $A$ with all $0$ on the main diagonal, every entry above the main diagonal $2$. Then $A$ is nilpotent, so that $\rho(A)= 0$. However $AA^{\ast}$ has trace $2n(n-1)$, so some eigenvalue of $AA^{\ast}$ has absolute value $2(n-1)$ or greater and $\rho(AA^{\ast}) \geq 2(n-1) \geq 2$. In fact we can find a nilpotent matrix $A$ such that $\rho(AA^{\ast})$ is as large as we like.


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