6. Linear Transformations

You will already be familiar with the use of functions in mathematics to study sets and the mapping from a set of inputs set to another set of outputs. The notation used to denote functions is of the form \(f: X \to Y\) where \(f\) is the name of the function, \(X\) is the set of inputs known as the domain and \(Y\) is a member of the set of outputs known as the codomain. The mapping which defines the relationship between the domain and codmain is defined using \(y = f(x)\) where \(x\) is a member of the domain and \(y\) is a member of the codomain.

In linear algebra we study the linear mapping of a set of vectors to another set of vectors so we define \(T: V \to W\) where \(V\) and \(W\) are vector spaces and the mapping from an input vector \(\vec{u} \in V\) to an output vector \(\vec{w} \in W\) is given by \(\vec{w} = T(\vec{u})\).

Linear transformations have lots of uses in mathematics and computing. A good example is in the field of computer graphics and computer games where they are fundamental to the manipulation and visualisation of three-dimensional objects.

We begin with the formal definition of a linear transformation.

Definition 6.1 (Linear transformation)

If \(V\) and \(W\) are two vector spaces over the same field \(F\) then by a linear transformation (or linear mapping) is a mapping \(T: V \to W\) that for any two vectors \(\vec{u}, \vec{v} \in V\) and any scalar \(\alpha \in F\) the following conditions hold

• addition operation: \(T(\vec{u} + \vec{v}) = T(\vec{u}) + T(\vec{v})\);

• scalar multiplication: \(T(\alpha \vec{u}) = \alpha T(\vec{u})\).

The result of applying a linear transformation to an object is known as the image.

For example, let \(V = \mathbb{R}^2\) and \(W = \mathbb{R}^3\) then \(T : V \to W\) defined by \(T : (x, y)^\mathsf{T} \mapsto (x, y, 0)^\mathsf{T}\) is a linear transformation. Let \(\vec{u} = (u_1, u_2)^\mathsf{T}, \vec{v} = (v_1, v_2)^\mathsf{T} \in \mathbb{R}^2\) and \(\alpha \in \mathbb{R}\) then

\[\begin{split} \begin{align*} T (\vec{u} + \vec{v}) &= T \begin{pmatrix} u_1 + v_1 \\ u_2 + v_2 \end{pmatrix} = \begin{pmatrix} u_1 + v_1 \\ u_2 + v_2 \\ 0 \end{pmatrix}, \\ T (\vec{u}) + T(\vec{v}) &= \begin{pmatrix} u_1 \\ u_2 \\ 0 \end{pmatrix} + \begin{pmatrix} v_1 \\ v_2 \\ 0 \end{pmatrix} = \begin{pmatrix} u_1 + v_1 \\ u_2 + v_2 \\ 0 \end{pmatrix}, \end{align*} \end{split}\]

so \(T(\vec{u} + \vec{v}) = T(\vec{u}) + T(\vec{v})\) and the addition condition is satisfied. Similarly

\[\begin{split} \begin{align*} T(\alpha \vec{u}) &= T \begin{pmatrix} \alpha u_1 \\ \alpha u_2 \end{pmatrix} = \begin{pmatrix} \alpha u_1 \\ \alpha u_2 \\ 0 \end{pmatrix}, \\ \alpha T(\vec{u}) &= \alpha \begin{pmatrix} u_1 \\ u_2 \\ 0 \end{pmatrix} = \begin{pmatrix} \alpha u_1 \\ \alpha u_2 \\ 0 \end{pmatrix}, \end{align*} \end{split}\]

so \(T(\alpha \vec{u}) = \alpha T(\vec{u})\) and the scalar multiplication condition is satisfied and combined with the addition condition we have shown that \(T\) is a linear transformation. We can combine the addition and scalar multiplication conditions to give a single condition.

Definition 6.2 (Linear transformation condition)

A transformation \(T : V \to W\) is a linear transformation if the following condition is satisfied for any \(\vec{u}, \vec{v} \in V\) and \(\alpha \in F\)

(6.1)\[ T(\vec{u} + \alpha \vec{v}) = T(\vec{u}) + \alpha T(\vec{v}). \]

Example 6.1

Determine which of the following transformations are linear transformations

(i)   \(T: \mathbb{R}^3 \to \mathbb{R}^2\) defined by \(T: (x, y, z)^\mathsf{T} \mapsto (x, y)^\mathsf{T}\)

Solution

Let \(\vec{u} = (u_1, u_2, u_3)^\mathsf{T}, \vec{v} = (v_1, v_2, v_3)^\mathsf{T} \in \mathbb{R}^3\) and \(\alpha \in \mathbb{R}\) then

\[\begin{split} \begin{align*} T(\vec{u} + \alpha \vec{v}) &= T\left( \begin{pmatrix} u_1 \\ u_2 \\ u_3 \end{pmatrix} + \alpha \begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix} \right) = T \begin{pmatrix} u_1 + \alpha v_1 \\ u_2 + \alpha v_2 \\ u_3 + \alpha v_3 \end{pmatrix} = \begin{pmatrix} u_1 + \alpha v_1 \\ u_2 + \alpha v_2 \end{pmatrix}, \\ T(\vec{u}) + \alpha T(\vec{v}) &= T \begin{pmatrix} u_1 \\ u_2 \\ u_3 \end{pmatrix} + \alpha T \begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix} = \begin{pmatrix} u_1 \\ u_2 \end{pmatrix} + \alpha \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \begin{pmatrix} u_1 + \alpha v_1 \\ u_2 + \alpha v_2 \end{pmatrix}. \end{align*} \end{split}\]

Since \(T(\vec{u} + \alpha \vec{v}) = T(\vec{u}) + \alpha T(\vec{v})\) then \(T: (x, y, z)^\mathsf{T} \mapsto (x, y)^\mathsf{T}\) is a linear transformation.

(ii)   \(T: \mathbb{R}^3 \to \mathbb{R}^2\) defined by \(T: (x, y, z)^\mathsf{T} \mapsto (x + 3, y)^\mathsf{T}\)

Solution

Let \(\vec{u} = (u_1, u_2, u_3)^\mathsf{T}, \vec{v} = (v_1, v_2, v_3)^\mathsf{T} \in \mathbb{R}^3\) and \(\alpha \in \mathbb{R}\) then

\[\begin{split} \begin{align*} T(\vec{u} + \alpha \vec{v}) &= T \begin{pmatrix} u_1 + \alpha v_1 \\ u_2 + \alpha v_2 \\ u_3 + \alpha v_3 \end{pmatrix} = \begin{pmatrix} u_1 + \alpha v_1 + 3 \\ u_2 + \alpha v_2 \end{pmatrix}, \\ T(\vec{u}) + \alpha T(\vec{v}) &= T \begin{pmatrix} u_1 \\ u_2 \\ u_3 \end{pmatrix} + \alpha T \begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix} = \begin{pmatrix} u_1 + 3 \\ u_2 \end{pmatrix} + \alpha \begin{pmatrix} v_1 + 3 \\ v_2 \end{pmatrix} = \begin{pmatrix} u_1 + v_1 + 3 + 3\alpha \\ u_2 + \alpha v_2 \end{pmatrix} \end{align*} \end{split}\]

Since \(T(\vec{u} + \alpha \vec{v}) \neq T(\vec{u}) + \alpha T(\vec{v})\) then \(T: (x, y, z)^\mathsf{T} \mapsto (x + 3, y)^\mathsf{T}\) is not a linear transformation.

Note that we could have shown this by a counterexample, e.g., let \(\vec{u} = ( 1, 0 , 0 )^\mathsf{T}, \vec{v} = (2, 0, 0)^\mathsf{T} \in \mathbb{R}^3\) then

\[\begin{split} \begin{align*} T(\vec{u} + \vec{v}) &= T \begin{pmatrix} 3 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 6 \\ 0 \end{pmatrix}, \\ T(\vec{u}) + T(\vec{v}) &= \begin{pmatrix} 4 \\ 0 \\ 0 \end{pmatrix} + \begin{pmatrix} 5 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 9 \\ 0 \end{pmatrix}. \end{align*} \end{split}\]

(iii)   \(T: P(\mathbb{R}) \to P(\mathbb{R})\) defined by \(T: p \mapsto p \dfrac{\mathrm{d}p}{\mathrm{d}x}\)

Solution

Let \(u = x \in P(\mathbb{R})\) then

\[\begin{split} \begin{align*} T(2u) &= T(2x) = 2x(2) = 4x, \\ 2T(u) &= 2T(x) = 2(x)(1) = 2x, \end{align*} \end{split}\]

therefore \(T(2u) \neq 2T(u)\) and \(T: p \mapsto p \dfrac{\mathrm{d}p}{\mathrm{d}x}\) is not a linear transformation.