I am struggling to solve an optimization problem of the following form:
$$\begin{array}{ll} \underset{A}{\text{maximize}} & \log \det (A) \\ \text{subject to} & a^T A^{-1} a \le b\end{array}$$
Is there any solution for it?
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$\begingroup$ Does $b$ have negative entries? $\endgroup$– fedjaAug 25, 2020 at 13:46
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$\begingroup$ @fedja Isn't $b$ a scalar? $\endgroup$– Rodrigo de AzevedoAug 25, 2020 at 13:50
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$\begingroup$ @RodrigodeAzevedo Ah, yes. Is it negative then? $\endgroup$– fedjaAug 25, 2020 at 13:50
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$\begingroup$ @fedja Since $A \succ 0$ (I assume), for $b < 0$ the feasible region is empty. $\endgroup$– Rodrigo de AzevedoAug 25, 2020 at 13:52
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$\begingroup$ @RodrigodeAzevedo And for $b>0$ you can multiply $A$ by a huge number and drive the determinant up without any bound. Actually, it is strange either way: WLOG $a=(1,0,\dots,0)$. Then you can take $A=diag(1/b,M,\dots,M)$ if $b>0$ and $diag(1/b,-M,M,\dots,M)$ if $b<0$. So, unless we are in dimension $1$, the answer is always $+\infty$. $\endgroup$– fedjaAug 25, 2020 at 13:56
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1 Answer
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I will presume you want $A$ to be constrained to be symmetric (hermitian) psd. In that case, this is a convex optimization problem which is a Linear Semidefinite Programming problem (SDP) a.k.a. Linear Matrix Inequality (LMI).
A convex optimization modeling tool, such as CVX, can formulate this as a standard Linear SDP and call a solver to solve it.
Here is the code for CVX.
cvx_begin
variable A(n,n) hermitian semidefinite
maximize(log_det(A))
subject to
matrix_frac(a,A) <= b
cvx_end
answered Aug 25, 2020 at 7:00
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$\begingroup$ If A can be complex, you can change the variable statment to "variable A(n,n) hermitian semidefinite, as I have now incorporated in the answer. $\endgroup$ Aug 25, 2020 at 12:12
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$\begingroup$ Sorry I am new here. How can I do it? Is it enough to click on that checkmark? $\endgroup$ Aug 26, 2020 at 11:18