Prove convexity of function over space of positive definite matrices 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.
 A: Since $X$ is positive definite, it admits a symmetric, invertible square root $X^{1/2}$. Then
$$\begin{aligned}
f(X)&=-\log\det(X+tV)\\
    &=-\log\det X^{1/2}(I+tX^{-1/2}VX^{-1/2})X^{1/2}\\
    &=-\log(\det X^{1/2})^2\det(I+tX^{-1/2}VX^{-1/2}) \\
    &=-\log\det X-\log\det(I+t\tilde{V})
\end{aligned}$$
where $\tilde{V}\triangleq X^{-1/2}VX^{-1/2}$. Let $Q\Lambda Q^T$ be a Schur decomposition of $\tilde{V}$, with $\Lambda=\mathop{\textrm{diag}}(\lambda_1,\lambda_2,\dots,\lambda_n)$. 
$$\begin{aligned}
f(X)&=-\log\det X-\log\det(I+tQ\Lambda Q^T) \\
    &=-\log\det X-\log\det Q(I+t\Lambda)Q^T \\
    &=-\log\det X-(\log\det Q)^2-\log\det (I+t\Lambda) \\
    &=-\log\det X-0-\log\prod_{i=1}^n(1+t\lambda_i) \\
    &=-\log\det X-\sum\log(1+t\lambda_i)
\end{aligned}$$
It is not difficult to show that $g_\lambda(t)=-\log(1+t\lambda)$ is a convex function of $t$ for any fixed $\lambda$. (For instance, $g_\lambda''(t)>0$ will be strictly positive for $t\neq 0$, and identically zero if $t=0$). 
Thus we have expressed $f(X)$ as the sum of a constant and $n$ convex functions of $t$.
A: 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. $trace(((HX^{-1})^2)<0$.
