Orthogonal projection and subspaces Consider the vector space $\mathbb{R}^m$ with usual inner product. Let $S_1$ and $S_2$ subspaces of $\mathbb{R}^m$ , $P_1\in\mathbb{M}_m(\mathbb{R})$ a orthogonal projection matrix on subspace $S_1$ and $P_2\in\mathbb{M}_m(\mathbb{R})$ a orthogonal projection on $S_2$. If $P_1P_2=P_2P_1=0$
Show that
i)$P_1+P_2$ is a orthogonal projection matrix
ii)$S_1$ and $S_2$ are orthogonal subspaces
iii)Show that $P_1+P_2$ is a orthogonal projection matrix on $W=S_1\oplus S_2$
What I think
i)$(P_1+P_2)^2=P_1^2+P_1P_2+P_2P_1+P_2^2=P_1^2+P_2^2=P_1+P_2$
ii)Here are starting problems, I think that is just show $\forall x\in S_1$ and $\forall y\in S_2$ $<x,y>=0$. But I do not know how to prove it formally
iii)I don't know how to proof

To show $M$ is an orthogonal projection matrix, it suffices that you demonstrate symmetry (for "orthogonal") and idempotency (for "projection"). For (i), you have done idempotency so you still need to do symmetry but that is straightforward (by assumption, both $P_1$ and $P_2$ are symmetric).
For (ii), suppose $x\in S_1$ and $y\in S_2$, then
$$
x'y=(P_1x)'(P_2y)=x'P_1'P_2y=x'P_1P_2y=x'0y=0.
$$
For (iii), again suppose $x\in S_1$ and $y\in S_2$. It is enough to show $(P_1+P_2)(x+y)=x+y$. This follows because:
\begin{align*}
(P_1+P_2)(x+y)&=P_1x+P_1y+P_2x+P_2y\\
&=P_1x+P_1(P_2y)+P_2(P_1x)+P_2y\\
&=x+0+0+y
\end{align*}
which of course is $x+y$.
