Proof that a space is the direct sum of a finite-dimensional subspace and its orthogonal complement

Proof

We use an orthonormal basis of $W$ to split each vector, and then show that $W \cap W^\perp = \set{0}$.

1. Let $V$ be an inner product space over $\mathbb{K} = \R$ or $\C$, with inner product $\ip{\cdot}{\cdot}$. Let $W$ be a finite-dimensional subspace of $V$, and let $W^\perp$ be its orthogonal complement.
2. For all $u, w \in V$ and $a \in \mathbb{K}$, conjugate symmetry and linearity in the first argument give $\ip{u}{a w} = \overline{a} \ip{u}{w}$. In the same way, $\ip{u}{\cdot}$ is additive.
3. $W^\perp$ is a subspace. Indeed, $\ip{0}{w} = 0$ for all $w$. If $x, y \in W^\perp$ and $a, b \in \mathbb{K}$, then $\ip{a x + b y}{w} = a \ip{x}{w} + b \ip{y}{w} = 0$ for all $w \in W$.
4. The restriction of $\ip{\cdot}{\cdot}$ to $W$ is an inner product on $W$. So $W$ is a finite-dimensional inner product space, and it has an orthonormal basis $e_1, \dots, e_k$.
5. Let $v \in V$. Set $w = \sum_{i=1}^{k} \ip{v}{e_i} e_i$, which lies in $W$, and $w' = v - w$. Then $v = w + w'$.
6. For each $j$, linearity in the first argument and orthonormality give $\ip{w}{e_j} = \ip{v}{e_j}$. So $\ip{w'}{e_j} = 0$.
7. Let $u \in W$. By the formula for coordinates in $W$, $u = \sum_j a_j e_j$ with $a_j \in \mathbb{K}$. By step 2, $\ip{w'}{u} = \sum_j \overline{a_j} \ip{w'}{e_j} = 0$.
8. Hence $w' \in W^\perp$. So $V = W + W^\perp$.
9. Let $x \in W \cap W^\perp$. Since $x \in W^\perp$ and $x \in W$, we get $\ip{x}{x} = 0$. By positivity of the inner product, $x = 0$.
10. Let $u \in W$ and $u' \in W^\perp$ with $u + u' = 0$. Then $u = -u'$ lies in $W$ and in the subspace $W^\perp$. By step 9, $u = 0$, and so $u' = 0$.
11. By steps 8 and 10 and the equivalent form of the definition of the direct sum, $V = W \oplus W^\perp$.