How to solve A=2BXB-Diag(BXB)

Question

Given $X,B,A \in \mathbb{R}^{p \times p}$ how do you solve

$A=2BXB - \text{Diag}(BXB)$

for $X$? Is there a closed-form? One can assume the following: $A,B$ and $X$ are symmetric positive definite.

I can iteravitly solve the problem as follows : $B^{-1}(A+\text{Diag}(BXB))B^{-1}/2=X_{+}$

Note: $Diag(BXB)$ is a diagonal matrix containing the diagonal elements of BXB.

Answer 1

Apologies, this has a trivial solution. On the diagonal, it is clear that $A_{ii}=(BXB)_{ii}$.

Thus $X=B^{-1}$(A + Diag(A))$B^{-1}/2$.

Answer 2

It may be easier for Readers to follow the solution if we decompose the problem into two phases.

Let $Y = BXB$, where $B$ is symmetric positive definite (and thus invertible). So if $Y$ is determined, then also will $X=B^{-1}YB^{-1}$ be determined. Furthermore $Y$ is symmetric if and only if $X$ is symmetric, and indeed $Y$ is symmetric positive definite if and only if $X$ is symmetric positive definite.

It remains to solve the simple problem:

$$ A = 2Y - \operatorname{Diag} (Y) $$

The entries of $Y$ are easily deduced. The diagonal entries of $Y$ are the same as the diagonal entries of $A$, e.g. by taking the diagonal entries on both sides of the above matrix equation. On the other hand the off-diagonal entries of $Y$ are half the corrresponding off-diagonal entries of $A$. Thus the solution:

$$ Y = \frac{1}{2} (A + \operatorname{Diag} (A)) $$

If $A$ is symmetric positive definite, so too is $A + \operatorname{Diag} (A)$. Hence $Y$ will be symmetric positive definite if $A$ is, but there are slightly weaker conditions that would allow $A$ to be not-quite diagonally dominant and yet give $A + \operatorname{Diag} (A)$ positive definite.

In any case $X$ is symmetric positive definite if and only if $A + \operatorname{Diag} (A)$ is symmetric positive definite.