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

1. 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 ?

2. 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.. ?

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*}