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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?
Consider the matrix $M=\begin{pmatrix}0&1\\ 0&0 \end{pmatrix}$. Then $M$ is not diagonalizable and $M^2=0$. Thus $k[M]=k[x]/(x^2)$ which is not isomorphic to $k^2$ because $k[x]/(x^2)$ has a nontrivial nilpotent element. In general, $k[M]=\{f(M): f\in k[x]\}$. In fact, the map $f\mapsto f(M)$ from $k[x]\to k[M]$ induces an isomorphism $$k[M]=k[x]/(p(M)),$$ where $p(M)$ is the minimal polynomial. If $p(M)$ has a repeated roots, then $k[M]$ has a nonzero nilpotent element. This implies that $M$ is diagonalizable when $k$ is algebraically closed.
