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I want to express an inequality of the form

$$\mbox{tr} (A^{-1}B)\leq t$$

as a matrix inequality, where $A$ is positive definite and $B$ is positive semidefinite. In particular, the matrix inequality should not invert $A$. Is this possible?

Similar to how: If $A$ is PD, the Schur complement of $A$ in $X$ is PSD if and only if $X$ is PSD.

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  • $\begingroup$ Solved this. You would need to take the eigendecomposition of B and then use the Schur complement. $\endgroup$ – user5016054 May 1 '17 at 3:38
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You can't directly write this as a matrix inequality, but you can constrain it to get a matrix inequality. First of all, note that $$ Tr(A^{-1}B) \leq t \iff Tr(A^{-1/2}BA^{-1/2}) \leq t $$ From there, $$ Tr(A^{-1/2}BA^{-1/2}) \leq t \Longleftarrow A^{-1/2}BA^{-1/2} \preceq \frac tn I \iff B \preceq \frac tn A $$ I don't think we can go in the other direction. That is, there is no $k$ independent of $A,B$ such that $$ Tr(A^{-1/2}BA^{-1/2}) \leq t \implies B \preceq kA $$

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