# Showing the spectral radius is smaller than the induced norm

The spectral radius of $$A \in \mathbb{R}^{n \times n}$$ denoted by $$\rho(A)$$ is the largest absolute value $$\lvert \lambda \rvert$$ of an eigenvalue of $$A$$. Show that $$\rho(A) \le \lvert \lvert A \rvert \rvert$$ for all induced matrix norms $$\lvert \lvert A \rvert \rvert$$.

### Attempt at solution

If $$\lambda$$ is an eigenvalue of $$A$$ then $$Ax = \lambda x$$ for $$x$$ an eigenvector. Then, for $$\rho(A) := \lvert \bar{\lambda}\rvert$$ the largest eigenvalue, \begin{align} \lvert \lvert A \rvert \rvert \overset{def.}{=} \max_{\lvert \lvert x \rvert \rvert=1 } \lvert \lvert Ax \rvert \rvert &= \max_{\lvert \lvert x \rvert \rvert =1} \lvert \lvert \lambda x \rvert \rvert \\ &= \lvert \lambda \rvert \underbrace{\max_{\lvert \lvert x \rvert \rvert =1} \lvert \lvert x \rvert \rvert}_{\color{red}{= 1 ?}} \\ &\le \lvert \bar{\lambda} \rvert\\ \therefore \lvert \lvert A \rvert \rvert &\le \rho(A) \end{align}

The inequality is the other way around, can someone help me find my mistake?

• The max over $\|x\|=1$ is greater than the max taken over such vectors that are $\lambda$-eigenvectors, since you're taking the max over a subset. Then, the rest are equalities, which give the desired result. Jan 17, 2021 at 17:09
• @AnthonySaint-Criq Sorry I am having trouble understanding. Are you saying that $\max_{\lvert \lvert x \rvert \rvert =1} \lvert \lvert Ax \rvert \rvert \ge \max_{\lvert \lvert x \rvert \rvert =1 } \lvert \lvert \lambda x \rvert \rvert$ ? In other words, $\lambda x \supseteq Ax$ ? Jan 17, 2021 at 17:14

Your mistake lies in $$\max_{\|x\|=1}\|Ax\|=\max_{\|x\|=1}\|\lambda x\|.$$ Did you fix the eigenvalue $$\lambda$$ beforehand ? Not all vectors are necessarily $$\lambda$$-eigenvectors.
What you want is fixing $$\lambda$$ with $$\rho(A)=|\lambda|$$, and notice that $$\{x\in E\;/\;\|x\|=1\text{ and }Ax=\lambda x\}\subset\{x\in E\;/\;\|x\|=1\},$$ so that : $$\max_{\|x\|=1}\|Ax\|\geqslant\max_{\|x\|=1\text{ and }Ax=\lambda x}\|Ax\|.$$ Now, the rest is easy :
\begin{align*}\||A\||&=\max_{\|x\|=1}\|Ax\|&\\&\geqslant\max_{\|x\|=1\text{ and }Ax=\lambda x}\|Ax\|\\&=\max_{\|x\|=1\text{ and }Ax=\lambda x}\|\lambda x\|\\&=\max_{\|x\|=1\text{ and }Ax=\lambda x}|\lambda|\|x\|\\&=\rho(A)\max_{\|x\|=1\text{ and }Ax=\lambda x}1\\&=\rho(A).\end{align*}