$\lambda^2$ 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

Hint:
$$0=\det (T^2-\lambda^2I) = \det(T-\lambda I)\det(T+\lambda I).$$
Or if you don't know $\det$ of a transformation, you can argue on the injectivity of $T-\lambda I$ and $T+\lambda I$.
