Product of two symmetric matrices with eigenvalues all 0 or 1 is idempotent Assume $A$, $V$ are symmetric and $V$ is positive definite. 
If $AV$ has eigenvalues that are all zero or one, show $AV$ is idempotent.
My proof so far (haven't used the symmetric property or the fact that $V$ is positive definite - where do those come in?):
If the eigenvalue $\lambda=0$, and assume there exists a nonzero $u$ such that 
$$AVu=\lambda u=0u=0$$
$$(AV)^2u=AVAVu=AV0=0$$
Then $(AV)^2u=AVu=0$ implies $(AV)^2=AV$ since $u\ne 0$.
If $\lambda=1$, and assume there exists nonzero $u$ such that 
$$AVu=\lambda u=u$$
$$(AV)^2u=AVAVu=AVu$$
$$\implies (AV)^2=AV$$
 A: As I stated in my comment, it is sufficient to prove that $AV$ is diagonalizable.  Note that if $M$ is a matrix whose eigenvalues are $0$ and $1$, then $M$ is idempotent if and only if $M$ is diagonalizable.
Proof that $A$ is idempotent $\implies$ $A$ is diagonalizable:
Suppose $A$ is idempotent. Then $A^2 - A = 0$.  It follows that the minimal polynomial of $A$ divides $x^2 - x = x(x-1)$. It follows that $A$ is diagonalizable (why?)
I will leave it to you to prove the converse, which is necessary for the proof of your statement (Hint: if $A = SDS^{-1}$, what does $A^2$ look like?  If $\lambda = 0,1$ are the only eigenvalues, what does $D$ look like?).
With that in mind...
Proof that $AV$ is diagonalizable
Because $A$ is positive definite, it has a unique positive definite square root given by $R = A^{1/2}$.  We note that
$$
R^{-1}AVR= A^{-1/2}AVA^{1/2} = A^{1/2}VA^{1/2} = RVR
$$
We note that $(RVR)^*=R^*V^*R^* = RVR$ (${}^*$ here denotes the conjugate transpose). It follows that $RVR$ is diagonalizable, i.e., that $RVR = S D S^{-1}$ for some invertible matrix $S$.
Thus, we have $R^{-1}(AV)R = S^{-1}DS$, which we rearrange to find 
$$
AV = (RS^{-1})D(RS^{-1})^{-1}
$$
Thus, we conclude that $AV$ is diagonalizable, as desired. Since $AV$ has only $0$ and $1$ as its eigenvalues, we may further conclude that it is idempotent.
