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In a finite dimensional vector space, if $0$ is an eigenvalue and the only eigenvalue of a linear operator, is that operator nilpotent?

There is this post which shows the other direction.

Prove that the only eigenvalue of a nilpotent operator is 0?

I would think the question would be posed as "iff" to the extent the answer to my question is affirmative.

To the extent that is not the case, I would please appreciate an example to that effect.

Thanks

Community
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    Which field are the eigenvalues allowed to come from? – Jonas Meyer Apr 05 '17 at 14:14
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    Take a look at: http://math.stackexchange.com/questions/256007/if-a-matrix-has-only-zero-as-an-eigen-value-then-it-is-nilpotent – StackTD Apr 05 '17 at 14:15
  • @JonasMeyer Thanks - didn't think of that –  Apr 05 '17 at 14:20
  • @JonasMeyer As I mentioned in a comment below: Algebraically closed seems to be key. In looking it up, over the reals, a characteristic polynomial such as $x^2+1$ is a problem. Thanks for the nice learning experience. With regards, –  Apr 05 '17 at 15:09

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If your field is algebraically closed (i.e. if we're including complex eigenvalues/eigenvectors), then the answer is yes.

If $0$ is the only eigenvector of the operator $A$, then $A$ has characteristic polynomial $p(x) = x^n$. By the Cayley-Hamilton theorem, $A^n = 0$.

On the other hand: if we're only including real eigenvalues, then we can say that the operator $$ \pmatrix{0&-1&0\\1&0&0\\0&0&0} $$ has zero as its only eigenvalue but fails to be nilpotent.

Ben Grossmann
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  • Do you mean "e.g." instead of "i.e.", or do you only want to mention complex rather than all algebraically closed fields? – Jonas Meyer Apr 05 '17 at 14:17
  • @JonasMeyer I thought about that. It is very likely that they're only working with real or complex numbers, so I'm leaving it as it is unless they ask for clarification. – Ben Grossmann Apr 05 '17 at 14:19
  • Actually the linked question was in Axler in the chapter of operators on a complex v.s., so I slept through that qualification when I asked my question. So I appreciate the distinction you and Jonas made. –  Apr 05 '17 at 14:28
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    @Andrew something that Axler doesn't spend much time with (which Jonas alludes to in his comment on my answer) is that there are fields besides the real numbers and complex numbers, hence my opening statement about "algebraically closed fields". – Ben Grossmann Apr 05 '17 at 14:30
  • Algebraically closed seems to be key. In looking it up, over the reals, a characteristic polynomial such as $x^2+1$ is a problem. –  Apr 05 '17 at 15:08
  • Looking it up... or looking at the second part of my answer – Ben Grossmann Apr 05 '17 at 16:00