If $A B=B A$ where $A$ is a square matrix and $B$ a square positive definite matrix.

$B$ has a Cholesky decomposition $$B=L L^*$$

Does $A$ commute with $L$ and $L^*$?

Naturally if $A$ commutes with $L$ and $L^*$, then $A$ commutes with $B$, but it doesn't seem anywhere near as obvious in the other direction.


1 Answer 1


Unfortunately even $B$ itself does not commute with its own Cholesky decomposition. A concrete counterexample is $$B = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$$ which does not commute with $L = \begin{pmatrix} \sqrt{2} & \sqrt{1/2} \\ 0 & \sqrt{3/2} \end{pmatrix}.$ (I am relying on wolframalpha's computation of the Cholesky decomposition here.)


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