Let $U,W$ be finite-dimensional subspaces. Prove that $P_U P_W =0$ if and only if $\langle u,w\rangle = 0$ for all $u\in U, w\in W.$

Question

This is question 6, page 201 of "Linear Algebra Done Right" by Axler (3rd Edition), which is a homework problem for my Linear Algebra course.

Suppose $U$ and $W$ are finite-dimensional subspaces of $V$. Prove that $P_U P_W =0$ if and only if $\langle u,w\rangle = 0$ for all $u\in U, w\in W.$

Proof. ($\implies$) Suppose $P_U P_W=0$, and let $v\in V.$ Then $P_U P_W v=0$ iff $P_W v \in U^\bot,$ and $P_W v \in U^\bot$ iff $W\subseteq U^\bot,$ which implies $\langle u,w\rangle =0$ for all $u\in U, w\in W.$

($\Longleftarrow$) Conversely, suppose $\langle u,w\rangle =0$ for all $u\in U, w\in W.$ Then by definition $W\subseteq U^\bot.$ Now let $v\in V,$ and then $v=w+w'$ for some $w\in W, w' \in W^\bot.$ Since $W\subseteq U^\bot$ therefore $w\in U^\bot$ so $P_U P_W v = P_U w = 0.$ Hence $P_U P_W = 0.$ $\Box$

Is this correct?

Answer 1

For the $\implies$ direction, when you say "let $v\in V$" it gives the impression you're talking about just one such vector, but then "$P_Wv\in U^\perp$ iff $W\subseteq U^\perp$" only makes sense when you're implicitly quantifying over all such vectors. Just knowing $P_Wv\in U^\perp$ for a single vector $v\in V$ is not sufficient to conclude $W\subseteq U^\perp$. Also you are implicitly using the fact the range of $P_W$ is all of $W$, so quantifying over all $v\in V$ can effectively be turned into a statement quantifying over all $w\in W$. This could be clearer.

Otherwise, correct.