If $A \in M(n , \mathbb R) $ be such that $A^2=A$ , then is it true that $rank (A)=Trace(A)$ ?

Question

If $A \in M(n , \mathbb R) $ be such that $A^2=A$ , then is it true that $rank (A)=Trace(A)$ ? What I have done is to just observe that all the eigenvalues must be either $1$ or $0$ ; but no headway any further . Please help