Convexity of a trace of matrices with respect to diagonal elements

Question

Can we prove that $\mbox{trace}({\bf A} ({\bf P}+{\bf Q})^{-1} {\bf A}^T)$ is a jointly convex function of positive variables $[q_1,q_i,...,q_N]$, where ${\bf Q}=\mbox{diag}(q_1,...,q_N)$, $q_i>0 , \forall i$, and ${\bf P}$ is a positive definite matrix.

Answer 1

Notice that the inverse of a symmetric positive definite matrix is convex (cf Is inverse matrix convex?) w.r.t. to the cone of symmetric positive semidefinite matrices $S^n_+$.

Edited Plain version:

For $t\in (0,1)$ and symmetric positive definite $X,\tilde X$ it follows the matrix $$ (1-t) X^{-1} + t\tilde X^{-1} - ((1-t) X + t \tilde X)^{-1} $$ is positive semidefinite. In particular, $$ A[(1-t) X^{-1} + t\tilde X^{-1} - ((1-t) X + t \tilde X)^{-1}]A^T $$ is positive semidefinite. Thus, $$ \operatorname{trace}(A[(1-t) X^{-1} + t\tilde X^{-1} - ((1-t) X + t \tilde X)^{-1}]A^T) \ge 0 $$ and \begin{align} \operatorname{trace}(A((1-t) X + t \tilde X)^{-1}A^T) &\le \operatorname{trace}(A[(1-t) X^{-1} + t\tilde X^{-1}]A^T) \\ &= (1-t)\operatorname{trace}(AX^{-1}A^T) + t\operatorname{trace}(A\tilde X^{-1}A^T). \end{align} Thus, $X\mapsto \operatorname{trace}(AX^{-1}A^T)$ is convex. Since, $(P,Q)\mapsto P+Q$ is affine, $(P,Q)\mapsto \operatorname{trace}(A(P+Q)^{-1}A^T)$ is also convex.

Composition rule with generalized inequality and monotonicity version:

Notice that $\operatorname{trace}(AXA^T)$ is convex (in fact linear) and monotone increasing in $X$ w.r.t. $S^n_+$. Thus, the composition $(P,Q)\mapsto \operatorname{trace}(A(P+Q)^{-1}A^T)$ is convex.

Answer 2

The epigraph of this function is $$\mathop{\textrm{epi}} f(P,Q) = \left\{ (P,Q,z) \,\middle|\, P+Q\succ 0,~\mathop{\textrm{Tr}}(A(P+Q)^{-1}A^T)\leq z\right\}$$ This is equivalent to $$\mathop{\textrm{epi}} f(P,Q) = \left\{ (P,Q,z) \,\middle|\, \exists Z ~~ P+Q\succ 0,~\begin{bmatrix} Z & A^T \\ A & P+Q \end{bmatrix} \succeq 0, ~ \mathop{\textrm{Tr}}(Z) \leq z \right\}$$ The epigraph is the intersection of linear matrix inequalities and a linear inequality, composed with the projection $(P,Q,Z,z)\rightarrow (P,Q,z)$, so it is convex.