Let $V$ be an inner product space of finite dimension. $P_1, P_2$ the orthogonal projections on sub-spaces $W_1, W_2$.
Prove that if $P_1 P_2 = P_2 P_1$ then $P_1 P_2$ is the orthogonal projection on $W_1 \cap W_2$
Hint (could be irrelelevant to this part of the question, there are more parts): $(W_1 \cap W_2)^{\bot} = W_1^{\bot}+W_2^{\bot}$
My attempt was to first define an orthonormal basis to $W_1 \cap W_2$, and then to "complete" it to an orthonormal basis of $W_1$, and to a orthonormal basis of $W_2$, then I used the formula $P(v)=\sum \left \langle v, x_i \right \rangle x_i$, but I somehow got that $P_1 P_2 = P_2 P_1$ is always true, which makes no sense unless the question is misleading.
So I'm looking for a hint on how to approach this question. If necessary I can show how I got to $P_1 P_2 = P_2 P_1$
Best Answer
I am not sure how exactly the hint is supposed to be used (if it is indeed meant for this question), but the following should suffice.
Verify that $P_1P_2$ is self-adjoint with $(P_1P_2)^2 = P_1P_2$; it follows that $P_1P_2$ is an orthogonal projection onto its image. The image of $P_1P_2$ lies in the image of $P_1$ and in the image of $P_2$, so the image of $P_1P_2$ is a subspace of $W_1 \cap W_2$. On the other hand, we see that for $x \in W_1 \cap W_2$, $P_1P_2 x = x$, so the image is indeed all of $W_1 \cap W_2$. The conclusion follows.