Consider the subspace $\mathbb{R}^m$ with usual inner product.Let $S_1$ and $S_2$ subspaces of $R^m$, $P_1\in M_m(\mathbb{R})$ the orthogonal projection matrix on $S_1$ and $P_2\in M_m(\mathbb{R})$ the orthogonal projection matrix on $S_2$ with the property that $P_1P_2=P_2P_1=0_m$
Show that $P_1+P_2$ is a orthogonal projection matrix and that $S_1$ and $S_2$ are orthogonal.
The first part imagine that there might be some special property of orthogonal matrix, but do not know how to show. In the second part I need to show that $\forall u\in S_1$ and $\forall v\in S_2$ $<u,v>=0?$
Best Answer
I think @Alamos already gave a good proof though I think you may still need to verify the condition that $P_1+P_2$ is orthogonal.
A matrix $P$ is an orthogonal projection iff $$P^T= P, P^2 = P.$$
Since $P_1,P_2$ are orthogonal projection and $P_1P_2=P_2P_1 = 0$, verifying $(P_1+P_2)^2 = P_1+P_2$ and $(P_1+P_2)^T= P_1+P_2$ should not be too hard.
You do only need to show $\forall u\in S_1,v\in S_2, u^Tv =0$. Using the fact that $u = P_1u, v= P_2v$ (Since $P_1,P_2$ projects $\mathbb{R}^m$ on to $S_1,S_2$) and $P_1 P_2= P_1P_2=0$ and compute $u^Tv =(P_1u)^T(P_2v)=u^TP_1^TP_2v=u^TP_1P_2v$. You should be able to get the result.