TITLE: How would one calculate the Jordan canonical form , if given the minimal polynomial first instead of the matrix?
QUESTION [0 upvotes]: Say we are given some $A \in Mat_n(F)$ where F is some field.
I know how to find the Jordan canonical form if I'm given the matrix first, as one need just find the characteristic polynomial, rather easily, and then it's a simple matter to find the minimal polynomial. 
I.e. The steps I would follow would be 
1)Find $c_A(x):=det(xI-A)$
2)from this write out the possible minimal polynomials 
3 ) Check to see which one satisfies $m(A)=0$
4) Calculate the possible Jordan canonical forms based on the roots and exponents of the characteristic polynomial and the minimal polynomial.
My question is what if instead of being given the matrix first , we are given the minimal polynomial first , then how does one calculate the Jordan canonical form ?  I think it would probably involve finding the characteristic polynomial based what we know about the field the matrix is over , the dimensions of the matrix, and the roots of the minimal polynomial. Beyond this however I couldn't guess at what one might do.

REPLY [2 votes]: If the minimal polynomial contains a factor $(x - \lambda)^m$, there must be at least one Jordan block of size $m$ for eigenvalue $\lambda$.  There may be more, or there may be smaller Jordan blocks for this eigenvalue.
Of course, the sum of all the sizes of the blocks is $n$.  But if the minimal polynomial has degree $< n$, it doesn't determine the Jordan canonical form.
For example, if the minimal polynomial is $x^2(x-1)$ and $n=4$, the JCF could be
$$ \pmatrix{0 & 1 & 0 & 0\cr 0 & 0 & 0 & 0\cr 0 & 0 & 0 & 0\cr 0 & 0 & 0 & 1}
\ \text{or} \pmatrix{0 & 1 & 0 & 0\cr 0 & 0 & 0 & 0\cr 0 & 0 & 1 & 0\cr 0 & 0 & 0 & 1} $$
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