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Is there a way (more efficient than the standard vectorization) to solve the following Sylvester equation in the skew-symmetric matrix $X$ $$AX+XA = C$$ where the matrix $A$ is symmetric positive semidefinite, and the matrix $C$ is skew-symmetric? Does this fact about $X$ follow from the statement?

Background: the matrices $C$ and $X$ are really bivectors, but I'm not sure if going the way of geometric algebra is helpful here.

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    $\begingroup$ If $A$ is invertible then the answer to "Does this fact about $X$ follow from the statement?" is "yes". To see this, note that the linear map $\Phi_A$ defined by $\Phi_A(X) = AX + XA$ is invertible if and only if $A$ is invertible. Doing some standard transposey stuff shows that $\Phi_A(X) = C$ implies $\Phi_A(X^T) = -C = -\Phi_A(X)$. Applying $\Phi_A^{-1}$ to both sides shows that $X^T = -X$. On the other hand, if $A$ is not invertible then this fact about $X$ clearly does not follow (e.g., if $A = C = 0$ then $X$ could be anything). $\endgroup$
    – Nathaniel Johnston
    Commented Jan 3 at 3:09
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    $\begingroup$ It's also worth noting that the Bartels-Sewart algorithm is a quicker method to solve this problem than vectorization ($O(n^3)$ versus $O(n^6)$). $\endgroup$
    – Nathaniel Johnston
    Commented Jan 3 at 3:14
  • $\begingroup$ If you are after a symbolic representation, then you can write $X= - \int_{0}^{\infty} \exp(-s A ) C \exp(-s A ) ds$, where $\exp(B)$ denotes the matrix exponential of $B$. $\endgroup$
    – Benji
    Commented Jan 3 at 9:29
  • $\begingroup$ Are $C$ and $X$ simple bivectors (i.e. have minimum rank, which is $2$ for an antisymmetric matrix), or general ones, in which case seeing them as bivectors does not restrict the problem further? $\endgroup$
    – Bruno Le Floch
    Commented Jan 3 at 10:21
  • $\begingroup$ @BrunoLeFloch I don't think I'm qualified enough to respond -- they represent angular velocity / momentum in physics in N-dimensions. Does that answer your question? $\endgroup$
    – Gabi
    Commented Jan 6 at 21:02

1 Answer 1

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Let $spectrum(A)=(\lambda_i)$, $\phi:X\mapsto AX+XA$ and $K$ be the condition number of $A$ for $||.||_2$.

We assume that $A$ is positive definite.

Then $spectrum(\phi)=(\lambda_i+\lambda_j)_{i,j}\subset (0,\infty)$ and $\phi$ is a bijection. It is not difficult to see that if $C$ is skew, then the unique solution $X$ too.

Now the condition number of $\phi$ is also $K$. Even if $K$ is great, the Schur decomposition -that gives here the spectral decomposition of $A$ ($P^TAP=D=diag(\lambda_i)$)- is stable and its complexity is in $O(n^3)$.

Putting $P^TXP=Y,P^TCP=F=[f_{i,j}]$, the considered equation becomes $DY+YD=F$, that is, for every $i,j$: $y_{i,j}=\dfrac{f_{i,j}}{\lambda_i+\lambda_j}$ (complexity in $o(n^3)$).

Finally $X=PYP^T$ (complexity in $O(n^3)$).

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