Exercise 8.A.13

Proof. It is easy when $𝔽=ℂ$, because then $V$ has a basis consisting of eigenvectors of $N$ and for each vector $v$ in this basis we have $0=N^{\dim <!--\; FUNCTION\; APPLICATION\; -->V}v={\lambda }^{\dim <!--\; FUNCTION\; APPLICATION\; -->V}v$ for the corresponding eigenvalue $\lambda$, which implies that $\lambda =0$.

More generally, without restricting $𝔽$ to $ℂ$, we will prove $N^{\dim <!--\; FUNCTION\; APPLICATION\; -->V-1}=0$ and this fact can be used to show $N^{\dim <!--\; FUNCTION\; APPLICATION\; -->V-2}=0$, which then can be used to show... and so on until $N^1$.

Let $N=N^{\dim <!--\; FUNCTION\; APPLICATION\; -->V-1}$. Note that $N$ is also normal and that $N^2=0$. Then, for all $v\in V$,

where the first equality comes from 7.20. Thus $N^{\text{∗}}N=0$. Therefore

which shows that $N=0$. □