The concept of linear mapping

Linear mappings: Linear mappings

The concept of linear 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 $L_A$ .

Matrix transformation The matrix mapping is also referred to as matrix transformation, certainly in case of a square matrix ( $m=n$ ).

If $m=n=1$ and $A=\matrix{a}$ , then $L_A$ is the mapping $\mathbb{R} \rightarrow \mathbb{R}$ defined by

$L_A(x) = a\cdot x$ Thus, left multiplication by a number $a$ is a linear mapping $\mathbb{R} \rightarrow \mathbb{R}$ .

Example 2

$A=\matrix{ 1 & -1 & 2 \\ 1 & -1 & 2 }$ then

the image of the vector $\cv{3\\ 1\\ 1}$ under $L_A$ equals $\matrix{ 1 & -1 & 2 \\ 1 & -1 & 2 }\cv{3\\ 1\\ 1}= \cv{4\\ 4}$ ;
the image of $5 \cdot \cv{3\\ 1\\ 1}$ under $L_A$ equals $5\cdot \cv{4\\ 4}=\cv{20\\ 20}$ thanks to linearity of $L_A$ .

The columns in a matrix $A$ are the images of the unit vectors $\vec{e_1}, \vec{e_2}, \ldots, \vec{e_n}$ , under the linear mapping determined by $A$ (check this for a small-sized example).

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.

These two properties follow directly from the properties of matrix calculus.

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 can generalize this definition to the following definition: A mapping $L$ from the vector space $V$ into the vector space $W$ is a linear mapping if it has the following two properties:

$L(\vec{x}+\vec{y})=L(\vec{x})+L(\vec{y})$ for all $\vec{x},\vec{y}\in V$ .
$L(\lambda\,\vec{x})=\lambda\,L(\vec{x})$ for any $\vec{x}\in V$ and any scalar $\lambda$ .

For example, let $V$ be the set of real functions that have a derivative. Then, differentiation is a linear mapping of $V$ to the vector space of real functions,

If the vector spaces $V$ and $W$ are finite dimensional, then the generalization adds little or nothing to the theory.

A linear mapping $L$ from $\mathbb{R}^n$ into $\mathbb{R}^n$ , or more generally from a certain vector space into itself, is also known as a linear transformation. Actually one even may omit the restriction of equal vector spaces.

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.

Substitute $\lambda=0$ in the second defining property $L(\lambda\,\vec{x})=\lambda\,L(\vec{x})$ of a linear mapping. Then you get $L(\vec{0})=\vec{0}$ .

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 $b=0$ .