Norm and inverse matrices I have to prove for $A \in \mathbb{C}^{n\times n}$ this statement:
If $x^{H}Ax\geq \gamma\left \| x \right \|_{2}^2$ for $x \in \mathbb{C}^n$ with $\gamma > 0$, then $\left \| A^{-1} \right \|_{2} \leq \gamma^{-1}$.
I really don´t know where to start or what to do exactly, any advice would be helpful.
 A: If $\langle x, Ax \rangle $ is real for all $x$ then $A$ is Hermitian.
To see this, note that $\langle x, Ax \rangle = \overline{\langle x, Ax \rangle} = \langle A x, x \rangle = \langle x, A^*x \rangle$ and
so $\langle x, (A-A^*)x \rangle = 0 $ for all $x$ and hence $A=A^*$.
If $A$ is Hermitian, it is unitarily diagonalisable, that is there is some unitary $U$
such that $A = U^T \Lambda U$, where
$\Lambda = \operatorname{diag}(\lambda_1,...,\lambda_n)$ and each $\lambda_k$ is real.
The Euclidean norm is invariant under unitary transformations, so the problem
reduces to showing that
if $\langle x, \Lambda x \rangle = \sum_k \lambda_k |x_k|^2 \ge \gamma \|x\|^2$
for all $x$ then $\| \Lambda^{-1} \| \le {1 \over \gamma}$.
By choosing $x=e_k$ we see that $\lambda_k \ge \gamma >0$ for all $k$.
Now choose $x$ and let $y_k={1 \over \sqrt{\lambda_k}}x_k$. Substituting $y$ into
$\sum_k \lambda_k |y_k|^2 \ge \gamma \|y\|^2$ we get
$\sum_k |y_k|^2 \ge \gamma \sum_k {1 \over \lambda_k} |y_k|^2= \|\Lambda^{-1} y\|^2$
from which the result follows.
A: *

*If $\,\|C\|_2 = \sqrt{\sum_{i,j=1}^n|c_{ij}|^2}\,$ denotes the Frobenius matrix norm, then the statement is wrong:
Consider for instance $\gamma=1$ and $A$ equal to the identity, because then $\|A^{-1}\|_2 = \sqrt{n}$.

*The statement holds true if $\|C\|_2=\max\big\{\|Cx\|_{2}\:\big|\: x \in \mathbb{C}^n,\|x\|_2=1\big\}$  denotes the derived matrix norm which is the spectral norm:
From the hypothesis $\gamma\|x\|_2^2\leqslant \langle x\,|Ax\rangle$ one obtains $\,\ker A=\{0\}$, thus $A^{-1}$ exists. 
Insert $x=A^{-1}z\,$ with a unit vector $z\in\mathbb C^n$, then
$$\begin{align}\gamma\,\|A^{-1}z\|_2^2 & \:\leqslant\:\langle A^{-1}z\,|z\rangle \\[1.3ex]
 & \:\leqslant\:\|A^{-1}z\|_2\quad\text{by  Cauchy–Bunyakovsky–Schwarz} \\[2ex]
\implies\;\|A^{-1}z\|_2 & \:\leqslant\:\gamma^{-1}
\end{align}$$
which yields the claim.

