If $X + X^T$ is positive definite, is $X^{-1} + X^{-T}$ also positive definite?

Question

Is it true or is there a counterexample?

Answer 1

$X + X^T$ is positive definite iff for every nonzero vector $v$, $v^T (X + X^T) v = 2 v^T X v > 0$. In particular, it implies $X$ is invertible. Taking $w = X v$, this is equivalent to $w^T X^{-1} w > 0$ for all nonzero vectors $w$, and thus to $X^{-1} + (X^{-1})^T$ being positive definite.

Answer 2

You can rewrite $$ X^{-1} + X^{-T} = X^{-T}(X+X^T)X^{-1}. $$ Then for a vector $x\ne0$ it follows $$ x^T(X^{-1} + X^{-T})x = x^TX^{-T}(X+X^T)X^{-1}x = (X^{-1}x)^T(X+X^T)(X^{-1}x)>0, $$ as $X^{-1}x\ne0$ and $X+X^T$ is positive definite.

Answer 3

Hint: $$\langle \,(X + X^{T}) v, v \rangle = \langle X v, v\rangle + \langle v, X v\rangle$$ For $w= X^{-1}w $ the above expression becomes $$ \langle w, X^{-1} w\rangle + \langle X^{-1}w, w\rangle= \langle \,(X^{-1} + X^{-T}) w, w \rangle$$

Obs: Both conditions are stronger than the equivalent conditions $\sigma(X),\, \sigma(X^{-1}) \subset\{z\ | \ \mathcal{Re}z > 0 \}$