Timeline for Efficiently solve the Sylvester equation $AX+XA = C$ where $X$ is skew-symmetric
Current License: CC BY-SA 4.0
15 events
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| Jan 6 at 21:04 | vote | accept | Gabi | ||
| Jan 6 at 21:02 | comment | added | Gabi | @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? | |
| Jan 4 at 18:53 | answer | added | loup blanc | timeline score: 4 | |
| Jan 3 at 10:21 | comment | added | Bruno Le Floch | 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? | |
| Jan 3 at 9:29 | comment | added | Benji | 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 $B$. | |
| Jan 3 at 9:22 | history | edited | Rodrigo de Azevedo | CC BY-SA 4.0 |
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| Jan 3 at 9:17 | history | edited | Rodrigo de Azevedo |
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| Jan 3 at 9:09 | history | edited | Rodrigo de Azevedo | CC BY-SA 4.0 |
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| Jan 3 at 3:14 | comment | added | Nathaniel Johnston | It's also worth noting that the Bartels-Sewart algorithm is a quicker method to solve this problem than vectorization ($O(n^3)$ versus $O(n^6)$). | |
| Jan 3 at 3:09 | comment | added | Nathaniel Johnston | 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). | |
| Jan 2 at 18:55 | history | edited | Rodrigo de Azevedo | CC BY-SA 4.0 |
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| Jan 2 at 18:37 | history | edited | Rodrigo de Azevedo | CC BY-SA 4.0 |
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| Jan 2 at 14:26 | history | edited | Rodrigo de Azevedo | CC BY-SA 4.0 |
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| S Jan 2 at 14:10 | review | First questions | |||
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| S Jan 2 at 14:10 | history | asked | Gabi | CC BY-SA 4.0 |