Given a matrix $A$ over a field $F$, does there always exist a matrix $B$ such that $AB = BA$? (except the trivial case and the polynomial ring?)
A square matrix $A$ over a field $F$ commutes with every $F$-linear combination of non-negative powers of $A$. That is, for every $a_0,\dots,a_n\in F$,

$$A\left(\sum_{k=0}^na_kA^k\right)=\sum_{k=0}^na_kA^{k+1}=\left(\sum_{k=0}^na_kA^k\right)A\;.$$

This includes as special cases the identity and zero matrices of the same dimensions as $A$ and of course $A$ itself.

Added: As was noted in the comments, this amounts to saying that $A$ commutes with $p(A)$ for every polynomial over $F$. As was also noted, there are matrices that commute only with these. A simple example is the matrix $$A=\pmatrix{1&1\\0&1}:$$ it’s easily verified that the matrices that commute with $A$ are precisely those of the form $$\pmatrix{a&b\\0&a}=bA+(a-b)I=bA^1+(a-b)A^0\;.$$ At the other extreme, a scalar multiple of an identity matrix commutes with all matrices of the same size.
