Under what conditions the following constraint is convex?

Question

We know that if a matrix $\mathbb{A}$ is positive definite then $tr~\mathbb{A}$ is a convex function. But how can we show that the following constraint results in a convex set $$-tr~\mathbb{AQ}\leq 0$$ where $\mathbb{Q}$ is some matrix with constant real values.

Answer 1

Like @Brian Borchers hinted at, it is a composition of two linear operations (multiply with a constant matrix, and trace), so the whole thing $Tr(AQ)$ is linear in $A$ given $Q$.

If you want to show it the long way, assume there are $A_1,A_2$ that satisfy the constraint, i.e. $-Tr(A_1 Q) \leq 0$,$-Tr(A_2 Q) \leq 0$. For any $\theta \in [0,1]$, let $A_3 = \theta A_1 + (1-\theta) A_2$ be the convex combination of the two matrices, then we have

\begin{equation} \begin{split} -Tr(A_3 Q) &= - Tr ((\theta A_1 + (1-\theta)A_2) Q) \\ &= -Tr(\theta A_1Q + (1-\theta) A_2 Q) \\ &= - \theta \cdot Tr(A_1Q) - (1-\theta) \cdot Tr(A_2Q) \\ &\leq 0 \end{split} \end{equation} where the last equality comes from the linear nature of trace operator and the inequality is from the simple fact that sum of two $\leq 0$ entities is still $\leq 0$. Therefore $A_3$ is still in the same set for arbitrary $\theta \in [0,1]$, hence the feasible set is convex.