[Physics] Commutation of vector operators

Question

I'm supposed to show that $\left[\mathbf A,\mathbf B\right]=0$ (for two vector operators $\mathbf A$ and $\mathbf B$) if and only if all components of $\mathbf A$ commute with all components of $\mathbf B$. What I have so far is this:
\begin{align*}
\mathbf A \otimes\mathbf B &= \mathbf A\mathbf B^T \Rightarrow \left[\mathbf A,\mathbf B\right] = \mathbf A\otimes \mathbf B – \left[\mathbf B \otimes \mathbf A\right]^T=\mathbf A\mathbf B^T – \left[\mathbf B \mathbf A^T\right]^T = \mathbf A\mathbf B^T – \mathbf B^T\! \mathbf A = 0.
\end{align*}
So much for the basic arithmetic… what am I supposed to do now? I can expand into components, but the next question "give an expression for $\left[\mathbf A,\mathbf B\right]_{ij}$ in Einstein notation" kind of suggests I should avoid saying $\mathbf A = a^i_j$ and go from there. I tried doing so anyway, with the obvious result of:
\begin{align*}
\mathbf A\mathbf B^T – \mathbf B^T\! \mathbf A = a^i_j\left(b^i_j\right)^T – \left(b^i_j\right)^T a^i_j = a^i_jb^j_i – b^j_i a^i_j = 0.
\end{align*}
Should I be interpreting the result as "One of these vectors will produce a matrix, the other will produce a scalar (an inner product), which means that this expression can only be zero when the matrix equals zero" (?). I'm at a loss as to how to approach this..

Answer 1

I think your confusion is arising from the fact that you are imagining operators as matrices. This is mostly fine, but in this case, the operator itself being a vector is what is causing the confusion - so let me elaborate.

${\bf A}$ is a vector of operators. For example $$ {\bf A} = \pmatrix{ A_1 \\ A_2 \\ A_3} $$ We can denote this collectively as $A_i$. Now, note that each of these $A_i$'s are themselves operators. In other words, they are matrices $(A_i)_{ab}$. Thus, each element of $A$ has three indices. One index is the vector index and the other two are the matrix operator indices.

Finally, the very definition of $[{\bf A} , {\bf B}]$ is the following $$ [ {\bf A} , {\bf B} ]_{ij} = [ A_i , B_j ] $$ The latter can further be written in matrix notation as $$ [ A_i , B_j ]_{ab} = (A_i)_{ac} (B_j)_{cb} - (B_j)_{ac} (A_i)_{cb} $$ Thus, $[{\bf A} , {\bf B} ] = 0$ is precisely the statement that all components of ${\bf A}$ commute with those of ${\bf B}$.

Let me finally point out that this question is really poorly worded. I would say that $[ {\bf A} , {\bf B} ] = 0$ is by definition the statement that the components commute. More clearly, (if it is not already) I mean that $[ {\bf A} , {\bf B} ] = 0$ already means that the components commute. There is nothing to derive here.