# Closed form solution of an SDP [closed]

Given symmetric positive definite matrices $$A$$, $$M_1$$ and $$M_2$$, is there any closed form solution for the following convex problem in $$X$$?

$$\begin{array}{ll} \text{maximize} & \mbox{tr}(AX)\\ \text{subject to} & \begin{bmatrix} M_1 & X\\ X^\top & M_2\end{bmatrix} \succeq 0\end{array}$$

## closed as off-topic by dantopa, Saad, Thomas Shelby, José Carlos Santos, Eevee TrainerApr 19 at 0:31

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Let $$Y=M_1^{-1/2} X M_2^{-1/2}$$, so $$X=M_1^{1/2} Y M_2^{1/2}$$. Then the objective can be written $$\mathop{\textrm{Tr}}(AX) = \mathop{\textrm{Tr}}(AM_1^{1/2} Y M_2^{1/2}) = \mathop{\textrm{Tr}}(M_2^{1/2}AM_1^{1/2} Y) = \mathop{\textrm{Tr}}(BY)$$ where $$B=M_2^{1/2}AM_1^{1/2}$$. The constraint becomes $$\begin{bmatrix} M_1 & M_1^{1/2} Y M_2^{1/2} \\ M_2^{1/2} Y^T M_1^{1/2} & M_2 \end{bmatrix} = \begin{bmatrix} M_1^{1/2} & 0 \\ 0 & M_2^{1/2} \end{bmatrix} \begin{bmatrix} I & Y \\ Y^T & I \end{bmatrix} \begin{bmatrix} M_1^{1/2} & 0 \\ 0 & M_2^{1/2} \end{bmatrix} \succeq 0$$ allowing us to drop the left and right terms to reduce this to $$\begin{bmatrix} I & Y \\ Y^T & I \end{bmatrix} \succeq 0 \quad\Longleftrightarrow\quad \|Y\|_2\leq 1$$ So the problem reduces to $$\begin{array}{ll} \text{maximize} & \mathop{\textrm{Tr}}(BY) \\ \text{subject to} & \|Y\|_2\leq 1 \end{array}$$ But this is just the definition of the dual norm for $$\|\cdot\|_2$$, applied to $$B$$ (technically, to $$B^T$$). So the optimal value is $$\sum_{i=1}^n \sigma_i(B) = \sum_{i=1}^n \sigma_i(M_2^{1/2}AM_1^{1/2})$$ Now we can read off the answer by inspection. If $$B=U\Sigma V^T$$ is the SVD of $$B$$, then $$Y=VU^T$$. To verify: $$\mathop{\textrm{Tr}}(BY)=\mathop{\textrm{Tr}}(U\Sigma V^TVU^T) =\mathop{\textrm{Tr}}(U^TU\Sigma V^TV)=\mathop{\textrm{Tr}}(\Sigma)=\sum_i^n \sigma_i(B)$$ Given $$Y$$, then, $$X=M_1^{1/2} UV^T M_2^{1/2}.$$