TITLE: Does it follow that the hermetian part of a matrix is positive definite, that the matrix itself is invertible?
QUESTION [5 upvotes]: 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!

REPLY [3 votes]: 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 $\mathcal{Re}M^{-1}\succ 0$