Proof of the formula for the dimension of the orthogonal complement

Proof

We combine the orthogonal decomposition with Grassmann's formula.

1. Let $V$ be a finite-dimensional inner product space over $\R$ or $\C$. Let $W$ be a subspace of $V$, and let $W^\perp$ be its orthogonal complement.
2. The subspace $W$ is finite-dimensional, since $V$ is. So $V = W \oplus W^\perp$.
3. By the definition of the direct sum, $W + W^\perp = V$ and the only way to write $0 = u + u'$ with $u \in W$, $u' \in W^\perp$ is $u = u' = 0$.
4. Let $x \in W \cap W^\perp$. Then $x + (-x) = 0$ with $x \in W$ and $-x \in W^\perp$. By step 3, $x = 0$. So $W \cap W^\perp = \set{0}$.
5. The set $W^\perp$ contains $0$ and is closed under linear combinations, by linearity of the inner product in the first argument. So $W^\perp$ is a subspace, and it is finite-dimensional by the same result as in step 2.
6. By Grassmann's formula, $\dim(W + W^\perp) + \dim(W \cap W^\perp) = \dim W + \dim W^\perp$, where $\dim$ is the dimension.
7. By steps 3 and 4, the left side is $\dim V + 0$. Hence $\dim W + \dim W^\perp = \dim V$.