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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.

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  • $\begingroup$ What is $\text{Diag}$ in this context? $\endgroup$ – Klaas van Aarsen Dec 19 '18 at 22:15
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    $\begingroup$ If "Diag" denotes the diagonal part of a matrix, then a solution exists if and only if $A$ is a symmetric matrix with a zero diagonal. In this case you may just take $X=B^{-1}(A+tI)B^{-1}$ for any $t>-\lambda_\min(A)$. $\endgroup$ – user1551 Dec 19 '18 at 22:35
  • $\begingroup$ @user1551 excuse me I made a mistake in the equation. It should read $A=2BXB−Diag(BXB)$ (instead of $A=BXB−Diag(BXB)$), this has been corrected. $\endgroup$ – user28958 Dec 20 '18 at 10:46
  • $\begingroup$ The user in the first comment asked you to clarify the meaning of "Diag". I think you should add a clarification in your question. $\endgroup$ – user1551 Dec 20 '18 at 10:49
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    $\begingroup$ @IlikeSerena "Diag(A)" is a diagonal matrix containing the diagonal of A. $\endgroup$ – user28958 Dec 20 '18 at 10:54
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Apologies, this has a trivial solution. On the diagonal, it is clear that $A_{ii}=(BXB)_{ii}$.

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

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  • $\begingroup$ Note, however, that the equation is not always solvable as $A+Diag(A)$ is not necessarily positive definite. $\endgroup$ – user1551 Dec 20 '18 at 11:31
  • $\begingroup$ @user1551 apparently we can assume that $X$ is positive definite, but it is not required. Without that requirement for the solution, it is always solvable. $\endgroup$ – Klaas van Aarsen Dec 20 '18 at 12:21
  • $\begingroup$ @user1551 Yes you are correct, but indeed all diagonal values of $A$ are positive, so $det(A+Diag(A)) \geq Det(A) + \Pi A_{ii} > 0$. Sorry, I didn't outline the question well enough. $\endgroup$ – user28958 Dec 20 '18 at 14:11
  • $\begingroup$ @IlikeSerena $X$ is always positive definite $Det(X)=Det(B)^{-2}Det(A+Diag(A)) \geq Det(B)^{-2}(Det(A)+ \Pi A_{ii}) > 0$ $\endgroup$ – user28958 Dec 20 '18 at 14:16
  • $\begingroup$ Consider $A=\begin{pmatrix}1&0&2\\0&1&0\\2&0&1\end{pmatrix}$. It has $\det A=-3$ and $\det(A+D)=0$. Both are certainly not positive definite. $\endgroup$ – Klaas van Aarsen Dec 20 '18 at 18:46
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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.

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