I am currently working on some properties of matrix products and their norms for $\mathbb{R}^{n \times n}$ matrices and i was wondering if there exists a completely multiplicative matrix norm, i.e. $\left\Vert AB \right\Vert = \left\Vert A \right\Vert \left\Vert B \right\Vert$, for a certain semigroup of matrices. I am aware of the fract that there does not exist a completely multiplicative matrix norm for all $M \in\mathbb{R}^{n \times n}$, since there exist nonzero matrices such that their product is the zero matrix. Are there any conditions for the semigroup to guarantee the existence or non-existence for such a norm?

To be more precise: Let

$$ A=\begin{pmatrix} 1 & \frac{1}{3} & \frac{1}{3} \\ 1 & 0 & 0 \\ 0 & 1 & 0 \\ \end{pmatrix},\,\,\,\,\,\,\,\,\, B=\begin{pmatrix} 1 & -\frac{1}{3} & -\frac{1}{3} \\ 1 & 0 & 0 \\ 0 & 1 & 0 \\ \end{pmatrix}, $$ and cosider the generated semigroup of these matrices, i.e. all matrices one can get by just multiplying those two matrices in any order and length. It would be a huge help if i could find such a completely multiplicative norm on that semigroup.

Thanks in advance!


Edit. As a semigroup $G\subseteq M_n(\mathbb C)$ is not necessarily closed under addition or scalar multiplication, it is not clear what is meant by a matrix norm on a semigroup $G$. In the following, I will consider instead a larger class of real-valued functions $\mathcal N$ such that $f\in\mathcal N$ iff

  • $f(cA)=|c|f(A)\ge0$ for every $c\in\mathbb C$ and every $A\in M_n(\mathbb C)$,
  • $\lim_{k\to\infty}f(A_k)=0$ for every sequence of matrices $\{A_k\}$ such that $\lim_{k\to\infty}\|A_k\|=0$,
  • $\lim_{k\to\infty}f(A_k)=\infty$ for every sequence of matrices $\{A_k\}$ such that $\lim_{k\to\infty}\rho(A_k)=\infty$.

Here $\|\cdot\|$ is any (not necessarily submultiplicative) matrix norm of choice. The spectral radius function $\rho$, as well as all matrix norms, are members of $\mathcal N$. Your question is now modified as follows:

Given a semigroup $G\subseteq M_n(\mathbb C)$, does there exist some $f\in\mathcal N$ that is completely multiplicative on $G$?

Gelfand's formula says that if $\|\cdot\|$ is any submultiplicative matrix norm on $M_n(\mathbb C)$, then $\rho(A)=\lim_{k\to\infty}\|A^k\|^{1/k}$. However, to prove the formula for any particular matrix $A$, it is possible to use only the fact that $\|\cdot\|\in\mathcal N$ (see the proof in Wikipedia for instance). The triangle inequality as well as strict positivity of for nonzero matrices are not needed. So, Gelfand's formula actually holds for every $f\in\mathcal N$.

Now it is a simple consequence of Gelfand's formula that if $f\in\mathcal N$ is completely multiplicative on a semigroup $G$, it must be equal to the spectral radius on $G$. Since $\rho$ itself is a member of $\mathcal N$, we conclude that there exists some $f\in\mathcal N$ that is completely multiplicative on $G$ if and only if $\rho$ is completely multiplicative on $G$ and $f=\rho$ on $G$.

For instance, $\rho\in\mathcal N$ is completely multiplicative on the semigroup $G$ of all doubly stochastic matrices because $\rho=1$ on $G$.

In your example, since $\rho(AB)<\rho(A)\rho(B)$, there does not exist any $f\in\mathcal N$ that is completely multiplicative on $G$. But that certainly doesn't preclude other $f\not\in\mathcal N$ from being completely multiplicative on $G$. For instance, $f:A\mapsto|\det(A)|$ is always completely multiplicative on any semigroup $G\subseteq M_n(\mathbb C)$, but this $f$ only satisfies the first two defining properties of $\mathcal N$.


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