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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

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    $\begingroup$ Which field are the eigenvalues allowed to come from? $\endgroup$ – Jonas Meyer Apr 5 '17 at 14:14
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    $\begingroup$ Take a look at: math.stackexchange.com/questions/256007/… $\endgroup$ – StackTD Apr 5 '17 at 14:15
  • $\begingroup$ @JonasMeyer Thanks - didn't think of that $\endgroup$ – user12802 Apr 5 '17 at 14:20
  • $\begingroup$ @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, $\endgroup$ – user12802 Apr 5 '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.

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  • $\begingroup$ Do you mean "e.g." instead of "i.e.", or do you only want to mention complex rather than all algebraically closed fields? $\endgroup$ – Jonas Meyer Apr 5 '17 at 14:17
  • $\begingroup$ @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. $\endgroup$ – Ben Grossmann Apr 5 '17 at 14:19
  • $\begingroup$ 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. $\endgroup$ – user12802 Apr 5 '17 at 14:28
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    $\begingroup$ @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". $\endgroup$ – Ben Grossmann Apr 5 '17 at 14:30
  • $\begingroup$ Algebraically closed seems to be key. In looking it up, over the reals, a characteristic polynomial such as $x^2+1$ is a problem. $\endgroup$ – user12802 Apr 5 '17 at 15:08

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