Suppose $M \geq 0$ is a symmetric, positive semi-definite matrix and $C \leq 0$ is a symmetric, negative semi-definite matrix, and suppose they are linked through the following equality:
$$
\mathrm{tr}(M C) = 0,
$$
that is, their product is traceless.
Can we say anything about the properties or structure of $M$ in this scenario?
For example, if $C < 0$ was strictly negative definite, then the only way the above equality can hold is if $M \equiv 0$.
However, can we make any simplifications in the original case?
Best Answer
Note that by replacing $C$ with $-C$ you might as well be asking about two PSD matrices $M, N$ so I will do this. Write $M = A^T A, N = B^T B$. Then
$$\text{tr}(MN) = \text{tr}(A^T A B^T B) =\text{tr}(BA^T B^TA) = \text{tr}(C^T C) = 0$$
where $C = B^T A$. Since $C^T C$ is also PSD, this implies $C^T C = 0$, so $C = 0$, so $MN = 0$.
This condition is equivalent to the condition $\text{Col}(N) \subseteq \text{Nul}(M) = \text{Col}(M)^{\perp}$, so equivalently the condition that the column spaces of $M$ and $N$ are orthogonal. This gives $\text{rank}(N) \le n - \text{rank}(M)$ where these are $n \times n$ matrices and the equality case can occur.