Minimize $\mbox{trace}(AX)$ over $X$ with a positive semidefinite $X$ I want to minimize $\mbox{trace}(AX)$ over $X$, under the constraint that $X$ is positive semidefinite. I guess the solution should be bounded only for a positive semidefinite $A$, and it's zero, or the solution should be minus infinity. If this is correct, can anyone tell me why? or if it is wrong, please tell me the correct solution. Thank you very much!!

Let $A\in M_n(\mathbb{R})$; there are a symmetric $S$ and a skew-symmetric $K$ s.t. $A=S+K$; thus $tr(AX)=tr(SX)+tr(KX)=tr(SX)$ (since $X$ is symmetric, $tr(KX)=0$). We may assume that $S=diag(\lambda_1,\cdots,\lambda_p,\mu_1,\cdots,\mu_q,0_r)$ where $\lambda_i>0,\mu_j<0,p+q+r=n$. 
If $q>0$, then take $X=diag(0_p,xI_q,0_r)$ with $x>0$. Therefore $\inf_X(tr(AX))=-\infty$.
If $q=0$, then $S\geq 0$ and the eigenvalues of $SX$ are $\geq 0$; therefore $\inf_X(tr(AX))=0$.
