# Trace constraints and rank-one positive semi-definite matrices.

Let $$C_1, C_2, \dots, C_N$$ be $$M \times M$$ Hermitian matrices and $$c > 0$$ be a given constant. Let $$W$$ be a positive (possibly semi) definite matrix such that

\begin{align} \text{trace}\{WC_1\} & \geq c \\ \text{trace}\{WC_2\} & \geq c \\ \vdots & \\ \text{trace}\{WC_N\} & \geq c \\ \end{align}

**Does this guarantee that there exists a vector $$w$$ such that

\begin{align} w^{H}C_1w &\geq c \\ w^{H}C_2w &\geq c \\ \vdots & \\ w^{H}C_Nw &\geq c \\ \end{align}

NOTE: $$w^{H}C_iw=\operatorname{trace}\{ww^HC_i\}$$. Thus, if the given $$W$$ is a rank-one matrix, the solution is straightforward.

• Do you agree with my edits? Commented Feb 27, 2023 at 14:27

Here is a counterexample: $$C_1=\begin{pmatrix}2&3\\3&-1\end{pmatrix} \quad C_2=\begin{pmatrix}2&-3\\-3&-1\end{pmatrix} \quad C_3=\begin{pmatrix}2&-3i\\3i&-1\end{pmatrix}\quad C_4=\begin{pmatrix}2&3i\\-3i&-1\end{pmatrix} \\ C_5=\begin{pmatrix}-1&3\\3&2\end{pmatrix} \quad C_6=\begin{pmatrix}-1&-3\\-3&2\end{pmatrix} \quad C_7=\begin{pmatrix}-1&-3i\\3i&2\end{pmatrix}\quad C_8=\begin{pmatrix}-1&3i\\-3i&2\end{pmatrix} \\ \\ W=\begin{pmatrix}1&0\\0&1\end{pmatrix}$$ The assumptions hold with $c=1$. Suppose that $w=\begin{pmatrix}a \\ b \end{pmatrix}\in\mathbb C^2$ is a vector such that $$w^HC_kw>0\ \text{ for all }\ k=1,\dots,8\tag1$$ This will lead to a contradiction.
Using (1) with $k=1,\dots,4$ we find that
$$2|a|^2-|b|^2 > 6\max(|\operatorname{Re}(a\bar b)|, |\operatorname{Im}(a\bar b)|)\tag2$$ while using (1) with $k=5,\dots,8$ yields $$-|a|^2+2|b|^2 > 6\max(|\operatorname{Re}(a\bar b)|, |\operatorname{Im}(a\bar b)|)\tag3$$ Since $a$ and $b$ are interchangeable in (2)-(3), we assume $|a|\le |b|$ from now on.
For any complex number $z$ we have $\max (|\operatorname{Re}z|, |\operatorname{Im} z|)\ge |z|/\sqrt{2}$. Therefore, (2) implies $$2|a|^2-|b|^2 > \frac{6}{\sqrt{2}}|ab| \tag4$$ Recalling that $|a|\le |b|$ and using (4), we arrive at $$|a|^2\ge 2|a|^2-|b|^2 > \frac{6}{\sqrt{2}}|ab| \ge \frac{6}{\sqrt{2}}|a|^2$$ which is a contradiction.