Is it possible for matrix trace?

Question

Suppose we have two positive semidefinite matrcies $X_1$ and $X_2$ of size $N\times N$ with elements from complex numbers. Suppose $Tr(X_1)>Tr(X_2)$ then is it possible that for any non-zero vector $s$ (of size $N\times 1$ with complex elements) we have $$Tr(ss^HX_1)=Tr(ss^HX_2)$$ where $s^H$ is the hermitian transpose of $s$. Any help in this regard will be much appreciated. Thank you.

Answer 1

Let $$ X_1=\begin{pmatrix}1&0\\0&2\end{pmatrix},\,X_2=\begin{pmatrix}1&0\\0&1\end{pmatrix},\,s=\begin{pmatrix}1\\0\end{pmatrix}. $$ Then $\mathrm{Tr}(X_1)=3>2=\mathrm{Tr}(X_2)$ and $\mathrm{Tr}(ss^H X_1)=1=\mathrm{Tr}(ss^H X_2)$.

Of course, one should expect an example like this: By the cyclicity of the trace, we have $\mathrm{Tr}(ss^H X_j)=s^H X_j s$. So the condition $\mathrm{Tr}(ss^H X_1)=\mathrm{Tr}(ss^H X_2)$ only controls the behavior of $X_1$ and $X_2$ on the subspace generated by $s$. This leaves us enough room to get any relation between the traces of the matrices by modifying them appropriately on the orthogonal complement of $s$.