Let $\|\cdot\|$ be an arbitrary vector norm in $\mathbb{C}^n$. For matrix $A\in \mathbb{C}^n$, we define $$\alpha(A) = \min_{\|x\|=1} \|Ax\|.$$ Prove that if $\alpha(A) > 0$ then $A$ is invertible and $\|A^{-1}\|\leq \frac{1}{\alpha(A)}.$
Can you give me hints on where to start?
Since $\alpha(A)>0$ then for every $x\neq 0$ we have $$0<\alpha(A)\leqslant||A\Big(\frac{x}{||x||}\Big)||=\frac{1}{||x||}||Ax||\Rightarrow ||Ax||>0$$ Therefore if $Ax=0$ then $x=0$ so $\ker{A}=\{0\}$ is trivial and hence $A$ is invertible matrix. Now for every $x\neq 0$ we also have $$||x||=||A^{-1}Ax||\leqslant ||A^{-1}||||Ax||\Rightarrow ||A^{-1}||\geqslant \frac{||x||}{||Ax||}$$ Hence $$||A^{-1}||\geqslant \max_{x\neq 0}\frac{||x||}{||Ax||}=\max_{||x||=1}\frac{1}{||Ax||}=\frac{1}{\min_{||x||=1}||Ax||}=\frac{1}{\alpha(A)}$$