Square root of an eigenvalue of linear map $T^2$ I was wondering if anyone could help me with this question, not really sure where to start:
Show that if $\lambda^2$ is an eigenvalue of $T^2 = T \circ T$, then either $\lambda$ or $-\lambda$ is an eigenvalue of T.

Let $\lambda_i$ the $n$ eigenvalues of $T$ (counted with their multiplicity), so that we have:
$$
Tx_i=\lambda_ix_i
$$
for the eigenvectors $x_i$.
Note that the $n$ eigenvalues of $T^2$ are $\alpha_i=\lambda_i^2$ because:
$$
T^2x_i=T(Tx_i)=T(\lambda_i x_i)=\lambda_i(Tx_i)=\lambda_i^2x_i
$$
so, if $\alpha_i=\lambda^2$ than $\lambda_i= \lambda$ or $\lambda_i= -\lambda$.
