Let $A$ be a square $n \times n$ matrix such that $A^2 = 2A$. Show that $A$ has only two distinct eigenvalues.

Question

I have tried the following:

$Ax = \lambda x$

$A^2x = \lambda A x $

$2Ax = \lambda A x$

But not sure where to go from here, or if this even is the right approach. Thanks to anyone who helps.

Answer 1

Let $P(A)=0$ where $P(X)=X^2-2X$.

This implies that the eigenvalues of $A$ are $0,2$

If $c$ is an eigenvalue associated to the vector $x$, $A(x)=cx, A^2(x)=A(cx)=cA(x)=c^2x=2A(x)=2cx$ implies that $(c^2-2c)x=0$.