3
$\begingroup$

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!

$\endgroup$
1
$\begingroup$

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 $\mathcal N$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.