How do you show that two $3 \times 3$ matrices with the same characteristic and minimal polynomials both conjugate to the same Jordan normal form, assuming no knowledge of the eigenspaces?
I know that it is possible to determine completely the Jordan normal form of a matrix only with its minimal and characteristic polynomial, up to dimension $6$, but only if one can compute the dimension of the eigenspace as well.
And why does this characterization fail for $4 \times 4$ matrices?
There are six cases for the characteristic polynomial $p(x)$ and for the minimal one $m(x)$:
$p(x)=(x-a)(x-b)(x-c)=m(x), \ a,b,c$ different. In this case the matrices are diagonalizable: $$\begin{pmatrix}a&0&0\\0&b&0\\0&0&c\end{pmatrix}$$
$p(x)=(x-a)^2(x-b)=m(x), \ a\neq b$. In this case, the JCF for both matrices is $$\begin{pmatrix}a&1&0\\0&a&0\\0&0&b\end{pmatrix}$$
$p(x)=(x-a)^2(x-b),\ m(x)=(x-a)(x-b),\ a\neq b$. Here, the JCF in both cases is $$\begin{pmatrix}a&0&0\\0&a&0\\0&0&b\end{pmatrix}$$
$p(x)=(x-a)^3$. Check the three cases for $m(x)$...
For each eigenvalue, its multiplicity as root of the characteristic polynomial gives the sum of the sizes of the Jordan blocks; its multiplicity as root of the minimal polynomial gives the size of the largest Jordan block.
Since the sizes of the Jordan blocks define a partition of their sum, the question is for which values of $n\geq m$ there is a unique partition of $n$ with largest part$~m$. This is true for all $m\leq n\leq3$, the partitions of$~3$ being $(3),(2,1),(1,1)$, all with different largest part. It fails first for $n=4,m=2$, which allows the two partitions $(2,2)$ and $(2,1,1)$. So an eigenvalue with multiplicity $4$ in the characteristic polynomial and multiplicity $2$ in the minimal polynomial would not allow determining the sizes of all Jordan blocks. This can happen with square matrices of size at least$~4$ (for some pair of characteristic and minimal polynomials).
