Given that the matrix $A$ is symmetric positive definite, and governed by the following set of ODEs for some matrix $B$:
$$\frac{dA}{dt}+AB+B^TA=0$$
is is possible to derive an evolution equation for the Cholesky factor $L$ where $A=LL^T$?
Evolution of Cholesky factor given evolution of symmetric positive definite matrix
2
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matrices
ordinary-differential-equations
derivatives
matrix-equations
matrix-decomposition
1 Answers
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$$\frac{dA}{dt} = \frac{dLL^\top}{dt} = L\frac{dL^\top}{dt} + \frac{dL}{dt}L^\top.$$
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0What is the connection with the differential equation given by the OP ? – 2017-02-20
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0This won't do the trick. It only tells you how the symmetric part of $\frac{dL}{dt}L^T$ evolves. It doesn't tell you everything about $\frac{dL}{dt}$. – 2017-02-20
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1I initially thought maybe this could be used to separate the ODEs into $L\frac{dL^T}{dt}+LL^TB=0$ and $\frac{dL}{dt}L^T+B^TLL^T=0$. These are essentially the same equation, $\frac{dL}{dt}+B^TL=0$. This would mean that the upper-triangle elements of $L$ turn non-zero over time, however. – 2017-02-20