# 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

## 2 Answers

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 $x+y$.

• This is sufficient to show that it is direct sum? $W=S_1\oplus S_2$ – Roland May 24 '15 at 16:21
• By (ii), you have shown $S_1$ is orthogonal to $S_2$ so that $S_1\oplus S_2=\{x+y:x\in S_1\text{ and }y\in S_2\}$. I should have added 1 additional thing: for any $v\in\mathbb{R}^m$, we can write $(P_1+P_2)v=x+y$ with $x=P_1v\in S_1$ and $y=P_2v\in S_2$. – Kim Jong Un May 24 '15 at 19:35

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.