Sorry for my bad English.
Let $k$ be a field (if necessary algebraically closed) and $M$ be a $n\times n$ matrix over $k$.
If $M$ is diagonalizable, then there is a $k$-algebra isomorphism $k[M]\cong k^m$ where $m\le n$ is the number of distinct eigenvalues of $M$.
Now is the converse true? i.e. if $k[M]\cong k^m$ for some $m\le n$, then is $M$ diagonalizable?