### Linear mappings: Linear mappings

### The concept of linear mapping

Matrix mapping Let \(m\) and \(n\) be natural numbers and let \(A\) be a real \(m\times n\) matrix. We write elements of \(\mathbb{R}^n\) and \(\mathbb{R}^m\) as column vectors.

Define the **matrix mapping** #L_A: \mathbb{R}^n \rightarrow \mathbb{R}^m# for matrix \(A\) by \[ L_A(\vec{x}) = A\vec{x}\] This mapping is linear.

We call \(L_A\) the **linear mapping determined by** **#A#**. Often we will call \(A\) a linear mapping, in which case we actually mean \(L_A\).

Linearity of a matrix mapping Denote the mapping determined by the \(m\times n\) matrix \(A\) from \(\mathbb{R}^n\) into \(\mathbb{R}^m\) as \(L_A\). Then this mapping has the following two properties

- \(L_A(\vec{x}+\vec{y})=L_A(\vec{x})+L_A(\vec{y})\) for all \(\vec{x},\vec{y}\in\mathbb{R}^n\).
- \(L_A(\lambda\,\vec{x})=\lambda\,L_A(\vec{x})\) for any \(\vec{x}\in\mathbb{R}^n\) and any scalar \(\lambda\).

We say that \(L_A\) is a *linear mapping*. We also use the term *linear transformation*.

Linear mapping A mapping \(L\) from \(\mathbb{R}^n\) into \(\mathbb{R}^m\) is characterized as a **linear mapping** or **linear transformation** if it has the following two properties.

- \(L(\vec{x}+\vec{y})=L(\vec{x})+L(\vec{y})\) for all \(\vec{x},\vec{y}\in\mathbb{R}^n\).
- \(L(\lambda\,\vec{x})=\lambda\,L(\vec{x})\) for any \(\vec{x}\in\mathbb{R}^n\) and any scalar \(\lambda\).

We can also replace the two defining characteristics of a linear mapping by one rule, namely: \[L(\lambda\,\vec{x}+\mu\,\vec{y})=\lambda\,L(\vec{x})+\mu\,L(\vec{y})\quad\text{for all }\vec{x},\vec{y}\in \mathbb{R}^n\text{ and all scalars }\lambda, \mu\]

We already know that a matrix image is a linear mapping. Reversely, we can describe any linear mapping as a matrix mapping. We give below an example.

Consider the mapping \(L:\mathbb{R}^2 \longrightarrow \mathbb{R}^2\) defined by \[L\cv{x\\y} = \cv{2x-y\\x+y}\] In matrix notation we can write this mapping as \[L\cv{x\\y}=\matrix{2&-1\\1&1}\cv{x\\y}\] So, this is a linear mapping.

We can also deduce from the definition that it is a linear mapping, but this is more pencil and paper work. \[\begin{aligned}L\left(\cv{x\\y}+\cv{\xi\\ \eta}\right) &= L\cv{x+\xi\\ y+\eta} \\ \\ &= \cv{2(x+\xi)-(y+\eta)\\ (x+\xi)+(y+\eta)} \\ \\ &=\cv{2x-y\\ x+y}+ \cv{2\xi-\eta\\ \xi+\eta} \\ \\ &= L\cv{x\\y} +L\cv{\xi\\ \eta} \\ \\ L\left(\lambda\cv{x\\y}\right) &= L\cv{\lambda\,x\\ \lambda y}\\ \\ &= \cv{2(\lambda\,x)-(\lambda\,y\\ (\lambda\,x)+(\lambda\,y)} \\ \\ &=\cv{\lambda(2x-y)\\\lambda(x+y)}\\ \\ &=\lambda \cv{2x-y\\x+y}\\ \\ &= \lambda\,L\cv{x\\y}\end{aligned}\]

The image of the zero vector under a linear mapping is the zero vector.

From this theorem follows immediately that the linear function \(x\mapsto a\,x+b\) is a linear transformation from \(\mathbb{R}\) to \(\mathbb{R}\) under the condition that \(b=0\).