Let $E$ be the set of the $n\times n$ symmetrics and $U$ the subset of those are positive definite.

Question

Then $f(X) = X^2$ is a diffeomorphism of $U$ onto itself. Ok, first of all, I am having troubles to show that $f(U) = U.$ How can I show that the image of this function is the set $U$?

For the second part, I did: $Df(X) = 2X.$ Then, if $v \neq 0, v \in \mathbb{R}^n$, $v^tDf(X)v = 2v^tXv > 0$ then $Df(x)$ is an isomorphism and the claim follows from the inverse function theorem. This is right?

To finish, how do I show that $U$ is convex? I have no clues...