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I need to prove that:

if a linear operator $\phi : V \rightarrow V$ on a vector space 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 an operator that satisfies $\phi^{n} = 0$ for some $n \geq 1$.

Rodrigo de Azevedo
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Mathlete
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  • Choosing a basis for $V$ allows you to identify $\phi$ with a matrix, but this isn't optimal. As eigenvalues and nilpotency are defined identically for matrices and operators, the exact same proof should work. – mdp Apr 25 '13 at 10:40
  • It just makes my proof look a little sloppy though? Should I start by saying something like 'Consider the nilpotent matrix $A$, where $A^{n} = 0$ for some $n \geq 1$' and then provide the proof for a nilpotent matrix? – Mathlete Apr 25 '13 at 10:41
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    Hint: suppose $\phi v=\lambda v$. – Ma Ming Apr 25 '13 at 10:41
  • @Mathlete I don't see why - I think it's much cleaner to not use matrices. As Rhys points out in his answer, this also works in the infinite dimensional case. – mdp Apr 25 '13 at 10:42
  • That does look like a better proof. – Mathlete Apr 25 '13 at 10:43

2 Answers2

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$\phi$ 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 $0$ is always an 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 largest 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 $0$.

Community
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Rhys
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    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 $0$. – wj32 Apr 25 '13 at 10:48
  • Yes, I probably should have added that part; I will do so, with a different wording. – Rhys Apr 25 '13 at 10:58
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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 $0$.

user26857
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Somaye
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