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

    \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

    \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

    \begin{equation} y = u + z \end{equation}

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

    \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