Commutativity condition forces diagonalization

$\textit{exercise}$ : Let $A,B,C\in{M_{n \times n}(\mathbb{R})}$ and suppose the characteristic polynomial of $A$ splits and separates in $\mathbb{C}$. Suppose also that $AB=BA$ and $AC=CA$. Prove that $BC=CB$

$\textit{proof}$: If the characteristic polynomial $p_A(t)$ splits and separates in $\mathbb{C}$ i.e. has no repeated roots in $\mathbb{C}$ then $A$ will diagonalize in $\mathbb{C}$. We first show that if $D$ is diagonal in $\mathbb{C}$ and $DB=BD$ then this implies that $B$ must also be diagonal. \\ Let $D\in{M_{n \times n}(\mathbb{C})}$ where $D$ is diagonal. Since $D$ separates in $\mathbb{C}$ then for the diagonal entries, $a_{ii} \neq a_{jj} \quad \forall i \neq j$. If $DB=BD$ then $B$ is diagonal since $$ [DB]_{ij}= \left\{ \begin{array}{lr} d_{ii}b_{ii} : i=j \\ d_{ii}b_{ij} : i \neq j \end{array} \right. $$ $$ [BD]_{ij}= \left\{ \begin{array}{lr} b_{ii}d_{ii} : i=j \\ b_{ij}d_{jj} : i \neq j \end{array} \right. $$ Since $DB=BD$, then $d_{ii}b_{ij}=b_{ij}d_{jj}$. But with each $d_{ii} \neq d_{jj}$ and $\mathbb{C}$ being a field (commutative, inverses exist, and also no zero divisors in case $d_{ii}=0$ for some $i$) this forces $b_{ij}=0$. Similarly, $b_{ji}=0$, so all off-diagonal entries of $B$ are zero i.e. $B$ is diagonal.

We have shown that if given a diagonal matrix with unique entries commuting with another matrix, then the other matrix must also be diagonal; regardless of the splitting field of the characteristic polynomial. This commutativity condition is once which can be employed in certain problems to create simultaneous diagonalization of linear operators. This technique is sometimes used in solving PDE's.

Now, given the conditions above, show that $BC=CB$.
$A$ diagonalizes in $\mathbb{C}$, so there exists an invertible matrix (change of basis matrix) $Q$ and a diagonal matrix $D$ s.t. $D=Q^{-1}AQ$. Since $AB=BA$ it follows that $$(Q^{-1}AQ)(Q^{-1}BQ)=Q^{-1}ABQ=Q^{-1}BAQ=(Q^{-1}BQ)(Q^{-1}AQ)$$ Thus, by the previous result, the matrix $Q^{-1}BQ$ is diagonal. Similarly, $Q^{-1}CQ$ must be diagonal. Using the fact that diagonal matrices commute with one another, we get, $$(Q^{-1}BQ)(Q^{-1}CQ)=(Q^{-1}CQ)(Q^{-1}BQ)$$ $$Q^{-1}BCQ=Q^{-1}CBQ \quad \therefore BC = CB$$