Submultiplicativity stronger than triangle inequality?

Question

I would like to ask a question about matrix norm.

Is the submiltiplicativity property always stronger than the triangle inequality? So, if i prove for a matrix norm that it's submultiplicative, i don't need to show that the triangle inequality holds too?

Thank you in advance!

Answer 1

The map $A \mapsto |\det A|^{1/n}$ satisfies all the desired property as a counter-example except for the nondegeneracy. So we perturb this map in the following way:

Let $\| \cdot \|$ be any submultiplicative matrix norm on $M_{n\times n}$, and define

$$ \| A \|' = \| A \| + t |\det A|^{1/n},$$

where $t \geq 1$ is a constant to be chosen later. Then

1. $\| A \|' \geq 0$ with equality holds if and only if $A = O$.
2. $\| \alpha A \|' = |\alpha| \| A \|'$.
3. Using $t \leq t^{2}$, we get \begin{align*} \| AB \|' &= \|AB\| + t |\det A|^{1/n} |\det B|^{1/n} \\ &\leq \|A\|\|B\| + t^{2} |\det A|^{1/n} |\det B|^{1/n} \\ &\leq ( \|A\| + t |\det A|^{1/n}) ( \|B\| + t |\det B|^{1/n}) = \|A\|' \|B\|'. \end{align*}

But if we choose two singular $A, B$ that add up to $I_{n}$, then

$$ \|I\| + t = \| I \|' = \| A+B \|'$$

while

$$ \|A\|' + \|B\|' = \|A\| + \|B\|. $$

Now choose $t$ sufficiently large so that $\| I \| + t > \| A \| + \| B \|$. This invalidates the triangle inequality.