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.
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.