Coordinate-Free Proof of the Orthogonal Decomposition Theorem 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?
 A: A coordinate-free version of the orthogonal decomposition theorem is stating that $V$ and $V^\bot$ are complementary, i.e. $\Bbb R^n=V\oplus V^\bot$.
I don't know what your definition of $V^\bot$ is, but if you defined it as the kernel of the orthogonal projection $\pi_v:\Bbb R^n\rightarrow V$ onto $V$, then it is known that the range and kernel of a projection are complementary, which proves the theorem. This is detailed a bit here: https://en.wikipedia.org/wiki/Projection_(linear_algebra)#Complementarity_of_range_and_kernel
A: 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 $V+V^\perp=\overline{V+V^\perp}=(V+V^\perp)^{\perp\perp}=\{0\}^\perp=H$. 
A: You might appreciate the proof strategy used in Pedersen's Analysis Now! First, he proves the following:

Lemma 3.1.6: If $\mathfrak C$ is a closed, non-empty, convex subset of a Hilbert space $\mathfrak H$, there is for each $y \in \mathfrak H$ a unique $x \in \mathfrak C$ that minimizes the distance from $y$ to $\mathfrak C$.

From there, we have the desired statement:

Theorem 3.1.7: For a closed subspace $\mathfrak X$ of a Hilbert space $\mathfrak H$, let $\mathfrak X = \{x^\perp \in \mathfrak H : x^\perp \perp \mathfrak X\}$. Then each vector $y$ in $\mathfrak H$ has a unique decomposition $y = x + x^\perp$, with $x \in \mathfrak X$ and $x^\perp \in \mathfrak X^\perp$. The element $x$ (respectively $x^\perp$) is the nearest point in $\mathfrak X$ (respectively $\mathfrak X^\perp$) to $y$. Moreover, $\mathfrak H = \mathfrak X \oplus \mathfrak X^\perp$ and $\mathfrak X^{\perp \perp} = \mathfrak X$.


The fact that we're proving these statements in $\Bbb R^n$ rather than in an arbitrary Hilbert space makes our job a bit easier. This wikipedia page gives the following proof of the lemma, as it applies to $\Bbb R^n$.  Note that $K$ (instead of $\mathfrak C$) denotes our convex set.
Let $\delta = \inf \{ \|x\| : x \in K \}.$ Let $x_j$ be a sequence in $K$ such that $\|x_j\| \to \delta$. Note that $(x_i + x_j)/2$ is in $K$ since $K$ is convex and so $\|x_i + x_j\|^2 \ge 4 \delta^2$. Since
$$
\|x_i - x_j\|^2 = 2\|x_i\|^2 + 2\|x_j\|^2 - \|x_i + x_j\|^2 \le 2\|x_i\|^2 + 2\|x_j\|^2 - 4\delta^2 \to 0
$$
as $i, j \to \infty$, $x_i$ is a Cauchy sequence and so has limit $x$ in $K$. It is unique since if $y$ is in $K$ and has norm $\delta$, then $\|x - y\|^2 \le 2\|x\|^2 + 2\|y\|^2 - 4\delta^2 = 0$ and $x=y$. $\square$
With that established, the result you're interested in can be proved as follows.
Given a point $v \in \Bbb R^n$, take $v_1$ to be the point of $V$ that is closest to $v$ (whose existence is guaranteed by the lemma). Let $v_2 = v - v_1$.  For any $w \in V$ and $\epsilon > 0$, we have
$$
\begin{align}
\|v_2\|^2 &= \|v - v_1\|^2 \leq \| v - (v_1 + \epsilon w)\|^2
\\ & = \|v_2 - \epsilon w\|^2 = \|v_2\|^2 - 2 \epsilon \,v_2^Tw + \epsilon^2 \|w\|^2.
\end{align}
$$
That is, we have 
$$
\|v_2\|^2 \geq \|v_2\|^2 - 2 \epsilon \,v_2^Tw + \epsilon^2\|w\|^2 \implies
2\epsilon\, v_2^Tw \leq \epsilon^2\|w\|^2
$$ 
for every $\epsilon > 0$. Thus, we must have $v_2^Tw \leq 0$ for every $w \in V$. Since $w \in V \implies -w \in V$, we conclude $v^Tw = 0$ for every $w \in V$. That is, $v_2 = v - v_1$ is an element of $V^\perp$, as was desired. $\square$
