So I need to find all dissimilar Jordan matrices J with the minimal polynomial: $x^3(x^2-1)^3$ and characteristic polynomial: $x^4(x^2-1)^4(x+1)^2$. So my question is, since the minimal polynomial does not have a $(x+1)^n$ term like in the characteristic polynomial, does this mean that the minimal polynomial is actually $x^3(x^2-1)^3(x+1)^0$ and then I find the dissimilar matrices using that or do I need to divide the minimal polynomial by the characteristic polynomial since the minimal polynomial always divides the characteristic polynomial by Cayley-Hamilton and then get the same eigenvalues on both terms and then find the dissimilar matrices that way? Hopefully that made sense, I'm pretty confused.
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- $\begingroup$ $(x^2-1)=(x-1)(x+1)$ $\endgroup$user91684– user916842015-03-27 10:16:35 +00:00Commented Mar 27, 2015 at 10:16
- $\begingroup$ @loupblanc oh duh oops $\endgroup$Sofia June– Sofia June2015-03-27 16:39:28 +00:00Commented Mar 27, 2015 at 16:39
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1 Answer
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If you want to reason about irreducible factors, make sure the factors you see are irreducible. Here the field is not specified, but $x^2-1=(x-1)(x+1)$ holds in any field. So your minimal polynomial is $x^3(x-1)^3(x+1)^3$ and your characteristic polynomial is $x^4(x-1)^4(x+1)^6$. The former does divide the latter, and every irreducible factor of the latter is also factor of the former; these two conditions ensure that the two candidate polynomials are compatible, and so that there exists some solution.
The exponent of $x-\lambda$ in the characteristic polynomial gives the sum of the sizes of the Jordan block associated to$~\lambda$, and its exponent in the minimal polynomial gives the size of the largest such Jordan block. If the characteristic is not$~2$, this leaves you some uncertainty about the sizes of the blocks only for$~\lambda=-1$, where partitions $(3,3)$, $(3,2,1)$ and $(3,1,1,1)$ are possible. In characteristic$~2$ where $1=-1$ the last two factors are in fact combined, and you have more choice ($5$ possibilities)
