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I would like to solve a matrix equation of the form $$ \mathbf{A} \mathbf{X} + \mathbf{X} \mathbf{A}^T = \mathbf{B} $$

where $\mathbf{A}$ and $\mathbf{B}$ are known $n \times n$ matrices, and $\mathbf{X}$ is an unknown $n \times n$ matrix.

  1. Is there a general way to isolate $\mathbf X$ in this expression?
  2. Is there a solution for various special cases, such as (1) the case where $\mathbf{A}$ and $\mathbf{B}$ are real, or (2) when $\mathbf{B}$ is real and diagonal?

My intuition is that the case where $\mathbf{A}$ and $\mathbf{B}$ are both real may be solved using the SVD of $\mathbf{A}$. Is there at least a way to compute an approximate solution for X using the pseudoinverse?

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    $\begingroup$ Bartels Stewart algorithm. $\endgroup$ – copper.hat May 26 at 0:44
  • $\begingroup$ incredible, thank you so much---I had not heard of the Sylvester equation but that looks like it's exactly it $\endgroup$ – wil3 May 26 at 0:56
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    $\begingroup$ As far as isolating $X$ goes, we can write $$ (I \otimes A + A \otimes I)\operatorname{vec}(X) = \operatorname{vec}(B), $$ where $I$ denotes an $n \times n$ matrix, $\otimes$ denotes a Kronecker product, and vec refers to the vectorization operator. $\endgroup$ – Omnomnomnom May 26 at 2:09
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The Bartel Stewart algo. works only when the function $f:X\mapsto AX+XA^T$ is one to one.

Let $spectrum(A)=(\lambda_i)_i$. Since $spectrum(f)=\{\lambda_i+\lambda_j;i,j\}$, B.S. works iff for every $i,j$, $\lambda_i+\lambda_j\not= 0$. Note that if the previous condition is not fufilled, then the equation may have no solutions; for example, take

$A=diag(1,-1),B=\begin{pmatrix}0&0\\1&0\end{pmatrix}$.

More precisely, for this value of $A$, the equation $AX+XA^T=B$ has no solutions except for rare values ​​of $B$.

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