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If $X\le Y$, that is, $Y-X$ is positive semi-definite, does it imply that $Tr\{XZ\}\le Tr\{YZ\}$?

If it's true, how should one go about proving it? Is it a result from some other theorem?

*edit: $Z$ is also positive, all matrices are Hermitian.

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    @sivaram-ambikasaran Thanks for pointing that out, $Z$ is also positive, updated. And are there properties that $Z$ should have to make it hold?2011-02-12

2 Answers 2

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Yes.

Hints:

  1. You can write $Y - X = S^2$, $Z = T^2$ with $S,T \geq 0$.
  2. Recall $\text{Tr}(AB) = \text{Tr}(BA)$ and $\text{Tr}(C^{\ast}C) \geq 0$.
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If the matrices are not hermitian, here is an example to prove that the statement is false.

The statement is equivalent to is $\text{Tr}((Y-X)Z) \geq 0$

For instance, take

$Y-X = \left(\begin{array}{c c c c} 1 & -1 & -1 & -1 \\ 0 & 1 & -1 & -1 \\ 0 & 0 & 1 & -1 \\ 0 & 0 & 0 & 1 \end{array} \right)$

$Z = \left(\begin{array}{c c c c} 1 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ 1 & 1 & 1 & 0 \\ 1 & 1 & 1 & 1 \\ \end{array} \right)$

Clearly, $Y>X$ and $Z>0$, however the product is

$(Y-X)Z = \left(\begin{array}{c c c c} -2 & -1 & 0 & 1 \\ -3 & -1 & 0 & 1 \\ -2 & -2 & 0 & 1 \\ -1 & -1 & -1 & 1 \\ \end{array} \right)$

and the trace of the above is negative and hence $\text{Tr}(XZ) \leq \text{Tr}(YZ)$ is false.

However, the fact that the matrices are Hermitian might turn out to be helpful in proving that the statement is true. My hunch is that the statement is true but I do not have a proof now.