The assertion is as follows:"If $\lambda$ is an eigenvalue of a matrix A for which $A = A^2$ then $\lambda = 0 $ or $ \lambda = 1$. Prove if true or use a counter example.
My intuition is to first find a matrix A that satisfies the constraint: $$ A = AA \\ A^{-1}A = A^{-1}AA \\ I = A $$
I solved for the eigenvalues and found that $\lambda = 1$, however I am not sure how to end off the proof by showing that $\lambda = 0$ is also an eigenvalue. Any help would be appreciated.
Let $x$ be an eigenvector associated with $\lambda$, then one has: $$Ax=\lambda x\tag{1}.$$ Multiplying this equality by $A$ leads to: $$A^2x=\lambda Ax.$$ But since $A^2=A$ and $Ax=\lambda x$, one has: $$Ax=\lambda^2x\tag{2}.$$ According to $(1)$ and $(2)$, one gets: $$(\lambda^2-\lambda)x=0.$$ Whence the result, since $x\neq 0$.
Let $v$ be an Eigenvector associated to $\lambda$.
$$Av=\lambda v$$
and
$$AAv=\lambda^2v=Av=\lambda v$$
so that $$\lambda^2=\lambda.$$
Another way to prove this is to use the fact that if $\lambda$ is an eigenvalue of $A$, then $f(\lambda)$ is an eigenvalue of $f(A)$. So take $f(x) = x^2 - x$. Then we have that $\lambda^2 - \lambda$ is an eignevalue of $f(A) = A^2 - A = 0$. But all eigenvalues of the zero matrix are $0$, hence we must have $\lambda^2 - \lambda = 0$, i.e. $\lambda = 1$ or $0$
