Does it follow that the hermetian part of a matrix is positive definite, that the matrix itself is invertible? I came across this in a paper and I was wondering whether it is true. We have a complex matrix $M$ such that $(M+M^H)$ is positive definite. Now, it is clear that $(M+M^H)$ is invertible, but does that hold for $M$?
Is there (in general) some connections between the hermetian part of a matrix and the hermetian part of its inverse?
Thanks guys!
 A: In fact, you can show that $\mathrm{Re}(z^*Mz) > 0$ for every non-zero complex vector $z$. Indeed, suppose that you can find a $z$ for which it doesn't hold:
$$\mathrm{Re}(z^*Mz) \leq 0.$$
Then you have
\begin{align*}
\mathrm{Re}(z^*M^*z) &= \mathrm{Re}((z^*Mz)^*)\\
&= \mathrm{Re}(z^*Mz)\\
&\leq 0.
\end{align*}
However, you then get
$$\mathrm{Re}(z^*(M + M^*)z) \leq 0, $$
which contradicts the positive-definiteness of the Hermitian part.
A: \begin{eqnarray}
\langle x , (M+M^*) x \rangle &=& \langle x , M x \rangle + \langle x , M^* x \rangle  \\
&=&  \langle x , M x \rangle + \overline{\langle M^* x , x \rangle}  \\
&=& \langle x , M x \rangle + \overline{\langle  x , M x \rangle}  \\
&=& 2 \operatorname{re} \langle x , M x \rangle
\end{eqnarray}
In particular, $\langle x , M x \rangle \neq 0 $ for all $x \neq 0$.
A: The implication $\mathcal{Re}M \succ 0 \implies \mathcal{Re}M^{-1} \succ 0$ appeared before on the site, might as well repeat the argument. 
First, if $v$ is an eigenvector for $M$ for the eigenvalue $\lambda$ then 
$v$ is an eigenvector for $\mathcal{Re}M$ for the eigenvalue $\mathcal{Re}\lambda$. 
Next, if $\mathcal{Re}M\succ 0$ then all the eigenvalues of $M$ have real part $>0$ ( from the above). In particular, all are nonzero, so $M$ is invertible.
Now, $\mathcal{Re}M \succ 0$ is equivalent to $\langle M v, v \rangle +\langle v, M v \rangle > 0$ for any nonzero vector $v$. We know from the above that $M$ is invertible. Plug in in the previous equation $v\colon = M^{-1} w$ and get 
$\langle  w, M^{-1}w \rangle +\langle M^{-1}w, w \rangle > 0$ for any nonzero vector $w$. Therefore $\mathcal{Re}M^{-1}\succ 0$
