Specifically, I'm trying to solve this problem.
Let $V$ be an $n$-dimensional inner-product space over $\mathbb{C}$, and let $U$ and $W$ be $m$-dimensional subspaces of $V$. Assume that $U$ contains a nonzero vector that is in $W^{\perp}$. Prove that $W$ contains a nonzero vector that is in $U^{\perp}$.
I thought I had a very simple solution, but I see now that it is incorrect. Below is my attempted solution and an explanation of why it's wrong.
Because $V$ is an inner-product space with finite dimensional subspaces $U$ and $W$, we have $V=U\oplus U^{\perp}=W\oplus W^{\perp}$. Thus, every vector in $V$ can be written uniquely (by definition of direct sum) as a sum of elements in $U$ and $U^{\perp}$ and as a sum of elements in $W$ and $W^{\perp}$. Thus, let $v:=x+y$, where $0\neq x\in U$, and $0\neq y\in U^{\perp}$. Furthermore, suppose also that $x\in W^{\perp}$. It follows from uniqueness that $y\in W$, as desired.
I started with a nonzero $x\in U\cap W^{\perp}$. Then I picked a nonzero $y\in U^{\perp}$ and let $v:=x+y$. Thus, I have that $x=\operatorname{proj}_U(v)$ and $y=\operatorname{proj}_{U^{\perp}}(v)$. I also have that $v=\operatorname{proj}_{W^{\perp}}(v)+\operatorname{proj}_W(v)$. This is where my original work got careless: I hastily reasoned that, since $x$ in an orthogonal projection of $v$ that is in $W^{\perp}$, it must be that $x=\operatorname{proj}_{W^{\perp}}(v)$. In other words, I made the following assumption: If $\operatorname{proj}_U(v)\in W^{\perp}$, then $\operatorname{proj}_U(v)=\operatorname{proj}_{W^{\perp}}(v)$. But this is not true in general.
Realizing my mistake, I'm now a bit lost on how to reach the desired conclusion.
Best Answer
Ed. Here is another proof which would prove directly $\operatorname{dim}(U∩W^{\perp})=\operatorname{dim}(W∩U^{\perp})$ if the dimension $n$ is a finite number. (You statement automatically follows)
Let $T=U\cap W^{\perp},V=W\cap U^{\perp}$, and decompose $U^{\perp}$ and $W$ as $U^{\perp}=V \oplus U_1$, $W=V \oplus V_1$. It's not hard to see that $U^{\perp}+W=U_1\oplus V \oplus W_1$, and $\operatorname{dim}(U^{\perp}+W)=n-\operatorname{dim}(V)$. Clearly $T$ is othogonal to $U^{\perp}+W$, so $U^{\perp}+W+T=(U^{\perp}+W)\oplus T$ is a subspace of $n$-dimensional space, so $\operatorname{dim}(U^{\perp}+W)+\operatorname{dim}(T) \le n$, or $n-\operatorname{dim}(V)+\operatorname{dim}(T) \le n$ $\implies$ $\operatorname{dim}(T) \le \operatorname{dim}(V)$. You can similarly prove that $\operatorname{dim}(T) \ge \operatorname{dim}(V)$. We therefore have $\operatorname{dim}(T) = \operatorname{dim}(V)$.
Below was my first/original proof, more an "elementary school style".
First, the following is true: if the intersection of two subspaces $X$ and $Y$ is $\{0\}$, assume ${a_1,...,a_m}$ and ${b_1,...,b_l}$ are the bases of $X$ and $Y$, then ${a_1,...,a_m,b_1,...,b_l}$ are linearly independent, i.e. if a linear combination of vectors ${a_1,...,a_m,b_1,...,b_l}$ is zero, then all coefficients must be zero. This implies that $\operatorname{dim}(X+Y)=\operatorname{dim}(X)+\operatorname{dim}(Y)$. Its proof should be trivial.
Second, if the statement is false, i.e. $W∩U^{\perp}=\{0\}$,then $\operatorname{dim}(W+ U^{\perp})=\operatorname{dim}(W)+\operatorname{dim}(U^{\perp})=\operatorname{dim}(U)+\operatorname{dim}(U^{\perp})=n$, so the bases of $W$ and $U^{\perp}$ should be linearly independent, and any vector can be expressed as the sum of a vector in W and a vector in $U^{\perp}$.
However, we know that $U$ contains a nonzero vector that is in $W^{\perp}$, say $z \in W^{\perp}$. We then can find $x \in W$ and $y \in U^{\perp}$ such that $z=x+y$. Let's check the inner product of $z$ with itself. $(z,z)=(x+y,z)=(x,z)+(y,z)$. We know that $(x,z)=0$ because $z \in W^⊥$ and $(y,z)=0$ because $z \in U$. Therefore $(z,z)=0$, which is clearly wrong.
Therefore $W$ contains at least one nonzero vector that is in $U^{\perp}$.
We can also prove that $\operatorname{dim}(U∩W^{\perp})=\operatorname{dim}(W∩U^{\perp})$ if the dimension $n$ is a finite number.