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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!

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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.

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\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 $x \neq 0$.

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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 $\mathcal{Re}M^{-1}\succ 0$

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