Convexity of $\operatorname{tr} (X^T P X)$

Question

Suppose $P\in S^n$ and $X\in \mathbb{R}^{m\times n}$ (we know that $n>m$). Is $\operatorname{tr}(X^T PX)$ a convex function of $X$?

Here, $\operatorname{tr}(\cdot)$ denotes the trace operator for square matrices.

Answer 1

No, it clearly isn't. Let $n=1$, $P=-1$, and you get $\mbox{tr}(X^{T}PX)=-x^{2}$.

However, If $P \in S^{n}_{+}$ ($P$ is positive semidefinite), then it's easy to show that the function is convex.

To show this, let $S$ be the symmetric matrix square root of $P$.

$ P = S^{2}=S^{T}S$

Then

$f(X) = \mbox{tr}(X^{T}PX)= \mbox{tr}(X^{T}S^{T}SX)$

$f(X) = \mbox{tr}((SX)^{T}(SX))$

$f(X)= \| SX \|_{F}^{2}$

Since $\| \cdot \|_{F}^{2}$ is convex and $SX$ is affine, $f$ is convex.