# How to express this inversion matrix in CVX form

Hi, everyone,

How to write {\rm{tr}}\left( {{\bf{R}}{{\left( {{\bf{XR}} + {\bf{I}}} \right)}^{ - 1}}} \right) in CVX form? where X is a semi-definite matrix that needs to be optimized, R and I are constant matrices.

Do you have a proof that it is convex? The argument of trace it not necessarily symmetric, let alone PSD. Do R and I have special properties you haven’t told us about?

\bf I is an identity matrix and \bf is a semi-definite matrix, so that we can rewrite {\rm{tr}}\left( {{\bf{R}}{{\left( {{\bf{XR}} + {\bf{I}}} \right)}^{ - 1}}} \right) as {\rm{tr}}\left( {{\bf{R}}{{\left( {{\bf{XR}} + {\bf{I}}} \right)}^{ - 1}}} \right) = {\rm{tr}}\left( {{{\bf{R}}^{1/2}}{{\left( {{\bf{XR}} + {\bf{I}}} \right)}^{ - 1}}{{\bf{R}}^{1/2}}} \right).

X*R is not necessarily symmetric. So what does your use of cyclic permutation invariance of trace prove? You use of R^{1/2} seems to imply you are assuming R is symmetric PSD, which you did not state.

If X*R were symmetric PSD, you could use the approach in Generalizing "trace_inv" for matrix quadratic forms .

Sorry, Mark. In fact, \bf R and \bf X are both positive semi-definite matrices. Thank you for your suggestions.

It is still the case that X*R need not be symmetric. So my solution from the previously linked thread does not apply.

X=[2 1;1 1];R=[1 1;1 2];
disp(X*R)
3     4
2     3

Yes, I fully agree with you. The suggestions inspire me a lot. Thanks.