# If $0$ is the only eigenvalue of a linear operator, is the operator nilpotent

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

• Which field are the eigenvalues allowed to come from? – Jonas Meyer Apr 5 '17 at 14:14
• Take a look at: math.stackexchange.com/questions/256007/… – StackTD Apr 5 '17 at 14:15
• @JonasMeyer Thanks - didn't think of that – user12802 Apr 5 '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, – user12802 Apr 5 '17 at 15:09

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.
• Algebraically closed seems to be key. In looking it up, over the reals, a characteristic polynomial such as $x^2+1$ is a problem. – user12802 Apr 5 '17 at 15:08