Is it true that if $\mbox{tr}({\bf A} {\bf X_1}) \geq \mbox{tr}({\bf A} {\bf X_2})$, then $\mbox{tr}({\bf X_1}) \geq \mbox{tr}({\bf X_2})$?

Question

I am not a student majoring in math, so I apologize if my question is stupid to you.

Assume we have ${\bf X_1}$, ${\bf X_2}$, and ${\bf A}$ that are all Hermitian and positive semidefinite matrices with rank $1$. Or, more precisely, $\bf A$, ${\bf X_1}$ and ${\bf X_2}$ are formed by

$${\bf A}={\bf aa}^H \qquad \qquad {\bf X_1}={\bf x}_1 {\bf x}_1^H \qquad \qquad {\bf X_2}={\bf x}_2 {\bf x}_2^H$$

Is it true that

1. if $\mbox{tr}({\bf A} {\bf X_1}) \geq \mbox{tr}({\bf A} {\bf X_2})$, then $\mbox{tr}({\bf X_1}) \geq \mbox{tr}({\bf X_2})$?

2. if $\mbox{tr}({\bf X_1}) \geq \mbox{tr}({\bf X_2})$, then $\mbox{tr}({\bf A} {\bf X_1}) \geq \mbox{tr}({\bf A} {\bf X_2})$?

I assume this is an extension of the linear case where we have y=a*x with a>0, and it is clearly true if we have x1>x2 then y1>y2, but I am not sure if this still holds for matrix case. I have tried to search for 'matrix function' and 'trace inequalities' but I still couldn`t get an answer.

Thanks in advance!

Answer 1

The first claim is false, consider for example $$A = \begin{pmatrix} 1 & 0 \\ 0 & 0\end{pmatrix} \qquad X_1 = \begin{pmatrix}0 & 0 \\ 0 & 0\end{pmatrix} \qquad X_2 = \begin{pmatrix}0 & 0 \\ 0 & 2\end{pmatrix}.$$

Answer 2

The answer to each of these questions is no. Take $$ X_1 = \pmatrix{20&0\\0&1}, \qquad X_2 = \pmatrix{10&0\\0&2}, \qquad A = \pmatrix{0&0\\0&1} \DeclareMathOperator{\tr}{tr} $$ It's clear that $\tr(X_1) \geq \tr(X_2)$, but $\tr(AX_1) < \tr(A X_2)$.

Likewise, take $$ X_1 = \pmatrix{10&0\\0&2}, \qquad X_2 = \pmatrix{20&0\\0&1}, \qquad A = \pmatrix{0&0\\0&1} $$ and find that the other assertion proves false.