Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm trying to solve some equations with arbitrary matrices. My Problem is that I don't know what way to solve the equation should be taken.


  1. Solve for $X$. $X,A$ are arbitrary matrices. $XA+A^T = I $

  2. Solve for $X$. $X,A,B$ are arbitrary matrices and $C-2A^T$ is invertible. $X^TC = 2 A(X+B)$

I solved the equations for some values (which I randomly chose), but cannot get the grip by doing it for generally.

share|improve this question
Interestingly the first matrix equation seems to appear rather often, see math.stackexchange.com/q/131393/7266. –  Fabian Apr 16 '12 at 21:37
Thanks alot. Yeah the Question appears alot because of a Homework Assignment. But its only the first of five Problems and im more interested in the way of solving this then the solutions itself :-) @Fabian I updated my Question –  GnrlKnowledge Apr 17 '12 at 7:59
so if it is a homework assignment, it would be good/fair to include the homework tag. You can find more information here (meta.math.stackexchange.com/q/1803/7266). –  Fabian Apr 17 '12 at 15:14
darn, thanks, i didnt know that a homework tag exists :-) –  GnrlKnowledge Apr 17 '12 at 20:34

1 Answer 1

up vote 1 down vote accepted

Since 1.) is already solved here, I will answer 2.):

$$ \begin{eqnarray} X^T C-2AX &=& B &(1)\\ C^T X - 2X^T A^T &=& B^T &(2)\\ \end{eqnarray} $$ Now add $(1)$ and $(2)$: $$ \begin{eqnarray} X^T\underbrace{\left(C-2A^T\right)}_{D}+\underbrace{\left(C^T-2A\right)}_{D^T}X&=&B+B^T \tag{1+2}.\\ \end{eqnarray} $$ Now let's multiply by $D^{-1}$ from the right: $$ X^T + D^T X D^{-1} = \left(B+B^T\right)D^{-1} $$ and use the following (to me known as superoperator formalism) representation of the problem: $$ \text{vec}(AXB) = (B^T \otimes A) \text{vec}(X). $$ (see here for a definition of $\text{vec}(X)$...). We get: $$ \hat{T}\text{vec}(X) + \left((D^{-1})^T\otimes D \right) \text{vec}(X)=\text{vec}\left((B+B^T)D^{-1}\right), $$ where $\hat{T}$ is the superoperator representation of the transposition operation (essentially a permutation matrix, that is not representable as product $A\otimes B$).

We finally get: $$ \text{vec}(X)= \left[\hat{T} + \left((D^{-1})^T\otimes D \right) \right]^{-1}\text{vec}\left((B+B^T)D^{-1}\right) $$

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.