Let $A$ be a given matrix and $X$ a symmetric positive definite matrix which shall be optimized:
min $tr (AX)$ subject to $X>0$
This optimization problem is convex if $A$ is symmetric as it is the standard form of Semidefinite Programming. But is the problem convex for non-symmetric $A$ as well?
Yes, the program is convex for any $A$.
Let $X \in S^n$ (the space of $n\times n$ dimensional symmetric matrices). Then,
\begin{align*} \textrm{trace}(AX) &= \textrm{trace}\left(A\left(\sum_{i=1}^n\nu_i x_i x_i^T\right)\right)\\ &= \textrm{trace}\left(\sum_{i=1}^n\nu_i A x_i x_i^T\right)\\ &= \sum_{i=1}^n\nu_i \textrm{trace}\left( A x_i x_i^T\right)\\ &= \sum_{i=1}^n\nu_i \textrm{trace}\left(x_i^T A x_i \right)\\ &= \sum_{i=1}^n\nu_i \left(x_i^T A x_i \right)\\ \end{align*} We will use the following identity: For any square matrix $A$ $$ x^TAx = \frac{1}{2}x^T(A+ A^T)x$$ Proof: \begin{align*} \frac{1}{2}x^T(A+ A^T)x &= \frac{1}{2}x^TAx + \frac{1}{2}x^TA^Tx\\ &= \frac{1}{2}x^T(Ax) + \frac{1}{2}(Ax)^Tx\\ &= \frac{1}{2}x^T(Ax) + \frac{1}{2}x^T(Ax)\\ &= x^TAx \end{align*} Thus, we get \begin{align*} \textrm{trace}(AX) &= \sum_{i=1}^n\nu_i \left(x_i^T \frac{1}{2}(A+ A^T)x_i \right)\\ &= \frac{1}{2}\textrm{trace}\left(\sum_{i=1}^n\nu_i x_i^T(A+ A^T)x_i \right)\\ & = \frac{1}{2}\textrm{trace}\left((A+ A^T)\sum_{i=1}^n\nu_i x_ix_i^T \right)\\ & = \frac{1}{2}\textrm{trace}\left((A+ A^T)X\right) \end{align*} The term $\textrm{trace}\left((A+ A^T)X\right)$ is in standard form as desired.
