The matrix of a linear map in coordinate space

Linear maps: Matrices of Linear Maps

The matrix of a linear map in coordinate space

Previously we encountered the linear map $L_A: \mathbb{R}^n\rightarrow \mathbb{R}^m$ determined by the $(m\times n)$ -matrix $A$ . Here we show that every linear map between two coordinate spaces has this form.

In order to see how this can be the case, note that the columns of a real $(m\times n)$ -matrix $A$ are the images under $L_ A$ of the vectors of the standard basis of $\mathbb{R}^n$ .

Let $m$ and $n$ be natural numbers and let $\varepsilon=\basis{\vec{e}_1,\ldots ,\vec{e}_n}$ be the standard basis for $\mathbb{R}^n$ .

Each linear mapping $L : \mathbb{R}^n \rightarrow\mathbb{R}^m$ is determined by the matrix $A = L_{\varepsilon}$ whose columns are ${ L (\vec{e}_1) ,\ldots , L (\vec{e}_n)}$

The matrix $L_{\varepsilon}$ is called the matrix of the linear map $L$ .

Write a vector $\vec{x}\in \mathbb{R}^n$ as a linear combination of the standard basis vectors: $x_1 \vec{e}_1 +\cdots + x_n\vec{e}_n$ . Then, due to the linearity of $L$ , $L(\vec{x})=x_1\cdot L (\vec{e}_1) +\cdots +x_n\cdot L (\vec{e}_n)$ Collect the $n$ vectors $L( \vec{e}_1) ,\ldots , L( \vec{e}_n)$ as columns in an $(m\times n)$ -matrix $A = L_{\varepsilon}$ . Then $L(\vec{x}) =A\vec{x}$ Here, the right hand side represents the matrix product of $A$ and $\vec{x}$ .

Determination of the image

The image of the vector $\vec{x}$ under $L$ is computed as the matrix product $L_{\varepsilon}\vec{x}$ .

For each linear map $L:V\to W$ and each basis $\alpha$ of $V$ and basis $\beta$ of $W$ , we will indicate later how you can find a matrix that describes the map.

We will then write ${}_\beta L_{\alpha}$ for the matrix and $L_{\alpha}$ if $\beta = \alpha$ . In the notation $L_\varepsilon$ used here the standard basis of $V$ appears. If both $V$ and $W$ are coordinate spaces, we sloppily also use $\varepsilon$ to refer to the standard basis of $W$ ; thus the notation $L_\varepsilon$ resembles the shorthand for ${}_\varepsilon L_\varepsilon$ .

Let $\vec{a}$ be a vector in an inner product space $V$ . Since the inner product $\dotprod{\vec{x}}{ \vec{y}}$ is linear in $\vec{y}$ , the map that assigns to $\vec{y}$ the inner product $\dotprod{\vec{a}}{\vec{y}}$ is a linear mapping $V\to\mathbb{R}$ . In the case where $V = \mathbb{R}^n$ with standard inner product, $\dotprod{\vec{a}}{\vec{y}}$ coincides with the matrix product $A\,\vec{y}$ , where $A$ is the $(1\times n)$ -matrix whose only row is the coordinate vector $\vec{a}=\rv{a_1,\ldots,a_n}$ , since then

$\dotprod{\vec{a}}{\vec{y}}=a_1\cdot y_1+\cdots+a_n\cdot y_ n = \matrix{a_1&\cdots & a_n}\, \matrix{y_1\\ \vdots\\ y_n} = A\, \vec{y}$

This way, we have found a new role for matrices.

Let $L:\mathbb{R}^2\to\mathbb{R}^2$ be the linear map defined by
$L\left(\rv{x,y}\right)=\rv{-9\cdot x-7\cdot y, -6\cdot x}$
Determine the matrix $L_{\varepsilon}$ of $L$ with respect to the standard basis.

$L_\varepsilon =$ $\matrix{-9& -7\\ -6 & 0}$

After all, $\begin{array}{rcl} L(\vec{e}_1) &=& \rv{-9\cdot 1 -7\cdot 0,-6\cdot 1 + 0\cdot 0} = \rv{-9,-6}\\ L(\vec{e}_2) &=&\rv{-9\cdot 0 -7\cdot 1,-6\cdot 0 + 0\cdot 1} =\rv{-7,0}\end{array}$ By the theorem Linear maps in coordinate spaces defined by matrices these image vectors, viewed as column vectors, are the columns of $L_\varepsilon$ , so
$L_\varepsilon = \matrix{-9& -7\\ -6 & 0}$

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