Extreme Rays Of A Spectrahedron

Let $\mathcal{S}_+^n$ be the cone of $n \times n$ semidefinite matrices and let $L$ be a linear subspace of the vector space of symmetric matrices. Define a spectrahedron $C$ by $C = \mathcal{S}_+^n \cap L$. Then a vector $Q$ spans an extreme ray of $C$ exactly when the kernel of $Q$ is maximal among all the matrices on $L$: if $\ker Q \subseteq \ker T$ for $T \in L$, then $T = \lambda Q$ for some $\lambda \in \mathbb{R}$.


Suppose that $Q$ spans an extreme ray and take $T \in L$ with $\ker Q \subseteq \ker T$. For small enough $t \in \mathbb{R}_+$, $Q \pm tT$ must belong to $C$ (since $\ker Q \subseteq \ker T$). We have $Q = \frac{1}{2} ((Q + tT) + (Q - tT))$ and also $T = \frac{1}{2t}((Q + tT) - (Q - tT))$, which together imply $T = \lambda Q$ for some $\lambda \in \mathbb{R}$ since $Q$ is extreme. So $\ker Q$ is maximal in the above sense.

Conversely, suppose that $\ker Q$ is maximal and write $Q = T_1+T_2$ with $T_1,T_2 \in C$. Since $T_1,T_2,Q$ are all semidefinite, the above equation implies $\ker Q \subseteq \ker T_1$ and $\ker Q \subseteq \ker T_2$. Since $\ker Q$ is maximal, we have $T_1 = \lambda_1 Q$ and $T_2=\lambda_2 Q$, and therefore $Q$ spans an extreme ray.

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