Composition of projections If $P$ in $L(X,Y)$ and $Q$ in $L(Y,Z)$ are projections, Can we conclude that $QP$  is also a projection ? Thank you !
 A: Even for orthogonal projections on a Hilbert space the answer is NO: Consider
in $\mathbb R^2$ the orthogonal projection $P$ on the $x$-axis and $Q$ the
orthogonal projection onto the diagonal. Draw a picture to see what happens with
the orbits $\lbrace (PQ)^n x: n\in\mathbb N_0\rbrace$.
A: As mentioned by Jochen, this is not always true. However, there is a nice condition that definitely ensures that a composition of two projections is a projection. Here it is (taken from http://www.polishedproofs.com/necessary-and-sufficient-conditions-for-equality-of-composition-of-projections-and-one-of-its-factors/ )
Let $V_1, V_2$ Be closed subspaces of the Hilbert space $H$ and let $p_{V_1}$ and $p_{V_2}$ denote the orthogonal projections on $V_1$ and $V_2$, respectively. Show that
$$p_{V_1}\cdot p_{V_2}=p_{V_1} \quad \text{if and only if} \quad V_1 \subset V_2$$
First assume that $p_{V_1}\cdot p_{V_2}=p_{V_1}$ $\forall x \in H$. Then for any $x_0 \in V_2^{\perp}$,
\begin{equation}
\begin{split}
p_{V_1}(x_0)
& = p_{V_1}\left(p_{V_2}(x_0)+p_{{V_2}^{\perp}}(x_0)\right)\\
& = p_{V_1}p_{V_2}(x_0)+p_{V_1}p_{{V_2}^{\perp}}(x_0)\\
& = p_{V_1}p_{V_2}(x_0)+p_{V_1}(x_0)\\
\end{split}
\end{equation}
Since $p_{V_1}\cdot p_{V_2}=p_{V_1}$ for all $x \in H$, this simplifies to
$$p_{V_1}(x_0)=0$$
Thus, $x_0 \perp V_1$. Since $x_0 \in V_2^{\perp}$ was arbitrary, that means that
$$V_2^{\perp} \perp V_1$$
Since $V_2$ is closed, and since orthogonal complement of an orthogonal complement of a subspace S of a Hilbert space is the closure of S, we can conclude that
$$V_1 \subset V_2^{\perp \perp}=\overline{V_2}=V_2$$
Conversely, assume that $V_1 \subset V_2$. Then we know that
$V_2^{\perp} \subset V_1^{\perp}$. Now let $x \in H$ be arbitrary. Then
\begin{equation}
\begin{split}
p_{V_1}(x)
& = p_{V_1}\left(p_{V_2}(x)+p_{{V_2}^{\perp}}(x)\right)\\
& = p_{V_1}p_{V_2}(x)+p_{V_1}p_{{V_2}^{\perp}}(x)\\
\end{split}
\end{equation}
But $p_{{V_2}^{\perp}}(x) \in {V_2}^{\perp} \subset V_1^{\perp}$. Since $p_{V_1}$ restricted to $V_1^{\perp}$ is zero, we know that $p_{V_1}p_{{V_2}^{\perp}}(x)=0$, and since $x$ was arbitrary,
$$p_{V_1}\cdot p_{V_2}=p_{V_1},$$
as desired.
