A calculus of the absurd

22.3.4 Subspaces

  • Definition 22.3.3 Let \(\textsf {V}\) be a vector space. The set \(\textsf {W}\) is a subspace of \(\textsf {V}\) if \(\textsf {W}\) is a vector space, and \(\textsf {W}\) is a subset of \(V\).

  • Technique 22.3.1 Showing that something is a subspace. Suppose we have a vector space \(\textsf {V}\), and we want to prove that \(\textsf {W}\) is a subspace of \(\textsf {V}\). The steps to do so are this

    • 1. Show that the zero vector is in the subspace in question.

    • 2. Show that \(W \subseteq V\) using the standard technique for showing that something is a subset of something else (as in Section TODO: write).

    • 3. Then we must show that \(W\) is closed under vector addition and scalar multiplication. The rest of the vector space axioms follow from the fact that \(W \subseteq V\) and \(V\) is a vector space.

This theorem is given as both an example of how to prove facts about vector spaces, but also because it is important in its own right.

  • Theorem 22.3.1 Let \(\textsf {V}\) be a vector space, and \(\textsf {U}\) and \(\textsf {W}\) be subspaces of \(\textsf {V}\). Prove that \(U \cup W\) is a subspace of \(V\) if and only if \(U \subseteq W\) or \(W \subseteq U\).

To prove this, first we will show the “if” direction, and then the only if direction.

  • • If. Without loss of generality, assume that \(U \subseteq W\), in which case \(U \cup W = W\), and this is a subspace of \(V\) as \(W\) is a subspace of \(V\). The proof for the other case follows by swapping \(U\) and \(W\) in the proof.

  • • Only if. This direction requires a bit more of an intuition about what directions to explore. First we will assume that \(U \cup W\) is a subspace, and then we will assume that the consequent is untrue (i.e. that \(U \subseteq W \text { or } W \subseteq U\) is not true), in which case there exist \(u, w \in \textsf {V}\) such that

    \begin{equation} u \in U \setminus W \text { and } w \in W \setminus U. \label {def of u and w} \end{equation}

    We can then ask (this is the core idea in the proof which is not immediately obvious – to me at least), about the status of \(u + w\). As \(u, w \in U \cup W\) and by assumption \(U \cup W\) is a subspace (and therefore by definition closed under vector addition) it must be that \(u + w \in U \cup W\). Then either \(u + w \in U\) or \(u + w \in W\) (by definition of the set union).

    • 1. If \(u + w \in U\), then also \(u + w + (-u) = w\) which is a contradiction as by definition of \(w\) (Equation ??) \(w \notin U\).

    • 2. If \(u + w \in W\), a very similar thing is the case; also \(u + w + (-w) = u \in W\) which is a contradiction as \(u \notin W\).

    Therefore, by contradiction this direction of the theorem must be true.