# What is the apex angle of the cone of positive semidefinite matrices?

Let $\def\S{\mathbf S}\S^n$ be the linear space of symmetric $n \times n$ matrices and $\S_+^n$ be the subset of positive semidefinite matrices. It is well-known that $\S_+^n$ is a convex cone in $\S^n$. In order to get a better geometric understanding of this object, I asked myself what the apex angle of this cone might be.

We use the inner product $\DeclareMathOperator{\tr}{tr}\def\<{\langle}\def\>{\rangle}\<A,B\>=\tr(AB)$, where $\tr(A)$ is the trace of $A$.

The apex angle $\theta$ of $\S_+^n$ is the biggest value of $\arccos\<A_1,A_2\>$ for $A_i\in\S_+^n$ with $\<A_i,A_i\>=1$.

My best result so far

Let $\def\E{\mathbf E}\E$ be some Euclidean space and $S\subset \E$ a proper subspace. Let $A_1\in\S_+^n$ be the orthogonal projection onto $S$ and $A_2\in\S_+^n$ the orthogonal projection onto the orthogonal complement $S^\bot$. Then $\<A_i,A_i\>=1$ but $A_1A_2=0$, hence $\<A_1,A_2\>=\tr(A_1A_2)=0$.

So we have that $\theta\ge 90^\circ$. Can we do better? Especially, can we have $\tr(A_1A_2)<0$?

It is impossible to have $\operatorname{tr}(A_1A_2) < 0$ if each $A_i$ is positive semidefinite.
A proof I like is as follows: note that (via the spectral theorem, for instance) we may decompose $A_2$ into $$A_2 = \sum_{k=1}^n x_kx_k^T, \qquad x_k \in \Bbb R^n$$ With that, we find that $$\operatorname{tr}(A_1A_2) = \operatorname{tr}\left(A_1\sum_{k=1}^n x_kx_k^T\right) = \sum_{k=1}^n \operatorname{tr}(A_1x_kx_k^T) = \sum_{k=1}^n \operatorname{tr}(x_k^TA_1x_k) = \\ \sum_{k=1}^n x_k^TA_1x_k \geq 0$$
• Another notable result is that the cone $S_+^n$ is self-dual – Omnomnomnom Jan 4 '18 at 16:06
• No problem. Off the top of my head, I believe that all self-dual (closed, convex) cones have a $90^\circ$ apex angle, but that's just a hunch. – Omnomnomnom Jan 4 '18 at 16:09