Idempotent Matrix: Eigenvalue Must be 0 or 1

[Dustinsfl]

An nxn matrix A is said to be idempotent if A^2=A. Show that if \lambda is an eigenvalue of an idempotent matrix, then \lambda must be 0 or 1.

The only reason I can think of is that it must 0 or 1 because if you square the values 0 and 1 don't change.

 

[Jakesaccount]

Alright so let x be an eigenvalue to A. That means that Av=xv for some vector v. Then xv=Av=A^2v=A(Av)=A(xv)=x(Av)=x^2v which implies x^2=x so x(x-1)=0 which implies that the eigenvalues to A are either 1 or 0