Proof of Matrix Norm (Inverse Matrix)

Question

Show for any induced matrix norm and nonsingular matrix A that $$ \left\|A^{-1}\right\| ≥ (\left\|A\right\|)^{-1} $$ where $$ \left\|A^{-1}\right\| = \max_{\left\|x\right\|=1}\{\left\|A^{-1}x\right\|\}\\ \left\|A\right\| = \max_{\left\|x\right\|=1}\{\left\|Ax\right\|\}. $$

I am not sure how to show that: \begin{equation} \begin{split} \left\|A^{-1}\right\| ≥ (\left\|A\right\|)^{-1}\\ \text{or}\\ \max_{\left\|x\right\|=1}\{\left\|A^{-1}x\right\|\} ≥ (\max_{\left\|x\right\|=1}\{\left\|Ax\right\|\})^{-1} \end{split} \end{equation}

Answer 1

Use $\lVert AB\rVert \leq \lVert A\rVert \lVert B\rVert$, as the induced norm is in particular submultiplicative. So that $\lVert I_n\rVert \leq \lVert A\rVert \lVert A^{-1}\rVert$.

Answer 2

Suppose $|y|=1$ is such that $|A|=|Ay|$. Then $x=Ay/|A|$ also has norm $1$ so it follows that $$ |A^{-1}|\geq |A^{-1}x|=|A^{-1}Ay/|A||=\frac{|y|}{|A|}=\frac{1}{|A|}=|A|^{-1}. $$

Answer 3

Proof: $$ \left\|A^{-1}\right\| ≥ (\left\|A\right\|)^{-1} $$ where $$ \left\|A^{-1}\right\| = \max_{\left\|x\right\|=1}\{\left\|A^{-1}x\right\|\}\\ \left\|A\right\| = \max_{\left\|x\right\|=1}\{\left\|Ax\right\|\}\\ \left\|A\right\|^{-1} = (\max_{\left\|x\right\|=1}\{\left\|Ax\right\|\})^{-1} = (\min_{\left\|x\right\|=1}\{\left\|A^{-1}x\right\|\}) $$ and $$ \max_{\left\|x\right\|=1}\{\left\|A^{-1}x\right\|\} ≥(\min_{\left\|x\right\|=1}\{\left\|A^{-1}x\right\|\}) $$ therefore: $$ \left\|A^{-1}\right\| ≥ (\left\|A\right\|)^{-1} $$