# Idempotent Matrix

1. Apr 18, 2010

### 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.

2. Apr 18, 2010

### 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