Eigenvectors of nilpotent $3 \times 3$ matrix

Question

Let $N$ be a $3\times3 $ matrix such that

1. $N^2 = O_3$, then how many linearly independent eigenvectors will $N$ have?
2. $N^3 = O_3$, then how many linearly independent eigenvectors will $N$ have?

I think if $N^2=O_3$ then there will only one eigenvector and if $N^3=O_3$ then it can have 1 or 2 linearly independent eigenvectors depending on whether $N^2= O_3$ or $N^2\ne O_3$.

Am I right?

Answer 1

1) From $N^2=0$ we have $\operatorname{Im}N\subset \ker N$ and from the rank-nullity theorem we get $\dim\ker N=2$ and $\dim\operatorname{Im}N=1$ (assuming $N\ne0$). So since $\ker N= E_0(N)$ the eigenspace associated to $0$ so $N$ has two linearly independent eigenvectors associated to $0$.

2) Now if $N^2\ne0$ then there is $e\in E$ such that $N^2e\ne0$. We prove that $(N^2e,N e)$ are linearly independent so it's basis of $\operatorname{Im}N$ and then $\dim \ker N=1$. Can you take it from here to achieve the proof?