If A is diagonal and A commutes with B, does A commute with the Cholesky decomposition of B? 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.
 A: 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.
