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?

  • $\begingroup$ 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. $\endgroup$
    – Anthony
    Jan 17, 2021 at 17:09
  • $\begingroup$ @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, $\lambda x \supseteq Ax$ ? $\endgroup$
    – hexaquark
    Jan 17, 2021 at 17:14

1 Answer 1


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 :

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


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