TITLE: Coordinate-Free Proof of the Orthogonal Decomposition Theorem
QUESTION [9 upvotes]: The Orthogonal Decomposition Theorem says that, if $V$ is a subspace of $\mathbb R^n$, then for any vector $v$ in $\mathbb R^n$,
$$
v = v_1 + v_2,
$$
where $v_1$ and $v_2$ are the projections of $v$ onto $V$ and $V^{\perp}$, respectively. The only proof that I have seen constructs a basis that can be partitioned into a basis for $V$ and a basis for $V^{\perp}$. Is there a proof that does not require constructing a basis but just uses coordinate-free geometric arguments?

REPLY [4 votes]: Here is a slightly different argument. Note that you need $V$ closed. We need two previous steps: 

$V+V^\perp$ is closed. Indeed, if $v_n+w_n$ is a Cauchy sequence with $v_n\in V$ and $w_n\in W$, we have 
$$
\|v_n-v_m\|^2\leq\|v_n-v_m\|^2+\|w_n-w_m\|^2=\|(v_n+w_n)-(v_m-w_m)\|^2. 
$$
So $\{v_n\}$ is Cauchy, and thus it converges to some $v\in V$. Similarly, $w_n\to w$ for some $w\in V^\perp$. So $v_n+w_n\to v+w$, and $V+V^\perp$ is closed. 
$(V+V^\perp)^\perp=\{0\}$. It is easy to check that $$(V+V^\perp)^\perp \subset V^\perp\cap V^{\perp\perp}=V^\perp\cap V=\{0\}.$$
It then follows that $V+V^\perp=\overline{V+V^\perp}=(V+V^\perp)^{\perp\perp}=\{0\}^\perp=H$.