Let $A$ and $B$ be two non-commuting linear operators on an $n-$dimensional vector space $V$ with fixed canonical basis $(e_j)$, such that $A$ is symmetric and $B$ is a projector into the basis element $e_j$.
Consider the sum $s=e_1+...+e_n$. Clearly $B(s)=e_j$, and assume $A(s)=0$.
I'm curious whether the expression $(A-B)^k(s)$ can be explicitly evaluated for any integer $k$ in terms of $e_j$ and the columns $C_j=A(e_j)$. The cases $k=1$ and $k=2$ are trivial: $$A(s)-B(s)=-e_j, \\ (A-B)^2(s)=A^2(s)-AB(s)-BA(s)+B^2(s)=0-A(e_j)-0+B(s)=e_j-C_j.$$ Of course as $[A,B]\ne 0$ the generalized binomial theorem doesn't hold, but is there another way to compute this?
There is a closed expression you could use to evaluate it when $k \geq 2$, but it's not pretty: $$ (A - B)^k(s) = (-1)^k \left( e_j - \sum_{i=0}^{k-2}(B - A)^i (C_j) \right).$$ (If you want to prove the above relationship, you can do it with induction, noting that $ (A-B)^{k+1}(s) = (A - B) (A - B)^{k}(s)$).
After playing with this expression, I don't think you'll be able to compute $(A-B)^k(s)$ in terms of only $e_j$ and $C_j$. The reason is because of that nasty summation term, $\left( \sum_{i=0}^{k-2}(B - A)^i \right)C_j$. This summation term will have summands like $A^{k-2} (C_j)$ and $(ABA) (C_j)$. There's just no getting around these terms.
I think the best way to evaluate an expression like this is probably iteratively, but it seems like you want a closed-form evaluation for any $k$, which I don't think is in the cards.