I was presented with this theorem: Let $\mathcal{S}$ be a subspace of $\mathbb{R}^n$. Then any $x\in\mathbb{R}^n$ can be uniquely written as the sum of some $x_1\in \mathcal{S}$ and $x_2 \in\mathcal{S}^\perp$ (where '$^\perp$' sign denotes orthogonal complement).
Proof presented: Let $\{v_i\}_{i=1}^k$ be a basis for $\mathcal{S}$ and $x\in\mathbb{R}^n$. Define $$x_1 = \sum\limits_{i=1}^k(x^Tv_i)v_i, \quad x_2 = x - x_1.$$ Then for each $1 \leq j \leq k$. $$x_2^Tv_j = x^Tv_j-x_1^Tv_j = x^Tv_j-x^Tv_j = 0$$
NOTE: This was only labeled as a 'partial proof' as what is left to show is the uniqueness of the decomposition.
What I am confused about is how does this show that our $x$ was able to be decomposed. I am a bit lost. Any help is appreciated! Thanks!
the idea of the space being a direct sum is A: Every element can be written as a sum of something in the first set and something in the second, and B: that way of writing it is unique.
For A: Note that since $v_i$ form a basis of $S$ and all of the $(x^Tv_i)$ are scalars, the definition of $x_1$ has it as a linear combination of vectors in $S$ and therefore $x_1\in S$. The second part shows that $x_2=x-x_1$ is in fact orthogonal to $x_1$, and therefore $x_2\in S^{\perp}$. This shows that there is A way of constructing something in $S$ and something in $S^\perp$ that add to $x$.
What is left is to assume you have another pair of elements, $x_3\in S$, $x_4\in S^\perp$ such that $x=x_3+x_4$ and then show that will force $x_3=x_1$ and $x_4=x_2$, which will show uniqueness