# Eigenspace, Jordan Canonical form

Let $$A\in \mathbb{C}^{n\times n}$$ with all eigenvalues equal to $$\lambda$$, i.e. the characteristic function $$c_A(z)=(z-\lambda)^n$$.

Denote $$F_i=N[(A-\lambda I)^i]$$ ($$N$$ represents the null space).

If $$dimF_1=2$$, show that $$$$dimF_i=\begin{cases} i+1, & \text{if \quad i

My attempt: It is obvious that $$F_1 \subset F_2 \subset \cdots \subset F_n$$ and $$dimF_n=n$$.

Also I tried to simplify and see what happens when $$n=3$$.

Let $$F_1=span \{x,y\}$$, $$F_3=span \{x,y,z\}$$,$$\quad$$ where $$x,y,z$$ are linearly independent. Then $$(A-\lambda I)^3z=0$$, thus $$(A-\lambda I)^2z \in F_1$$ and $$(A-\lambda I)z \in F_2$$. I feel here we need to assume that $$dimF_2=2$$ and contradict it using the fact $$x,y,z$$ are linearly independent.

For example: $$$$A=\begin{bmatrix} 2&1&0\\ 0&2&0\\ 0&0&2 \end{bmatrix},$$$$ then $$\quad dimF_1=2 \quad F_1=span\{x=\begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix},y=\begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix}\}, \quad dimF_2=dimF_3=3$$. $$z=\begin{bmatrix}0 \\ 1 \\ 0\end{bmatrix}$$, and note that $$(A-\lambda I)^2=0$$, thus $$dimF_2=3$$, and $$(A-\lambda I)z=x$$.

Let $$A = \begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ \end{bmatrix}$$, $$\lambda = 0$$.
Then $$\dim F_1 = 2$$ and $$\dim F_2 =\dim F_3 = \cdots = 4$$.