I need to prove that the induced matrix norm satisfies $$\|A\| = \max_{\|x\| = 1} \|Ax\|$$
Here's what I've done so far, and I'm not sure how to make the connection.
By definition, $$\|A\| = \max_{x\neq 0}{\|Ax\| \over \|x\|}.$$ Because $x\neq0$, $\|x\| \neq 0 \implies \|x\| = \alpha > 0$ for any nonzero $x\in \mathbb{R}^n$ by definition of a vector norm. So, $$\begin{align}\|x\| = \alpha &\iff {1 \over \alpha}\|x\| = 1 \\ &\iff\left\|{1 \over \alpha}x\right\| = 1 \\ &\iff \|\hat{x}\| = 1. \tag{Let ${1 \over \alpha}x = \hat{x}$}\end{align}$$ I need to show that $$\max_{\|x\| = 1}\|Ax\| = \max_{x\neq0}{\|Ax\| \over \|x\|}.$$ Well I've shown that $$x \neq 0 \implies \|\hat x\| = 1.$$ So does it suffice to say that since $\|\hat x\| = 1$, $$\|A\| = \max_{x \neq 0}{\|Ax\| \over \|x\|} = \max_{\|\hat x\| = 1}{\|A\hat x \| \over \|\hat x\|} = \max_{\|\hat x\| = 1} {\|A\hat x\| \over 1} = \max_{\|\hat x\| = 1} \|A\hat x\|?$$
Thanks so much!
Best Answer
Just to add to the previous answer: a clearer argument (I think) is to proceed in two steps. You certainly have $$ \max_{\|x\| = 1} \|Ax\| = \max_{\|x\| = 1} \frac{\|Ax\|}{\|x\|} \le \max_{x \not= 0} \frac{\|Ax\|}{\|x\|}. $$
For the reverse inequality, fix $x \not= 0$. Then $$ \frac{\|Ax\|}{\|x\|} = \left\|A\frac{x}{\|x\|}\right\| \le \max_{\|y\| = 1} \|Ay\|, $$ and so taking the maximum over nonzero $x$ gives the result.