To find orthogonal complement of a subset in inner product space.

Question

Let $V$ be a inner product space and $S$ be a subset of $V$, then how do I show the intersection of S with its orthogonal complement in V is $\{0\}$. If I take $V = \mathbb{R}$ and $S=\{1,2\}$, then the orthogonal complement of $S$ is $\{0\}$ and so, the intersection between them is empty, so how do I show the intersection of $S$ with its orthogonal complement in $V$ is $\{0\}$.

Answer 1

Every element $v$ in a subspace $S$ can be represented by its basic vectors $s_1,\ldots,s_N$ as

$$v=\sum_{i=1}^N a_i s_i$$

Equivalently, if $v$ lies in $S^{\perp}$, then

$$v=\sum_{j=1}^N b_j s^{\perp}_j$$

holds. Thus, $$\sum_{i=1}^N a_i s_i=\sum_{j=1}^N b_j s^{\perp}_j$$

In other words, each $s_i$ can be generated by the sum of $s^{\perp}_i$. More precisely,

$$s_i=\sum_{j=1}^N \langle s_i, s^{\perp}_j\rangle s^{\perp}_j$$

But because $S^\perp$ is orthogonal to $S$, the inner product is zero, thus to have the equality, all $a_i$ and $b_i$ are zero.

Combining all steps, this means $$v=\sum_{i=1}^N 0 \cdot s_i = 0$$ thus the intersection is only the $0$ vector.