Jordan Normal Form of $A_{7 \times 7}$ with a singal eigenvalue $\lambda$

Question

Let $A \in M^{\Bbb C}_{n \times n}$ a square matrix $7 \times 7$. A has one eigenvalue, $\lambda \in \Bbb C$.

In addition, rank $(A-\lambda I)=2$, and rank$(A-\lambda I)^2=1$.

Find the Jordan Normal Form and the minimal polynomial of A.

Since rank $(A-\lambda I)=2$ I can conclude that the geometric multiplicity of $\lambda$ is $7-2=5$. Therefore I have two options for the Jordan Normal Form:

1. $J_{(3)}(\lambda), J_{(1)}(\lambda),J_{(1)}(\lambda),J_{(1)}(\lambda),J_{(1)}(\lambda)$
2. $J_{(2)}(\lambda), J_{(1)}(\lambda),J_{(1)}(\lambda),J_{(1)}(\lambda),J_{(1)}(\lambda),J_{(1)}(\lambda) $

Where $J_{(3)} (\lambda)$ means a size 3 Jordan Block:

$$\begin {bmatrix} \lambda & 1 & 0 \\ 0 & \lambda & 1 \\ 0 & 0 & \lambda \end {bmatrix}$$

I know that since these are the options from the geometric multiplicity I have.

However, I don't know how to choose from my two options. I assume I have to use the other data I have - rank$(A-\lambda I)^2=1$. But How?

Thanks,

Alan

Answer 1

Here's what I think.

If the Jordan Normal Form is either $$J_2(\lambda),J_2(\lambda),J_1(\lambda),J_1(\lambda),J_1(\lambda)$$

OR

$$J_2(\lambda),J_1(\lambda),J_1(\lambda),J_1(\lambda),J_1(\lambda),J_1(\lambda)$$

It means that $\left(J_A-\lambda I \right) ^2 =\left \{ 0\right \}$.

Since $J_A$ is similar to $A$, $J_A-\lambda I is $ similar to $A- \lambda I$. Therefore one of these two forms leads us to a contradiction.

Hence the minimal polynomial is $(t-\lambda)^3$ and the Jordan Normal Form is $J_3(\lambda), J_1(\lambda),J_1(\lambda),J_1(\lambda),J_1(\lambda)$.

Thanks @GitGod for your help.