Jordan Normal form — Complex matrices

Suppose we are given the  characteristic polynomial and minimal polynomial of a matrix $(x-a)^4(x-b)^2$ and $(x-a)^2(x-b)$ say. Then I can tell what the largest Jordan blocks are, and hence work out the possible forms the JNF can take. However, my question specifies that the matrix is "complex" whereas $a,b\in \mathbb R$. Is there some catch?

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Matrices with real entries are a subset of matrices with complex entries - are you sure you're supposed to have non-real entries? –  Matt Pressland Mar 16 '12 at 10:37
@MattPressland: Thant's what I thought, but I just want to make sure that there are no caveats :) –  Chump Mar 16 '12 at 10:39
Since the characteristic polynomial factors into linear factors, there should be no problems - after all, all eigenvalues are real. –  Johannes Kloos Mar 16 '12 at 10:41
@Chump Fair enough! :) I don't think it's an issue, I regularly got given exercises to do with theorems about complex matrices in which everything could be done with integers, let alone general real numbers! –  Matt Pressland Mar 16 '12 at 10:44
$(x-a)^2(x-b)$ cannot be a minimal polynomial. Minimal polynomials are products of distinct irreducible factors. These factors are always linear over $\mathbb{C}$ or any closed field, but can be linear or quadratic over $\mathbb{R}$. The minimal polynomial tells you which eigenvalues the matrix has (in the field). The characteristic polynomial tells you, additionally, their algebraic multiplicity, which is the sum of the sizes of all Jordan blocks corresponding to each eigenvalue. However, neither of these gives you the geometric multiplicity, which equals the number of such blocks. –  bgins Mar 16 '12 at 15:40

In general, a matrix $A$ with coefficients in a field $F$ has a Jordan Canonical Form (over $F$) if and only if the characteristic polynomial of $A$ splits over $F$.
We usually work in the context of the complex numbers because then we know that the characteristic polynomial will definitely split, no matter what $A$ is.
But in fact you can work over any field if you happen to know that your specific matrix $A$ has a characteristic polynomial that splits. That is the case here: you are told what the characteristic polynomial is, so you know that it splits. Therefore, you know for sure that $A$ has a Jordan Canonical From (over the reals; or in fact, over $\mathbb{Q}(a,b)$).