A calculus of the absurd

22.4 Linear transformations

22.4.1 Introduction to linear transformations

  • Definition 22.4.1 Let \(\textsf {V}, \textsf {W}\) be vector spaces over a field \(\mathbb {K}\). We say a function \(T : \textsf {V} \to \textsf {W}\) is linear if it satisfies these two properties

    • 1. For every \(x, y \in \textsf {V}\),

      \begin{align} & T (x + y) = T(x) + T(y) \label {linear transformation homomorphism} \\ & T (\alpha x) = \alpha T(x) \label {linear transformation distributes} \end{align}

  • Theorem 22.4.1 Let \(\textsf {V}, \textsf {W}\) be vector spaces over a field \(\mathbb {K}\) and let \(x,y \in \textsf {V}\) and \(\alpha \in \mathbb {K}\). The map/function/ whatever you want to call it \(T : \textsf {V} \to \textsf {W}\) is linear if and only if

    \begin{equation} T(\alpha x + y) = \alpha T(x) + T(y) \label {simpler linear condition} \end{equation}

Proof: there are two directions to show.

  • 1. Only if. We assume that \(T\) is a linear transformation. Therefore, \(T\) satisfies 22.106, so we can write

    \begin{align} T(\alpha x + y) &= T(\alpha x) + T(y) \\ \end{align}

    As \(T\) is linear it also satisifies 22.107, so by this property,

    \begin{align} T(\alpha x + y) &= \alpha T(x) + T(y) \end{align}

  • 2. If. We assume that 22.108 holds, and therefore (this is just a restatement of the equation from the theorem)

    \begin{align} T(\alpha x + y) = T(\alpha x) + T(y) \end{align} We then set \(\alpha = 1\), so it follows that

    \begin{equation} T(x + y) = T(x) + T(y) \end{equation}

    What remains to show is that for all \(\alpha \) and \(x\) we have

    \begin{equation} T(\alpha x) = \alpha T(x) \end{equation}

    We obtain this by fixing \(y = 0\) from which the result for all \(\alpha \) and \(\mathbb {K}\) follows.