2
$\begingroup$

Matrix norms are equivalent and can bound each other like some examples on Wikipedia. I was wondering if there is a matrix norm $|| \cdot ||$ that can be upper bounded by $||\cdot ||_{\infty}/n$ ? That is, $||A|| \le \frac{||A||_{\infty}}{n}$, where $A$ is an $n \times n$ square matrix.

Since norms are equivalent, perhaps we can define a matrix norm by ourselves, only if it satisfies the definition of a matrix norm. A naive idea is just letting $|| \cdot || = ||\cdot ||_{\infty}/n$ but it is not sub-multiplicative (the last condition in the definition).

$\endgroup$

2 Answers 2

2
$\begingroup$

Since $\|I\|=\|I^2\|\le\|I\|^2$, we have $\|I\|\ge1$. So, if $\|I\|\le\frac{\|I\|_\infty}n$, then ...

$\endgroup$
1
  • $\begingroup$ Thank you. I also found the same counter example and gave up finding such a norm... $\endgroup$
    – D. Chen
    Commented Jan 21, 2014 at 21:25
1
$\begingroup$

Such a matrix norm cannot exist because let $A$ be any nonzero matrix. Then consider $e^{tA}$. We have that $||e^{tA}||\le ||e^{t/k A}||^k$ for all $t, k>0$. Now take $A=I$. Then we have that $||e^{t/k A}||^k\le e^t/n^k$ for all $k$. Take $k \rightarrow \infty$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .