This is a follow-up question to this question.

Let $A$ be a diagonal matrix, and $B$ a symmetric positive-definite matrix. Suppose $A$ commutes with $B$. $B$ has a Cholesky decomposition $B=L L^*$. Does $A$ commute with $L$?

There are no $2\times 2$ counterexamples:

$$B=\left[\begin{matrix}b_1 & b_2 \\ b_2 & b_3 \end{matrix}\right]$$ $$A=\left[\begin{matrix}a_1 & 0 \\ 0 & a_2 \end{matrix}\right]$$ $A B=B A$ if either $a_1=a_2$, in which case $A$ is scalar and commutes with everything including $L$, or $b_2=0$, in which case $L$ is diagonal and commutes with $A$.

I have used computer algebra software to exhaustively try all possibilities, and found that there are no $3\times 3$, $4\times 4$ or $5\times 5$ counterexamples either.


1 Answer 1


Yes. Use induction. The case is obvious for 1x1 matrices. On the other hand, for (n+1)x(n+1) matrices, we can express A and L in block form as

$$ A = \begin{bmatrix} a && \mathbf 0^*\\ \mathbf 0 && A_- \end{bmatrix} $$

$$ L = \begin{bmatrix} x && \mathbf 0^*\\ \mathbf y && L_- \end{bmatrix} $$

where $A_-$ is n-by-n diagonal, x is nonzero and $L_-$ is n-by-n lower triangular with positive diagonal entries. Hence $L_-L_-^*$ is the Cholesky decomposition of an n-by-n symmetric positive definite matrix. Our induction hypothesis then tells us that if $A_-$ commutes with $L_- L_-^*$, then $A_-$ commutes with $L_-$. Now, given A commutes with $B = LL^*$, compute:

$$ ALL^* = \begin{bmatrix} ax^2 && ax\mathbf y^*\\ xA_-\mathbf y && A_-\mathbf y \mathbf y^* + A_-L_-L_-^* \end{bmatrix} $$

$$ LL^*A = \begin{bmatrix} ax^2 && x\mathbf y^* A_-\\ xa\mathbf y && \mathbf y \mathbf y^* A_- + L_-L_-^*A_- \end{bmatrix} $$

Setting these two matrices to be equal, then since x is nonzero, we find from the lower-left and upper-right that $A_-\mathbf y = a\mathbf y$ and $\mathbf y^*A_- = a \mathbf y^*$. So since the lower-right entries are equal, this implies that $A_-L_-L_-^* = L_-L_-^*A_-$. Hence, by the induction hypothesis, $A_-L_- = L_-A_-$. Finally,

$$ AL = \begin{bmatrix} ax && \mathbf 0^*\\ A_-\mathbf y && A_- L_- \end{bmatrix} $$

$$ LA = \begin{bmatrix} ax && \mathbf 0^*\\ a\mathbf y && L_- A_- \end{bmatrix} $$

Since we already found that $A_-\mathbf y = a\mathbf y$ and $A_-L_- = L_-A_-$, these two matrices are equal.


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