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.
$$
\frac{ \partial K_{ij}}{\partial {K_{kl}}} = ...
4
votes
Accepted
Solving underdetermined Lyapunov equation?
Since $A$ is symmetric, it has an eigendecomposition $A = QDQ^*$ with $Q$ orthogonal. Then
$$
M = A \otimes I + I\otimes A = (Q\otimes Q)(D\otimes I + I \otimes D)(Q\otimes Q)^*
$$
is an ...
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-...
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}}
...
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