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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 \begin{equation} dimF_i=\begin{cases} i+1, & \text{if $\quad$ $i<n-1$}.\\ n, & \qquad i\geq n-1. \end{cases} \end{equation}

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: \begin{equation} A=\begin{bmatrix} 2&1&0\\ 0&2&0\\ 0&0&2 \end{bmatrix}, \end{equation} 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 $(A-\lambda I)z=x$.

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It is not true without extra conditions.

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 $\dim F_2 =\dim F_3 = \cdots = 4$.

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  • $\begingroup$ yes, I found that we need to know the degree of minimal polynomial in these case. $\endgroup$ – Lee Nov 14 '18 at 2:16

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