# Eigenvalues of a linear transformation

So I've been given a question that basically looks like this: Let $$T:V \to V$$ be a linear transformation with $$T^3 = T \circ T \circ T = T$$, Show that $$T$$ can only have eigenvalues from the set $$\{-1,0,1\}$$.

So I understand the question and how to calculate eigenvalues 'normally' but in this case the problem I have is what to actually write. I get the idea of the answer (I think) which is basically that only the values of $$-1$$, $$0$$ and $$1$$ can be cubed and get the same value back ($$-1$$ cubed is $$-1$$, $$0$$ cubed is $$0$$, etc.). I'm pretty much just not sure how I would 'prove' this to show what I need to show. Thanks in advance.

Let $$\lambda,x$$ be an eigenvalue/vector pair of $$T$$. Then $$T x = \lambda x$$. Then note that $$\lambda x = Tx = T^3 x = T^2 (Tx) = T^2 (\lambda x) = \lambda T(Tx) = \lambda^2 T x = \lambda^3 x$$.
Therefore $$\lambda^3 = \lambda$$ or equivalently $$\lambda(\lambda^2-1)=0$$ with solutions $$\lambda = -1,0,1$$.
Your idea is correct. The underlying idea is that if $$\lambda$$ is an eigenvalue of $$T$$ then $$\lambda^3$$ is an eigenvalue of $$T^3$$. Can you prove this?