When looking for Jordan canonical form of a matrix lets say $\begin{pmatrix} 2 &0 &0\\0 &5 &0\\1&2&2 \end{pmatrix} $ for simplicity.
It's easy to calculate the characteristic polynomial $c_A(x):=\det(xI-A)=\begin{pmatrix}(x-2)&0&0 \\0 &(x-5)&0\\-1&-2&(x-3)\end{pmatrix}=(x-2)^2(x-5)$.
so we know the possible minimal polynomials are $m_a(x)=(x-2)^d(x-5)$, where $1\leq d \leq2$.
and that the unique minimal polynomial must satisfy $m(A)=0$. This is a pretty straightforward matrix so it's easy enough to check both cases. But for larger matrices checking all the cases by process of elimination can be a long process. Is there anyway to make an educated guess on what the minimal polynomial will be without just checking all the possible cases ?
P.S. The matrix I've given here was only a basic example I used to elucidate what I meant. I'm seeking tips and tricks that could apply to any general matrix.
The minimal polynomial is the product of the minimal polynomials of each restriction to a generalised eigenspace. So the problem comes down to finding the minimal polynomial of a matrix with a single eigenvalue $\lambda$.
One way to find the structure of the associated Jordan matrix is to consider the following sequence of subspaces: $$\{0\}\subset\ker(A-\lambda I)\subset\ker(A-\lambda I)^2\subset\dots\subset\ker(A-\lambda I)^r\subset\dotsm$$
This sequence first increases (strictly) and eventually stabilises. We have the following result: $$\dim\ker(A-\lambda I)^i-\dim\ker(A-\lambda I)^{i-1}= \text{number of Jordan blocks of size}\ge i.$$ In particular,$\dim\,\ker(A-\lambda I)$ is the total number of Jordan blocks.