Convexity of log det X??? In Boyd's book on convex optimization he proves convexity of log det X by proving it to be concave along a line i.e. he proves that the Hessian of the function $g(t) = f(Z+tV)$ is negative therefore this function is concave. He assumes that $Z \in S_{++}^n$ and $V \in S^n$ to ensure $Z+tV>0$. I understand that to ensure that the domain of this function is positive he assumes $Z \in S_{++}^n$ but why does he assume that $V \in S^n$. In the videos he says that V is a direction so it can either be positive or negative hence only the condition of symmetric is needed for V. I am not sure what that means.Can someone please explain?
I even tried an example of $Z = \bigl(\begin{smallmatrix} 1 & 1\\ 1 & 2 \end{smallmatrix}\bigr)$ and $V = \bigl(\begin{smallmatrix} 0 & 1\\ 1 & -1 \end{smallmatrix}\bigr)$ where $Z \in S_{++}^n$ and $V \in S^n$. Assume t = 1, we get $Z+tV = \bigl(\begin{smallmatrix} 1 & 2\\ 2 & 1 \end{smallmatrix}\bigr)$ which is not positive definite.
 A: Basically Boyd is taking the approach of showing what happens with $\log \det$ in a neighborhood of $t=0$. You're quite correct in that you can make $Z+tV$ into something that's not positive definite, but that isn't really the point here. In fact if you let $t=0.1$, then the sum is still p.d.
What this approach is doing is to show that if you have a line through any $Z$, then you can show that it's concave along that line. That doesn't mean that you have to follow that line out of the domain under scrutiny. It just gives you a way to show convexity locally.
As for $V$ being a "direction": this might be a little confusing until you realize that matrices can be vectors, and that $S^n$ is just a vector space. $Z+tV$ is just the expression of the fact that $Z$ is a vector, and by adding $tV$ (also a vector), you're moving in a line through $Z$. $V$ can be any symmetric matrix, because the important thing is that $Z$ is a member of the domain you're looking at ($S^n_{++}$, the domain of $\log \det$). 
So assuming $t$ is small enough, you can answer the question of whether $g$ is concave on a line through $Z$. The fact that $S^n_{++}$ is an open subset of $S^n$ lets us do that.
