# A calculus of the absurd

##### 22.7.7 Orthogonal complement
• Definition 22.7.2 Let $$\textsf {V}$$ be an inner product space, and $$E$$ be a subspace of $$\textsf {V}$$. We define the orthogonal complement of $$E$$ as the set

$$E^{\bot } = \{x \in \textsf {V} : \forall v \in E \big ( x \bot e \big ) \}$$

• Theorem 22.7.3 Let $$V$$ be a vector space, and $$W$$ be a subspace of $$V$$. In this case wouldn’t it be nice if $$W \cap W^{\bot } = \{0\}$$?

Yes, it would. We can prove separately that $$W \cap W^{\bot } \subseteq \{0\}$$ and $$\{0\} \subseteq W \cap W^{\bot }$$.

• • $$\{0\} \subseteq W \cap W^{\bot }$$. This direction is the easier one, because we know both $$W$$ and $$W^{\bot }$$ are subspaces and therefore they both contain at least the $$0$$ vector.

• • $$W \cap W^{\bot } \subseteq \{0\}$$. Let us suppose that $$x \in W$$ and $$x \in W^{\bot }$$, and to prove this by contradiction that $$x \notin \{0\}$$, i.e. $$x \ne 0$$. Because $$x \in W^{\bot }$$ we know that for all $$w \in W$$ that $$\langle x, w \rangle = 0$$ (this follows directly from the definition of $$W^{\bot }$$). As it is also true that $$x \in W$$, it follows from this that therefore also

$$\langle x, x \rangle = 0$$

which is true if and only if $$x = 0$$, which contradicts our earlier assumption that $$x \ne 0$$, and thus this direction is true.

Therefore, the theorem is true.

$$\Box$$

• Theorem 22.7.4 (The very important resolving theorem) Let $$V$$ be a vector space, of which $$W$$ and $$W^{\bot }$$ are subspaces, then for every vector $$y \in V$$ there exist unique vectors $$u, z$$ such that

$$y = u + z$$

• Theorem 22.7.5 Let $$V$$ be a vector space, of which $$W$$ is a subspace. Then

$$\dim (W) + \dim (W^{\bot }) = \dim (V).$$

This theorem follows mostly from the definitions. First, let $$\dim (W) = k$$ and $$\dim (W^{\bot }) = m$$, then our goal is to show that $$\dim (k + m) = \dim (V)$$.

First, fix a basis $$\alpha = \{w_1, ..., w_k\}$$ for $$W$$ and a basis for $$W^{\bot }$$, $$\beta = \{w_{k+1}, ..., w_{k + m}\}$$. Then we will prove that $$\alpha \cup \beta$$ is a basis for $$V$$.

• • Generating. Let $$x \in V$$, then $$x = v_1 + v_2$$ for $$v_1 \in W$$ and $$v_2 \in W^{\bot }$$ using