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

characteristic polynomialjordan-normal-formlinear algebramatricesminimal-polynomials

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!

Best Answer

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.

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