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.
Best Answer
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 $\lambda = -1,0,1$.