TITLE: Jordan Canonical Form determined by characteristic and minimal polynomials in dimension $3$, but not beyond
QUESTION [14 upvotes]: Why and how is the Jordan Canonical form of a matrix in $M_3(\mathbb C)$ fully determined by its characteristic and minimal polynomials? And why does it fail for $n >3$? 
Thanks.

REPLY [6 votes]: In $M_3(\mathbb{C})$, if your characteristic polynomial gives three distinct complex  roots your matrix will be diagonalisable. If you have one complex root and another double root, we can analyse it as follows:
Say your roots of the characteristic equation are $\lambda$ and $\lambda_1$ (twice). Then the block corresponding to the first eigenvalue is already completely determined. Then we only need to concentrate on the second block, which will be $2 \times 2$. This is because if the characteristic polynomial of your matrix is $(t - \lambda)(t - \lambda_1)^2$, there must be the factor $(t - \lambda)$ when $\lambda \neq 0$ in your minimal polynomial, otherwise there is no way to eliminate the block corresponding to $\lambda$ in the Jordan Form of the matrix when $\lambda \neq 0$ (Recall the definition of the minimal polynomial). To see this, observe that if your matrix $M$ is block diagonal, multiplication is done as follows:
$M = \left[\begin{array}{c|c} A & 0 \\ \hline 0 & B \end{array}\right]$, $M^2$ will be equal to $\left[\begin{array}{c|c} A^2 & 0 \\ \hline 0 & B^2 \end{array}\right].$
Having understood this, let us look at the factor in the minimal polynomial corresponding to that block. It will either be $(t - \lambda_1)$ or $(t - \lambda_1)^2$, if $\lambda_1 \neq 0$. Now the first case corresponds to the $2 \times 2$ block being 
$\left[\begin{array}{cc} \lambda_1 & 0 \\ 
0 & \lambda_1  \end{array}\right]$ (why ?) while the second case to corresponds to a second block of the form $\left[ \begin{array}{cc} \lambda_1 & 1 \\ 0 & \lambda_1 \end{array} \right]$ (why)? 
In the case when $\lambda$ or $\lambda_1$ is zero, you should know how the Jordan Form will look like given the minimal and characteristic polynomial just by looking at their definitions.
In the last case that you have three repeated complex roots, $\chi(t)$ will look like $(t - \lambda)^3$. Now we know that the minimal polynomial divides the characteristic polynomial, so that it will either be $(t - \lambda), (t - \lambda)^2$ or $(t - \lambda)^3$. This will then completely the determine the Jordan Form of your matrix, because in the first case, there are no $1's$ on the superdiagonal, the second case only one $1$ on the super diagonal up to rearrangement, and the third case two ones on the super diagonal.
From this we can see that in the case of $3 \times 3$ complex matrices, the characteristic and minimal polynomial of a matrix completely determines its Jordan Form. I believe Jonas has already told you to consider the $4 \times 4$ complex matrices he has given to see why this does not work when $n > 3$.