Finding Jordan Canonical form given the minimal polynomial and the dimension of the kernel How can we find the Jordan Canonical form given the minimal polynomial and the dimension of the kernel? I know how we could find it if either the matrix or the characteristic polynomial was given, but how does knowing the dimension of the kernel help us?
$$
B: \mathbb{R}^4\rightarrow \mathbb{R}^4
$$
$$
m_B(x) = (x-2)^2*x
$$
$$\newcommand{\Ker}{{\operatorname{Ker}}}
\dim_{\mathbb{R}}\Ker(B)=2
$$
Using the minimal polynomial I found that the Jordan form should be either
$$
\begin{pmatrix}
2 & 1 & 0 & 0\\
0 & 2 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
\end{pmatrix}
$$
or
$$
\begin{pmatrix}
2 & 1 & 0 & 0\\
0 & 2 & 0 & 0 \\
0 & 0 & 2 & 0 \\
0 & 0 & 0 & 0 \\
\end{pmatrix}
$$
I suppose that the statement about the kernel should tell us which of this matrices is the right one. (assuming I found the right matrices in the first place :) )
Any help would be appreciated!
Thanks
 A: Focus on this case, the minimal polynomial implies that the elementary factors includes $(x-2)^2$ and $x$.
Plus we know $\dim\ker B=2$, which means $0$ is the double root of the characteristic polynomial.
So the Jordan block is in correspondence with $x$, $x$, $(x-2)^2$. Hence, the first one is right.
In general cases, it's customary to consider the minimal polynomial to get the biggest invariant factors to control other factors. Second, calculate the determinant factors. Calculate the geometric multiplicity of any eigenvalues is also helpful, which implies the numbers of Jordan block of each eigenvalues.
Hope my answer will be helpful. I can offer some examples using these methods if you want.
A: Knowing that $X(X-2)^2$ is the minimal polynomial gives you two kinds of information: (1) the (only) eigenvalues are $\lambda=0$ and $\lambda=2$ (the roots of the minimal polynomial), and (2) the part of the Jordan normal form for the eigenvalue $\lambda=0$ is diagonal (i.e., the all blocks for $\lambda=0$ are of size$~1$, the multiplicity of the root$~0$), while for the eigenvalue $\lambda=2$ the largest Jordan block has size$~2$ (the multiplicity of the root$~2$). This indeed leaves open the two options you mentioned, but if it is given that the kernel (which is none other than the eigenspace for $\lambda=0$) is of dimension$~2$, then it becomes clear that there must be two Jordan blocks (of size$~1$) for this eigenvalue, your first option.
