I want to show that the function $f(X) = -log \ det(X)$ is convex on the space $S$ of positive definite matrices.
What I have done:
It seems like this problem could be tackled by considering the restriction of $f(X)$ to a line through a given point $X \in S$ so that $g(\alpha) = f(X + \alpha V)$ for some $V \in S$. So now I am trying to show that g is convex.
More generally, let $f(X)=-log(|\det(X)|)$, let $\Omega$ be a convex subspace of $GL_n(\mathbb{R})$ and, if $X\in\Omega$, let $T_X$ be the tangent space in $X$ to $\Omega$. If $f$ is convex over $\Omega$, then, for every $X\in\Omega,H\in T_X$, $f''_X(H,H)\geq 0$. The converse is true if $f''_X(H,H)>0$ when $H\not=0$.
$f'_X(H)=-trace(HX^{-1})$ and $f''_X(H,K)=trace(HX^{-1}KX^{-1})$; finally $f''_X(H,H)=trace((HX^{-1})^2)$.
$\Omega$ is the set of symmetric $>0$ matrices. Then $T_X$ is the space of symmetric matrices, $X^{-1}H$ is diagonalizable and $spectrum(X^{-1}H)\subset \mathbb{R}$; we easily conclude.
One has the same result when $\Omega$ is the set of symmetric $<0$ matrices.
Yet, if $X$ is invertible symmetric not $>0$ and not $<0$, then one can show that there is a symmetric matrix $H$ s.t. $trace(((HX^{-1})^2)<0$.