Trace of the product of a rank-one and an indefinite matrix, subject to semidefinite constraints

Question

Let $Q$ be a Hermitian (indefinite) matrix. Is it true that \begin{equation} \operatorname{tr}(QX)\geq0 \quad\text{and}\quad X-xx^T\succeq0 \quad\Rightarrow\quad \operatorname{tr}(QX) \overset{?}{\geq} x^TQx=\operatorname{tr}(Qxx^T) \overset{?}{\geq} 0 \end{equation} In my case $Q$ is usually not full-rank. In fact, the most interesting case is when it is of rank 2, with exactly one positive and one negative eigenvalue (independent of the problem size).

Answer 1

Both inequalities are not true in general.

Example: take $X=I$ (identity) and denote $\lambda_+>0$ and $\lambda_-<0$ the positive and the negative eigenvalues of $Q$, respectively. Assume $\lambda_++\lambda_-\ge 0$. Then $I-xx^T\succeq0$ $\Leftrightarrow$ $\|x\|\le 1$ and we have \begin{align} \text{tr}(QX)&=\text{tr}(Q)=\lambda_++\lambda_-\ge0,\\ \text{tr}(Qx^Tx)&=x^TQx. \end{align} Choosing $x_\pm$ as normalized eigenvectors for $Q$ with $\lambda_\pm$ gives

• $x_+^TQx_+=\lambda_+>\lambda_++\lambda_-$ which contradicts the first inequality.
• ${x_-}^TQx_-=\lambda_-<0$ which contradicts the second inequality.