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5 votes

Confusion about matrix differentiation in a nonlinear matrix equation

If a matrix is differentiated with respect to itself, the result should be a fourth order tensor. The easist way to see this is to work with components. $$ \frac{ \partial K_{ij}}{\partial {K_{kl}}} = ...
NNN's user avatar
  • 760
4 votes
Accepted

Solving underdetermined Lyapunov equation?

Since $A$ is symmetric, it has an eigendecomposition $A = QDQ^*$ with $Q$ orthogonal. Then $$ M = A \otimes I + I\otimes A = (Q\otimes Q)(D\otimes I + I \otimes D)(Q\otimes Q)^* $$ is an ...
Federico Poloni's user avatar
2 votes

Solving for $X$ in $\sum_{a,b} b a^T b^T X a = Y$

Conjugate gradient (in a matrix-less implementation) might be a good idea to try. You have a $d^2\times d^2$ symmetric positive semidefinite matrix for which you can compute the action of the matrix-...
Federico Poloni's user avatar
1 vote

Confusion about matrix differentiation in a nonlinear matrix equation

$ \def\R#1{{\mathbb R}^{#1}} \def\o{{\large\tt1}} \def\D{{\cal D}} \def\k{\otimes} \def\h{\odot} \def\bR#1{\big(#1\big)} \def\BR#1{\Big(#1\Big)} \def\LR#1{\left(#1\right)} \def\op#1{\operatorname{#1}} ...
greg's user avatar
  • 604

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