The concept of linear mapping

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 \(L_A\).

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

Example 1 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

If \[A=\matrix{ 1 & -1 & 2 \\ 1 & -1 & 2 }\] 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).

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.

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

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 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.

Transformation 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 \(b=0\).