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Prove that for any vector norm and its subordinate matrix norm, and for any $n\times n$ matrix $A$, there corresponds a vector $x\neq 0$ such that $\|Ax\|=\|A\|\|x\|$

I know that $\|Ax\|\leq \|A\|\|x\|$ for all $x\in \mathbb{R}^n$, but I do not know how to find $x\neq 0$ in such a way that the other equality is fulfilled, could someone help me please? Thank you very much.

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2 Answers 2

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Hint: the map $f(x)=\|A(x)\|$ defined on $S_n=\{x:\|x\|= 1\}$ is continuous, since $S_n$ is compact, there exists $x\in B_n$ such that $f(x)=sup_{y\in S_n}f(y)$.

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$$\|A\| = \sup_{\|x\| =1}\|Ay\|=\sup_{\|x\| \le1}\|Ay\| $$

Since $\{ x: x \le 1\}$ is compact, and $\|Ax\|$ is continuous, then the maximum is well defined.

Hence there exists $\|x\|=1$ such that $\|A\|= \|Ax\|$.

$$\|A\|\|x\|=\|Ax\|$$

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