$\lambda^2$ is an eigenvalue of $T^2$ Let $T:V\rightarrow V$ be a linear map. If $\lambda^2$ is an eigenvalue of $T^2$, then $\lambda$ or $-\lambda$ is an eigenvalue of T.
I have then $(T\circ T )(v) = T^2 (v) = \lambda^2 v$, but $\lambda^2 = (-\lambda)(-\lambda)$ or $(\lambda)(\lambda)$.
Any hints

Suppose $T^2 v = \lambda^2 v$, then
$(T-\lambda I)(T+\lambda I) v = 0$.
If $(T+\lambda I) v = 0$ then $-\lambda$ is an eigenvalue corresponding
to eigenvector $v$, otherwise
$\lambda$ is an eigenvalue corresponding to eigenvector $(T+\lambda I) v$.
