[Math] Natural matrix norm of an inverse matrix

Question

Let $\left\|\cdot\right\| : \text{GL}(n,\mathbb{R})\to\mathbb{R}_{\ge 0}$ denote the natural matrix norm, i.e. $$\left\|A\right\|:=\max_{x\ne 0}\frac{\left\|Ax\right\|}{\left\|x\right\|}=\max_{\left\|x\right\|=1}\left\|Ax\right\|$$
is induced by a vector norm $\left\|\cdot\right\| : \mathbb{R}^n\to\mathbb{R}_{\ge 0}$. I want to show, that it holds $$\left\|A^{-1}\right\|=\left(\min_{\left\|x\right\|=1}\left\|Ax\right\|\right)^{-1}$$
Proof:
\begin{equation}
\begin{split}
\left\|A^{-1}\right\|&=\max_{\left\|x\right\|=1}\left\|A^{-1}x\right\|\\
&=\max_{\left\|Ay\right\|=1}\left\|y\right\|\\
&=\left(\min_{\left\|Ay\right\|=1}\left\|y\right\|^{-1}\right)^{-1}\\
&=\left(\min_{\left\|x\right\|=1}\left\|Ax\right\|\right)^{-1}
\end{split}
\end{equation}
How does the last step work?

Answer 1

The last step in your proof can be finished as follows. The mapping $y=\frac{x}{\|Ax\|}$ from $X=\{x:\|x\|=1\}$ to $Y=\{y:\|Ay\|=1\}$ is bijective. Its inverse is $x=\frac{y}{\|y\|}$. Therefore \begin{aligned} \left(\min_{\|Ay\|=1}\|y\|^{-1}\right)^{-1} =\left(\min_{y\in Y}\frac{\|Ay\|}{\|y\|}\right)^{-1} =\left(\min_{x\in X}\frac{\left\|A\frac{x}{\|Ax\|}\right\|}{\left\|\frac{x}{\|Ax\|}\right\|}\right)^{-1} =\left(\min_{\left\|x\right\|=1}\left\|Ax\right\|\right)^{-1}. \end{aligned} However, as pointed out by another user in a comment under your question, it is much easier to prove the original identity in another way: \begin{aligned} \|A^{-1}\|&=\max_{y\ne0}\frac{\|A^{-1}y\|}{\|y\|} =\max_{x\ne0}\frac{\|x\|}{\|Ax\|} =\left(\min_{x\ne0}\frac{\|Ax\|}{\|x\|}\right)^{-1} =\left(\min_{\|x\|=1}\|Ax\|\right)^{-1}. \end{aligned}