How can I prove this property about eigenvalue?

Question

Let $T$ be an operator in B(X). If $T^3=T$ and $\lambda$ is an eigenvalue of T, then can we prove that $\lambda^3=\lambda$?

thank you for your help.

Answer 1

You can get from the equality that the polynomial $p(x)=x^3-x = x(x-1)(x+1)$ getting the value zero when $T$ is set.

Therefore, $T$'s minimal polynomial, $m$, is a divisor of $p$, and since every eigenvalue of $T$ nullifies $m$, you can get that every eigenvalue of $T$ also nullifies $p$ which mean that every eigenvalue $\lambda$ of $T$ is from the set ${0,1,-1}$, which anyway satisfies $\lambda ^3 = \lambda.$

Answer 2

Take an eigenvector $v$ and compute: $$T^3 v = T(T(T(v))) = T(T(\lambda v) = T(\lambda^2 v) = \lambda^3 v$$ and $$T v = \lambda v$$ Since $T^3 = T$ you get $\lambda^3 = \lambda$.