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The following optimization is convex: \begin{align} \max_{X\succ0,Y} &\ \ \log \det (I + X +Y + Y^T)\\ & \text{s.t.} \begin{pmatrix} X& Y \\ Y^T & Z \end{pmatrix}\succeq 0, \ \ \ \mathbf{Tr}(X)\le P \end{align} where $Z\succ0$ and $P>0$ are given.

My question is whether the decision variable $Y$ can be transformed or replaced with a be positive semi-definite decision variable. My motivation is to write the optimization as a standard max-det problem or in a nicer form in order to show that the maximizer is unique.

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  • $\begingroup$ Perhaps it was not clear from my formulation, but $Y$ is not necessarily positive $\endgroup$ – Morad Nov 25 '20 at 3:49
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    $\begingroup$ I see. If $I+X+Y+Y^T$ is not positive definite over the entire feasible region, the objective is not strictly concave (and not defined everywhere). $\endgroup$ – LinAlg Nov 25 '20 at 4:03
  • $\begingroup$ I can show/assume that it is positive definite in the domain, but is it sufficient for a unique maximizer? My idea was to convert it first to a standard max-det problem by changing the variable $Y$ to an LMI or linear equality. $\endgroup$ – Morad Nov 25 '20 at 21:14
  • $\begingroup$ The log-det function is strictly concave on the set of positive definite matrices, but $f(X,Y) = I+X+Y+Y^T$ is not, as it is constant along the ray $X+Y+Y^T=0$ (nevertheless the objective is concave and the maximal value is unique). $\endgroup$ – LinAlg Nov 25 '20 at 22:06

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