I recently thought of a problem in linear algebra that seems relatively straightforward:
Suppose we have an $n\times n$ matrix $A$ over a field $k$, which is a zero divisor in the matrix ring $M_n(k)$. Does there exist a matrix $B$ that is a two-sided zero divisor of $A$, i.e. $AB=BA=0$?
The answer to this is yes, which I proved using ring theory as follows:
If we let $R$ be the centraliser of $A$ in $M_n(k)$, then $R$ is a $k$-subspace of $M_n(k)$, and also a subring, containing the identity $I_n$. So since $R$ is a finite dimensional, associative, unital $k$-algebra, it follows that $R$ is artinian, and hence every element of $R$ is either a unit or a zero divisor.
Clearly $A$ lies in $R$, since $A$ commutes with itself, and since $A$ is not a unit in $M_n(k)$ and $I_n\in R$, it follows that $A$ is not a unit in $R$, and hence is a zero divisor in $R$. Therefore there exists a matrix $B\in R$ with $AB=0$, and by the definition of the centraliser, $AB=BA$, so we are done.
This proof can be extended to the case where $k$ is a non-commutative division ring, finite dimensional over its centre.
However, this proof is not accessible to anyone who doesn't have knowledge of ring theory, even though the problem itself is grounded in standard linear algebra.
Can anyone think of a proof that only involves techniques in linear algebra?
Not sure why this old question has to be bumped up, but for sake of having an answer, let $A=MDN$, where $M,N$ are invertible and $D$ is diagonal (one may obtain this decomposition using Smith normal form or even elementary row/column operations). Let $S$ be any nonzero diagonal matrix such that $DS=0$. Then $B=N^{-1}SM^{-1}$ is a nonzero solution to $AB=BA=0$.