I have to prove for $A \in \mathbb{C}^{n\times n}$ this statement:
- If $\left \| Ax \right \| \geq \gamma \left \| x \right \|$ for all $x \in \mathbb{C}^n$ with $\gamma > 0$ and any vector norm $\left \| . \right \|$, then exists $A^{-1}$ and $\left \| A^{-1} \right \| \leq \gamma^{-1}$ for the matrix norm belonging to the vector norm.
Edit 1:
I tried this way:
I assumed $A$ is singular, which means that $detA = 0$ and from that, we conclude that there is an eigenvalue $\lambda = 0$. And because of it, there is an Eigenvector $x \in \mathbb{C}^{n}$ to $\lambda$ with $Ax=\lambda x = 0$. From that follows $$ \gamma \left \| x \right \| > 0 =\left \| Ax\right \| $$ which leads to contradiciton. I hope this one is correct.
Here's another way to establish invertibility. Note that $x \neq 0$ implies that $\|Ax\| \geq \gamma \|x\| > 0$, which means that $Ax \neq 0$. Thus, $A$ is a square matrix with a trivial nullspace, which means that it must be invertible.
A nice way to get the inequality: note that for all $x$, we have $$ \|x\| = \|A(A^{-1}x)\| \geq \gamma \|A^{-1}x\|. $$ Rearranging this inequality yields $\|A^{-1}x\| \leq \gamma^{-1}\|x\|$. Now, by definition we have $$ \|A^{-1}\| = \sup_{\|x\| = 1} \|A^{-1}x\| \leq \sup_{\|x\| = 1} \gamma^{-1}\cdot \|x\| = \gamma^{-1}. $$