# Invariant factors and Jordan reduction : how to find the adapted basis?

I have a question about the Jordan reduction using the module theory and especially the invariant factors. If we have a vector space $$E$$ of dimension $$n$$ over a field $$k$$, and $$f \in End(E)$$ then it's also a $$k[X]$$-module for the operation : $$P.v = P(f)(v)$$. And of course it's a finitely generated module, and then :

$$E \cong \frac{k[X]}{(P_1)} \times ... \times \frac{k[X]}{(P_s)}$$

as a $$k[X]$$-module, and $$f$$ is the multiplication by $$X$$ for $$\frac{k[X]}{(P_1)} \times ... \times \frac{k[X]}{(P_s)}$$. Then, $$E=E_1 \bigoplus ... \bigoplus E_n$$ and on a good basis of E, the matrix of $$f$$ is the compagnon matrix of $$P_i$$ on each subspace $$E_i$$ which is stable by $$f$$.

($$P_i$$ can be find as the invariant factors of $$\det(Mat(f) - XId)$$)

And then, if $$P_i$$ have the form $$(X-a)^r$$, then we can find a good basis of E when the matrix of $$f$$ has the Jordan form.

But, this is my problem : how to find the good basis ? Cause on $$\frac{k[X]}{(P_1)} \times ... \times \frac{k[X]}{(P_s)}$$, we know the basis on which the matrix of the multiplication by $$X$$ have the Jordan form, but actually, to determine it on $$E$$, we should know, in the explicit way, the isomorphism between $$E$$ and $$\frac{k[X]}{(P_1)} \times ... \times \frac{k[X]}{(P_s)}$$, and it's seems very complicated, in a practical way...

When we want to apply this principle in a concrete way (for example for the Linear difference equation), we make a change of basis to transform the matrix in order to obtain a matrix which have the jordan form, but we should also know the change of basis if we want to solve the problem, finally...

So, how to do ? And, are we able, from this application of invariant factors, to find the basis adapted to the jordan form ?