The spectral radius of a matrix is the max eigenvalue $\lambda_{max}$ of it.Matrix norm is a kind of norm satisfying some axioms.
For any matrix A,is it true that $\lambda_{max}\le \Vert A\Vert$ for any matrix norm on A?
It is easy to do with specially induced matrix norms ,which are induced from vector norms.When a matrix norm is induced by a vector norm,what we need to do is to notice the consistence of the matrix norm and the vector norm.
But what's about the others?What's about the matrix norms which can not be induced by vector norms,such as the Frobenius norm?(The Frobenius norm is defind by $\Vert A \Vert =\sqrt{trac(A^TA)}$.You can consider the $I_n$ to find that this kind of norm can't be induced by any vector norms.)
A proof or a counterexample is expected for this question.
Thank you.
Best Answer
As someone has pointed out in a comment, if $\|\cdot\|$ is a norm, then so is $c\|\cdot\|$ for any $c>0$. So, given any $A$, we will have $\rho(A)>c\|A\|$ when $c$ is sufficiently small. The answer to your question is therefore negative.
It is true, however, that $\rho(A)\le\|A\|$ provided that $\|\cdot\|$ is a submultiplicative matrix norm (i.e., a norm such that $\|AB\|\le\|A\|\|B\|$ for any $A$ and $B$). Let $\lambda$ be the largest-sized eigenvalue of $A$ and $x$ be a corresponding eigenvector. Pick any nonzero vector $y$. As $\|\cdot\|$ is submultiplicative, we have $$ \rho(A)\|xy^T\|=|\lambda|\|xy^T\|=\|Axy^T\|\le\|A\|\|xy^T\|. $$ Since $xy^T$ is a nonzero matrix, $\|xy^T\|>0$. Therefore $\rho(A)\le\|A\|$.