Determine the Jordan normal form from the given information. Question
$1. ~~ch_A(x)=(x-2)^7,~~m_A(x)=(x-2)^4,~~\dim(V_1(2))=3$
$2. ~~ch_A(x)=(x-3)^9,~~m_A(x)=(x-3)^4,~~\dim(V_1(3))=4$
Attempt


*

*This can be determined uniquely. From the minimal polynomial we know there must be a Jordan block of the form $J_4(2)$ and since $\dim(V_1(2))=3$ we know there is also a jordan block of the form $J_3(2)$. Thus:
$$A=\begin{bmatrix}J_4(2)&&\\&&J_3(2)\end{bmatrix}$$
Can someone please confirm my reasoning ?

*I don't think this can be uniquely determined since $m_A(x)=(x-3)^4$ and $\dim(V_1(3))=4$ give us the same information, which leaves us 5 dimensions to play with and allocate.. ?
 A: I'm unfamiliar with the notation $V_1(\lambda)$. I'm going to interpret it as $V_{\color{red}{1}}(\lambda)=\DeclareMathOperator{null}{null}\null(A-\lambda I)^{\color{red}{1}}$. This seems reasonable as its natural generalization is $V_\color{red}{k}(\lambda)=\null(A-\lambda I)^{\color{red}{k}}$. That is, $V_k(\lambda)$ is the space of all rank $k$ generalized eigenvectors of $A$. 
In particular, $\dim V_1(\lambda)$ is the geometric multiplicity of $\lambda$. Recall that the geometric multiplicity of an eigenvalue $\lambda$ of $A$ is the number of Jordan blocks associated to $\lambda$ in the Jordan form of $A$.
For $d\geq 1$ let $J_\lambda^{(d)}$ be the $d\times d$ matrix
$$
J_\lambda^{(d)}=
\begin{bmatrix}
\lambda & 1 & \\
& \lambda & 1 & \\
&& \lambda & 1 & \\
&&& \ddots & \ddots & \\
&&&& \lambda & 1 \\
&&&&& \lambda & 1 \\
&&&&&& \lambda 
\end{bmatrix}
$$
where all the unmarked entries are $0$.
For square matrices $A_1,\dotsc,A_n$ let $A_1\oplus\dotsb\oplus A_n$ be the block-diagonal matrix
$$
A_1\oplus\dotsb\oplus A_n
=
\begin{bmatrix}
A_1 \\
& A_2 \\
&&\ddots \\
&&& A_n
\end{bmatrix}
$$
where again the unmarked entries are $0$.
Now, we are given that $A$ has characteristic polynomial and minimal polynomial 
\begin{align*}
\chi_A(t) &= (t-2)^7 & \mu_A(t) &= (t-2)^4
\end{align*}
respectively. We are also given that $\dim\null(A-2\,I)=3$.
This tells us that the Jordan form of $A$ is of the form
$$
J=J_2^{(d_1)}\oplus J_2^{(d_2)}\oplus J_2^{(d_3)}
$$
where


*

*$d_1\geq d_2\geq d_3$

*$d_1+d_2+d_3=\deg\chi_A(t)=7$

*$\max\{d_1,d_2,d_3\}=d_1=\deg\mu_A(t)=4$


Thus the possible number of Jordan forms is the number of positive integer solutions to
$$
d_2+d_3=3
$$
The only possible solution is given by 
$$
2+1=3
$$
Hence the only possible Jordan form is
$$
J=J_2^{(4)}\oplus J_2^{(2)}\oplus J_2^{(1)}
=
\left[\begin{array}{rrrr|rr|r}
2 & 1 & 0 & 0 & 0 & 0 & 0 \\
0 & 2 & 1 & 0 & 0 & 0 & 0 \\
0 & 0 & 2 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 2 & 0 & 0 & 0 \\
\hline
 0 & 0 & 0 & 0 & 2 & 1 & 0 \\
0 & 0 & 0 & 0 & 0 & 2 & 0 \\
\hline
 0 & 0 & 0 & 0 & 0 & 0 & 2
\end{array}\right]
$$
Similar reasoning shows that the only possible Jordan forms of a matrix $A$ satisfying your second conditions are 
\begin{align*}
J_3^{(4)}\oplus J_3^{(3)}\oplus J_3^{(1)}\oplus J_3^{(1)} 
&&
J_3^{(4)}\oplus J_3^{(2)}\oplus J_3^{(2)}\oplus J_3^{(1)} 
\end{align*}
