# Let V be an inner product space, S and $S_0$be subsets of V, and W be a finite-dimensional subspace of V.

Let V be an inner product space, S and $$S_0$$be subsets of V, and W be a finite-dimensional subspace of V.

Prove:$$S_0 \subset S$$ implies $$S^{\bot} \subset S_0^{\bot}$$

Pf: Let $$x \in S^{\bot}$$, for all $$y \in S$$, the inner product $$\langle x,y \rangle$$ is given $$\langle x,y \rangle=0$$, which implies $$y \in S_0$$ Then $$S_0 \subset S$$, $$x \in S_0^{\bot}$$.Proved.

Prove: $$W=(W^{\bot})^{\bot}$$

Pf: Let $$x \in W$$ and $$y \in W^{\bot}$$. By definition, for all $$y \in W^{\bot}$$, the inner product is given by $$\langle x,y \rangle=0$$. Using $$x \in W$$, this gives $$W \subset (W^{\bot})^{\bot}$$ How to prove the other side?

The converse is trickier, and not necessarily true if $$W$$ is not finite-dimensional (actually, it is true if and only if $$W$$ is topologically closed in $$V$$). So, any proof will have to make use of the finite-dimensionality assumption.
Suppose $$(e_1, \ldots, e_n)$$ is an orthonormal basis for $$W$$. Suppose $$x \in (W^\perp)^\perp$$. Consider the projection $$w$$ of $$x$$ onto $$W$$, specifically: $$w = \langle x, e_1\rangle e_1 + \ldots + \langle x, e_n \rangle e_n \in W.$$ Then, $$x - w$$ is perpendicular to each $$e_i$$, since \begin{align*} \langle x - w, e_i\rangle &= \langle x, e_i \rangle - \sum_{j=1}^n\Big\langle \langle x, e_j \rangle e_j, e_i \Big\rangle \\ &= \langle x, e_i \rangle - \sum_{j=1}^n \langle x, e_j \rangle \langle e_j, e_i \rangle \\ &= \langle x, e_i \rangle - \sum_{j=1}^n \langle x, e_j \rangle \delta_{ij} \\ &= \langle x, e_i \rangle - \langle x, e_i \rangle = 0. \end{align*} Since the linear map $$\langle \cdot, x - w\rangle$$ is constantly $$0$$ on the basis of $$W$$, it is constantly zero on all of $$W$$, hence $$x - w \in W^\perp$$.
But then, since $$x \in (W^\perp)^\perp$$, we have $$\langle x - w, x \rangle = 0$$, and since $$\langle x - w, w \rangle = 0$$, $$\|x - w\|^2 = \langle x, x - w \rangle - \langle w, x - w\rangle = 0,$$ i.e. $$x = w \in W$$, completing the proof.
Suppose that $$w\in (W^\perp)^\perp$$ then $$\left=0$$ for all $$x\in W^\perp$$.
If $$w\not\in W$$, then $$w=v+v'$$ where $$v\in W$$ and $$v'\in W^\perp\setminus \{0\}$$. Then $$0=\left=\left+\left=\left$$ for all $$x\in W^\perp.$$ However, $$v'\in W^\perp$$ and $$\left\neq 0$$ so it is a contradiction.
• can you explain the logic? I don't follow really well. Starting from $w=v+v'$ – spruce Mar 15 '20 at 4:08
• @spruce So we note that $V=W+W^\perp$. Since $w\in V$ anyway, we should have such decomposition. And, especially, since $w\not\in W$, the decomposition should have a nonzero element in $W^\perp$. And the rest of them is just a calculations – Lev Bahn Mar 15 '20 at 4:11