  Let ${\bf M}_n(\mathbb{F})$ be the algebra of $n\times n$ matrices over an
arbitrary field $\mathbb{F}$. We consider linear maps $\Phi: {\bf
M}_n(\mathbb{F}) \rightarrow {\bf M}_r(\mathbb{F})$ preserving matrices
annihilated by a fixed polynomial $f(x) = (x-a_1)\cdots (x-a_m)$ with $m\ge 2$
distinct zeroes $a_1, a_2, \ldots, a_m \in \mathbb{F}$; namely, $$ f(\Phi(A)) =
0\quad\text{whenever} \quad f(A) = 0. $$
  Suppose that $f(0)=0$, and the zero set $Z(f) =\{a_1, \dots, a_m\}$ is not an
additive group. Then $\Phi$ assumes the form \begin{align}\label{eq:standard} A
\mapsto S\begin{pmatrix} A \otimes D_1 &&\cr & A^{T} \otimes D_2& \cr &&
0_s\cr\end{pmatrix}S^{-1}, \tag{$\dagger$} \end{align} for some invertible
matrix $S\in {\bf M}_r(\mathbb{F})$, invertible diagonal matrices $D_1\in {\bf
M}_p(\mathbb{F})$ and $D_2\in {\bf M}_q(\mathbb{F})$, where
  $s=r-np-nq\geq 0$. The diagonal entries $\lambda$ in $D_1$ and $D_2$, as well
as $0$ in the zero matrix $0_s$, are zero multipliers of $f(x)$ in the sense
that $\lambda Z(f) \subseteq Z(f)$.
  In general, assume that $Z(f) - a_1$ is not an additive group. If $\Phi(I_n)$
commutes with $\Phi(A)$ for all $A\in {\bf M}_n(\mathbb{F})$, or if $f(x)$ has
a unique zero multiplier $\lambda=1$,
  then $\Phi$ assumes the form \eqref{eq:standard}.
  The above assertions follow from the special case when $f(x) = x(x-1)=x^2-x$,
for which the problem reduces to the study of linear idempotent preservers.
  It is shown that a linear map $\Phi: {\bf M}_n(\mathbb{F}) \rightarrow {\bf
M}_r(\mathbb{F})$ sending disjoint rank one idempotents to disjoint idempotents
always assume the above form \eqref{eq:standard} with $D_1=I_p$ and $D_2=I_q$,
unless ${\bf M}_n(\mathbb{F}) = {\bf M}_2(\mathbb{Z}_2)$.

Functional Analysis