# Is the Jordan normal form uniquely determined by the characteristic and minimal polynomial?

I was looking into this answer to a question about obtaining the Jordan normal form given the characteristic and minimal polynomials of a matrix. In this answer, it is stated that

"The multiplicity of an eigenvalue as a root of the characteristic polynomial is the size of the block with that eigenvalue in the Jordan form. The size of the largest sub-block (Elementary Jordan Block) is the multiplicity of that eigenvalue as a root of the minimal polynomial".

I was then thinking of examples of matrices to apply this to, and I came up with the example of a matrix with characteristic polynomial $$f(x) = (x-1)^4(x+1)$$ and minimal polynomial $$m(x) = (x-1)^2(x+1)$$. Using the method described in the answer, I know that the largest elementary Jordan Block for the eigenvalue $$1$$ should be of size $$2$$. But given this, I can make $$2$$ distinct Jordan blocks for the eigenvalue $$1$$: $$\begin{pmatrix} 1&1&0&0\\ 0&1&0&0\\ 0&0&1&0\\ 0&0&0&1\\ \end{pmatrix} \qquad \text{and} \qquad \begin{pmatrix} 1&1&0&0\\ 0&1&0&0\\ 0&0&1&1\\ 0&0&0&1\\ \end{pmatrix}$$ where the first Jordan block has one elementary block of size $$2$$ and $$2$$ elementary blocks of size $$1$$, and the second Jordan block is made up of $$2$$ elementary blocks, each one of size $$2$$.

Do the characteristic and minimal polynomial always uniquely determine the Jordan normal form? In which case my understanding is wrong, and I would ask if someone could tell me what am I missing.

Or alternatively, when do the characteristic and minimal polynomial uniquely determine the Jordan normal form? Thank you!

Generally knowing only the characteristic polynomial and the minimal polynomial is not enough to determine uniquely the Jordan normal form, like you showed in the question.
I think that the only times where just knowing these two polyomials gives you also the Jordan normal form is when the degree of the minimal polynomial is very low or very high.
For example, if you know that

1. $$f(x) = (x-\lambda)^n$$ and $$m(x)=(x-\lambda)$$ you know that the Jordan normal form is the diagonal one
2. $$f(x) = (x-\lambda)^n$$ and $$m(x)=(x-\lambda)^n$$ you know that the Jordan normal form is the one made up by only one Jordan block of dimension $$n$$
3. $$f(x) = (x-\lambda)^n$$ and $$m(x)=(x-\lambda)^{n-1}$$ you know that the Jordan normal form is the one made up by one Jordan block of dimension $$n-1$$ and one Jordan block of dimension $$1$$.

I think that this cases, and the ones where every eigenvalue behave like one of these cases, are the only one where the two polynomials determines uniquely the Jordan normal form.

First, your terminology is not standard: a "Jordan block" is traditionally defined as what your block-quote calls an "Elementary Jordan Block"; note that the quote avoids the unadorned term "Jordan block" but speaks of "the size of the block with that eigenvalue in the Jordan form", which amounts to the sum of the sizes of all the (elementary) Jordan blocks associated to the eigenvalue. (Personally I would prefer however to say "dimension of the generalised eigenspace for that eigenvalue", as it is not at all necessary to do a full Jordan decomposition to find this number.)

Anyway, from the quote you get that in terms of the sizes of the Elementary Jordan Blocks, which could be any multiset of positive integers (that is, any finite list of such numbers in which permutation of terms is taken as equivalence), you are given their sum (as multiplicity in the characteristic polynomial) and their largest term (as multiplicity in the minimal polynomial). It seems rather obvious that this information does not in general determine the entire multiset; the example in your question show that a sum of $$4$$ with maximal term $$2$$ allows for two possibilities: $$\{\!\!\{2,2\}\!\!\}$$ and $$\{\!\!\{2,1,1\}\!\!\}$$. I'm not sure why after establishing this counterexample you could still ask whether characteristic and minimal polynomial always uniquely determine the Jordan normal form; you've just convincingly shown that this is not true. Moreover, if you realize that the sizes of the blocks in general determine a partition of the dimension of the generalised eigenspace, and the the number of partitions of $$n$$ grows quite fast with $$n$$ (there are just $$5$$ partitions of $$4$$, but there are $$42$$ partitions of $$10$$ and $$190569292$$ partitions of $$100$$), then you can see that it is somewhat naive to suppose that the Jordan form could ever in general be entirely determined by a few natural numerical statistics, even though this might sometimes be the case in certain low dimensions.

The question about when exactly the sum $$s$$ and the maximal term $$m$$ determine the entire partition has already been answered by NotPhiQuadro, but here is my approach: after you have put aside one copy of the maximal term, the remaining terms will form a partition of $$s-m$$ into parts that are at most $$m$$; the only way this can have a unique solution is when one of these two numbers $$s-m,m$$ is$${}\leq1$$ (since if not, one always has choices to either take all remaining parts $$1$$ or alternatively have one remaining part $$2$$ and the others $$1$$), which happens when $$m\in\{s,s-1,1\}$$. The smallest case where this is not so is $$s=4$$ and $$m=2$$, which gives precisely your counterexample.