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
\(\seteqnumber{0}{22.}{170}\)
\begin{equation} E^{\bot } = \{x \in \textsf {V} : \forall v \in E \big ( x \bot e \big ) \} \end{equation}

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
\(\seteqnumber{0}{22.}{171}\)
\begin{equation} \langle x, x \rangle = 0 \end{equation}

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
\(\seteqnumber{0}{22.}{172}\)
\begin{equation} y = u + z \end{equation}

Theorem 22.7.5 Let \(V\) be a vector space, of which \(W\) is a subspace. Then
\(\seteqnumber{0}{22.}{173}\)
\begin{equation} \dim (W) + \dim (W^{\bot }) = \dim (V). \end{equation}

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