Orthogonal projection and subspaces

Question

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

Answer 1

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 $x+y$.

Answer 2

Hint: The kernel of $P_2$ is $S_2^\perp$. Let $x\in S_1$ and $y\in S_2$. Then $0=P_2 P_1 x=P_2 x$, so $x\in Ker P_2=S_2^\perp$. In particular, $x$ is orthogonal to $y$.

Using the same logic, $P_1+P_2$ leaves $S_1\oplus S_2$ fixed and sends everything orthogonal to $S_1$ and $S_2$ to zero.