Linear transformation satisfies $T^n=T$; has eigenvalues? I have a linear transformation $T:V\to V$ over a (finite, is needed) field $F$, which satisfies $T^n=T$.
Prove that $T$ has a eigenvalue, or give a counter example.
Thanks

This is false. The matrix $$A = \left(\begin{matrix}0 & -1\\1 & 0 \end{matrix}\right)$$ satisfies $A^4 = 1$ (thus $A^5 = A$) but its characteristic polynomial $X^2+1$ has no real roots, so $A$ has no real eigenvalues.
For an example over a finite field, consider 
$$A = \left(\begin{matrix}0 & 1\\1 & 1 \end{matrix}\right)$$
over $\mathbb F_2$. It satisfies $A^3 = 1$ (thus $A^4 = A$) but the characteristic polynomial $X^2+X+1$ has no root in $\mathbb F_2$.
