Change of basis for the Frobenius normal form

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$M$ is an $n \times n$ matrix over a field $k$. I try to find a basis with respect to which $M$ is block diagonal with blocks of the form:

$$ \begin{matrix} 0 & 0 & \cdots & 0 & a_0 \\ 1 & 0 & \ddots & 0 & a_1 \\ 0 & \ddots & \ddots & \ddots & \vdots \\ 0 & 0 & \ddots & 0 & a_{d-2} \\ 0 & 0 & \cdots & 1 & a_{d-1} \\ \end{matrix} $$

(This is the Frobenius normal form, right?)
I have been given the following hint: M acts naturally on some n-dimensional k-vector space $V$. Consider $V$ as a $k[x]$-module via $f.v$ = $f(M)(v)$.

I have no idea where to start, and in particulary I don't see how to use the hint.