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?
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 :
$\begin{align*}\||A\||&=\max_{\|x\|=1}\|Ax\|&\\&\geqslant\max_{\|x\|=1\text{ and }Ax=\lambda x}\|Ax\|\\&=\max_{\|x\|=1\text{ and }Ax=\lambda x}\|\lambda x\|\\&=\max_{\|x\|=1\text{ and }Ax=\lambda x}|\lambda|\|x\|\\&=\rho(A)\max_{\|x\|=1\text{ and }Ax=\lambda x}1\\&=\rho(A).\end{align*}$