# Can we characterize the submultiplicative matrix norm such that $\| A\| \le \rho(A) + \varepsilon$?

We know for $A \in \mathcal M_n$ and every $\epsilon > 0$, there exists a submultiplicative matrix norm $\| \cdot\|_m$ such that $\|A\|_m \le \rho(A) + \varepsilon$. Is this norm vector norm induced? In other words, could we define some norm on $\mathbb C^n$ such that $\|A\|_m = \sup_{\|x\| = 1} \|Ax\|$ for $x \in \mathbb C^n$? Thanks.

Indeed! Let $S$ be the matrix for which $A=SJS^{-1},$ where $J$ is the Jordan canonical form of $A.$ Let $E=\mathrm{diag}(1,\varepsilon,\varepsilon^{2},\ldots,\varepsilon^{n-1}).$ Then $A=SEJ'E^{-1}S^{-1},$ where $J'$ is like the Jordan canonical form of $A$, but wherever $J$ had a one on its first superdiagonal, $J'$ has a $\varepsilon.$ Let $V=SE.$ Then if we let $\|x\|:=\|V^{-1}x\|_{\infty},$ we see that \begin{align*} \|A\|&=\sup_{\|x\|=1}\|Ax\|\\ &=\sup_{\|V^{-1}x\|_{\infty}=1}\|V^{-1}Ax\|_{\infty}\\ &=\sup_{\|y\|_{\infty}=1}\|V^{-1}AVy\|_{\infty}\\ &=\sup_{\|y\|_{\infty}=1}\|J'y\|_{\infty}=\rho(A)+\varepsilon. \end{align*}
I leave it to you to show that $\|V^{-1}\cdot\|_{\infty}$ defines a valid vector norm on $\mathbb{C}^{n}$, and the last equality.
• Okay, I don't think I was understanding before. In the chapter on norms, there should be an exercise asking you to prove that for a nonsingular matrix $S$ and vector norm $\|\cdot\|$, $\|S^{-1}\cdot\|$ defines a vector norm. – RideTheWavelet Feb 8 '18 at 20:40