If $V$ and $W$ are subspaces of the same dimension such that $V$ meets $W^\perp$, then $W$ meets $V^\perp$ I'm having a little difficulty understanding the proof for orthogonal complements.  I kind of understand orthogonal complements, but I cannot seem to find a logic to this.  I'm trying to follow along with theorems and definitions, but I'm not far.  Any advice on how to go about his proof problem?

Let V and W be two subspace of $\mathbb{R}^n$ of the same dimension, $m$.  Assume $\exists \mathbf{v} \in V \setminus \left \{ 0 \right \}$ such that $\mathbf{v} \in W^{\bot}$.
(a) Prove that $\exists \mathbf{w} \in W \setminus \left \{ 0 \right \}$ such that $\mathbf{w} \in V^{\bot}$.
(b) Show by an example that the conclusion of part (a) may not hold if V and W do not have the same dimension.


We are given two subspaces $V$ and $W$ of $\Bbb R^n$ such that $V\cap W^\perp\neq\{\vec 0\}$ and $\dim V=\dim W$. It follows that
\begin{align*}
\dim\left(W\cap V^\perp\right)
&= \dim\left(\left(W\cap V^\perp\right)^{\perp\perp}\right) \\
&= \dim\left(\left(W^\perp+V\right)^\perp\right) \\
&= n-\dim\left(W^\perp+V\right) \\
&= n-\left(\dim\left(W^\perp \right)+\dim(V)-\dim\left(W^\perp\cap V\right)\right) \\
&= n-\left(\dim\left(V^\perp \right)+\dim(V)-\dim\left(W^\perp\cap V\right)\right) \\
&= n-\left(n-\dim\left(W^\perp\cap V\right)\right) \\
&= \dim\left(W^\perp\cap V\right) \\
&>0
\end{align*}
Do you see why each equality holds?
