2013-04-25 10:38:30 8 Comments

I need to prove that if $\phi : V \rightarrow V$ is nilpotent, then its only eigenvalue is 0.

I know how to prove that this for a nilpotent matrix, but I'm not sure in the case of an operator. How would I be able to relate $\phi$ to a matrix?

**Note:** A nilpotent operator $\phi$ has been defined as $\phi^{n} = 0$, for some $n \geq 1$

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## 2 comments

## @Somaye 2013-04-25 12:10:07

Suppose that $\phi$ has another eigenvalue $\lambda \ne 0$ so that $\phi(x)=\lambda x $ ($x$ is an eigenvector corresponding to $\lambda$)

Then, $ \phi^n (x)= \phi^{(n-1)}(\phi(x))=\phi^{(n-1)}(\lambda x)=\lambda \phi^{n-1}(x)=\cdots=\lambda ^{n}x\ne 0$.

We have a contradiction, so $\phi$ can't have another eigenvalue except $0$.

## @Rhys 2013-04-25 10:41:16

$\phi$ is nilpotent, so $\phi^n = 0$ for some $n$. Now let $v$ be an eigenvector: $\phi v = \lambda v$ for some scalar $\lambda$. Now we get $$ 0 = \phi^n v = \lambda^n v ~\Rightarrow~ \lambda=0 ~. $$

Note that this works in the infinite-dimensional case as well; there is no need to relate $\phi$ to a matrix.

Edit: As suggested in the comments, we can also show that $0$ isalwaysan eigenvalue; in other words, $\phi$ always has at least one eigenvector. Take any $v \neq 0$; we know that $\phi^n v = 0$, so let $k$ be the smallest integer such that $\phi^k v \neq 0$. Then $\phi(\phi^k v) = 0$, so $\phi^k v$ is an eigenvector of $\phi$, with eigenvalue $0$.## @wj32 2013-04-25 10:48:29

For completeness, we might want to show that $0$ is actually an eigenvalue. If $v$ is injective then $\phi v \ne 0$ for every $v \ne 0$. If we choose any nonzero $v$ then $\phi^n v \ne 0$, which is a contradiction. Therefore $\ker\phi \ne \{0\}$, and any element of $\ker\phi$ is obviously an eigenvector with eigenvalue $0$.

## @Rhys 2013-04-25 10:58:08

Yes, I probably should have added that part; I will do so, with a different wording.