If $p$, $q$ are two orthogonal projections, then all eigenvalues of $q \circ p$ are in $[0,1]$ Let $E$ be an euclidean vector-space.

Prove that if $p$, $q$ are two orthogonal projections, then all
  complex eigenvalues of $q \circ p$ are in $[0,1]$.

A projector is orthogonal iff it is symmetric.
If $p\circ q$ =  $q\circ p$, then the eigenvalues are in $\{0,1\}$.
Do you have hint for the general case? Thanks
 A: Suppose $PQ x= \lambda x$, since $P$ is a projection we have $PQx = \lambda Px= \lambda x$.
Hence either $\lambda = 0$ or $x=Px$, in which case we have $PQPx = \lambda x$.
Since $P,Q$ are orthogonal, we have $P^*=P,Q^*=Q$ and so $(PQP)^* = PQP$ and
hence any eigenvalue is real.
It follows that any eigenvalue of $PQ$ is real.
Since any non zero eigenvalue is also an eigenvalue of $PQP=PQQP \ge 0$, we see that
the eigenvalues are non negative.
Since $\|PQ\| \le \|P\| \|Q\| \le 1$ we see that all the eigenvalues are in $[0,1]$.
A: Let $P$ and $Q$ be two orthogonal projections and let $\langle\cdot,\cdot\rangle$ be the usual inner product of a finite dimensional Euclidean space i.e. if $x,y\in E$ then 
$$\langle x,y\rangle:=x_1\overline{y}_1+...+x_n\overline{y}_n$$
where $\overline{y}_k$ denotes the complex conjugate of $y_k$ for each $k$. Let $x\in E$ be a eigenvector of the operator $PQ$ i.e. $PQx=\lambda x$. Then 
$$|\lambda|^2||x||^2=|\lambda|^2\langle x,x\rangle=\langle\lambda x,\lambda x\rangle=\langle PQx,PQx\rangle$$
Using the fact that both $P$ and $Q$ are idempotent ($P^2=P, Q^2=Q$) and self-adjoint then we get
$$\langle PQx,PQx\rangle=\langle Qx,P^*PQx\rangle=\langle Qx,P^2Qx\rangle=\langle Qx,PQx\rangle=\langle Qx,\lambda x\rangle=\overline{\lambda}\langle Qx,x\rangle$$
Analogue we obtain
$$\langle PQx,PQx\rangle=\langle P^*PQx,Qx\rangle=\langle P^2Qx,Qx\rangle=\langle PQx,Qx\rangle=\langle \lambda Q^*x,x\rangle=\lambda \langle Qx,x\rangle$$
Therefore we get $$|\lambda|^2||x||^2=\overline{\lambda}\langle Qx,x\rangle=\lambda \langle Qx,x\rangle$$
Therefore it must be the case that $\lambda\in\mathbb{R}$. Next notice that any $x\in E$ can be written as $x=Qx+(I-Q)x$ (the orthogonal decomposition of a vector) hence 
$$\langle Qx,x\rangle=\langle Qx,Qx+(I-Q)x\rangle=\langle Qx,Qx\rangle=||Qx||^2$$ this in turn implies $\lambda\geqslant 0$ since $0\leqslant|\lambda|^2||x||^2=\lambda \langle Qx,x\rangle=\lambda||Qx||^2$. From this last equation another thing we obtain is $$\lambda ||x||^2=||Qx||^2$$ whenever $\lambda>0$. Hence
$$\lambda=\frac{||Qx||^2}{||x||^2}\leqslant \sup_{x\neq 0}\frac{||Qx||^2}{||x||^2}=||Q||^2=1^2=1$$
Note any orthogonal projection operator has operator norm equal to unity. Thus if $\lambda$ is an eigenvalue of $PQ$ then $0\leqslant\lambda\leqslant 1$.
A: This actually holds for two orthogonal projections $p,q$ in a $C^*$-algebra.
Namely, we have
$$\sigma(pq) = \sigma(pq^2) = \sigma((pq)q) \subseteq \sigma(q(pq)) \cup \{0\} \subseteq [0,+\infty\rangle$$
since $qpq = qppq = q^*p^*pq = (pq)^*(pq) \ge 0$ and hence has spectrum in $[0,+\infty\rangle$.
On the other hand, we have
$$\|pq\| \le \|p\|\|q\| \le 1$$
so $\sigma(pq)$ is contained in the unit disk around the origin. We conclude that $\sigma(pq) \subseteq [0,1]$.
