Linear maps determined by matrices

Linear maps: Linear maps

Linear maps determined by matrices

The following example of a linear map between coordinate spaces is essential for the rest of this chapter.

Let $m$ and $n$ be natural numbers and $A$ a real $(m\times n)$ -matrix. We write elements of $\mathbb{R}^n$ and $\mathbb{R}^m$ as columns.

Define the map $L_A: \mathbb{R}^n \rightarrow \mathbb{R}^m$ by

$L_A(\vec{x}) = A\vec{x}$ This map is linear.

We call $L_A$ the linear map determined by $A$ .

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

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

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 the linearity of $L_A$ .

We will see later that any linear map between two coordinate spaces can be written as the linear map of a matrix.

If $\vec{x}$ and $\vec{y}$ are two elements of $\mathbb{R}^n$ , then the rules of matrix multiplication show that for all scalars $\alpha$ and $\beta$ the following holds:

$\begin{array}{rcl}L_A(\alpha \vec{x} + \beta \vec{y}) &=& A(\alpha \vec{x} + \beta \vec{y}) \\ &&\phantom{xxx}\color{blue}{\text{definition of }L_A}\\ &=&\alpha A\vec{x} + \beta A\vec{y}\\ &&\phantom{xxx}\color{blue}{\text{linearity of matrix multiplication}}\\ &=& \alpha L_A(\vec{x}) + \beta L_A (\vec{y}) \\ &&\phantom{xxx}\color{blue}{\text{definition of }L_A}\end{array}$

This shows that $L_A$ is linear.

The standard dot product, which we define below for all vector spaces $\mathbb{R}^n$ , allows us to obtain a good interpretation of matrices with a single row.

The standard dot product on $\mathbb{R}^n$ is defined as the map that adds to two vectors $\vec{x}=\rv{x_1,\ldots,x_n}$ and $\vec{y}=\rv{y_1,\ldots,y_n}$ the number

$\dotprod{\vec{x}}{\vec{y}} = x_1\cdot y_1+\cdots +x_n\cdot y_n$

The vectors $\vec{x}$ and $\vec{y}$ are called orthogonal if $\dotprod{\vec{x}}{\vec{y}} =0$ . The length of the vector $\vec{x}$ is the number $\sqrt{\dotprod{\vec{x}}{\vec{x}}}$ .

Now choose $\vec{a}\in\mathbb{R}^n$ fixed.

The map that assigns to $\vec{x}$ in $\mathbb{R}^n$ the number $\dotprod{\vec{x}}{\vec{a}}$ is a linear map $\mathbb{R}^n\to\mathbb{R}$ .
If $\vec{a}$ is not the origin, then there is a unique linear map $P_\ell$ which maps each point of the line $\ell=\linspan{\vec{a}}$ through the origin onto itself, and each vector orthogonal to $\vec{a}$ onto the origin. The map $P_\ell$ is called the orthogonal projection on $\ell$ and has the mapping rule
$P_\ell(\vec{x}) =\frac{\dotprod{\vec{x}}{\vec{a}}}{\dotprod{\vec{a}}{\vec{a}}}\cdot\vec{a}$

If $n=2$ or $n=3$ then $\dotprod{\vec{x}}{\vec{y}}$ is the familiar standard dot product of $\vec{x}$ and $\vec{y}$ .

The map that assigns to $\vec{x}$ in $\mathbb{R}^n$ the number $\dotprod{\vec{x}}{\vec{a}}$ is the linear map $L_A$ determined by the $(1\times n)$ -matrix $A = \matrix{a_1&a_2&\cdots &a_n}$ .

In order to derive the mapping rule for $P_\ell$ we assume $\vec{x}\in\mathbb{R}^n$ . Suppose we can write $\vec{x}$ as the sum of a scalar multiple $\lambda \cdot\vec{a}$ of $\vec{a}$ and a vector $\vec{u}$ perpendicular to $\vec{a}$ :

$\vec{x} = \lambda\cdot\vec{a}+\vec{u}$

Linearity of the standard dot product with $\vec{a}$ then gives

$\dotprod{\vec{a}}{\vec{x}}=\lambda\cdot(\dotprod{\vec{a}}{\vec{a}})+\dotprod{\vec{a}}{\vec{u}}=\lambda\cdot(\dotprod{\vec{a}}{\vec{a}})$ so the assumption $\vec{a}\neq\vec{0}$ allows us to write

$\lambda = \frac{\dotprod{\vec{a}}{\vec{x}}}{\dotprod{\vec{a}}{\vec{a}}}$

Choosing this value for $\lambda$ and $\vec{u}=\vec{x}-\lambda\cdot\vec{a}$ indeed allows us to write $\vec{x}$ as a sum of the scalar multiple $\lambda\cdot\vec{a}$ of $\vec{a}$ and a vector $\vec{u}$ orthogonal to $\vec{a}$ . This means that we can determine the mapping rule for $P_\ell$ :

$\begin{array}{rcl} P_\ell(\vec{x}) &=& P_\ell(\lambda\cdot\vec{a}+\vec{u})\\&&\phantom{xxx}\color{blue}{\text{found decomposition of }\vec{x}}\\ &=& \lambda\cdot P_\ell(\vec{a})+P_\ell(\vec{u})\\ &&\phantom{xxx}\color{blue}{\text{linearity of }P_\ell}\\ &=& \lambda\cdot \vec{a}\\&&\phantom{xxx}\color{blue}{\vec{a}\in\ell\text{ and }\vec{u} \text{ is perpendicular to }\vec{a}}\\&=&\dfrac{\dotprod{\vec{a}}{\vec{x}}}{\dotprod{\vec{a}}{\vec{a}}}\cdot \vec{a}\\&&\phantom{xxx}\color{blue}{\text{found expression for }\lambda}\\\end{array}$

The definition of $P_{\ell}$ does not depend on the choice of $\vec{a}$ on the line $\ell$ as long as it is unequal to the origin: replacing $\vec{a}$ by another vector $\mu\vec{a}$ on $\ell$ where $\mu$ is a scalar unequal to $0$ shows

$\frac{\dotprod{\vec{x}}{(\mu\vec{a})} }{\dotprod{(\mu\vec{a})}{(\mu\vec{a})}}\cdot(\mu\vec{a})=\frac{\dotprod{\vec{x}}{\vec{a}}}{\dotprod{\vec{a}}{\vec{a}}}\cdot\vec{a}=P_\ell(\vec{x})$

In particular, we may choose $\dotprod{\vec{a}}{\vec{a}}$ (which is positive when $\vec{a}$ is not the origin) to be equal to $1$ , in which case $P_\ell(\vec{x})=\left(\dotprod{\vec{x}}{\vec{a}}\right)\vec{a}$ .

A similar expression exists for the orthogonal projection $P_U$ on an arbitrary linear subspace $U$ of $\mathbb{R}^n$ , the linear mapping sending each vector in $U$ to itself and each vector in $U^\perp$ to the zero vector. If $\basis{\vec{a}_1, \ldots , \vec{a}_k}$ is an orthonormal basis of the subspace $U$ (meaning $\dotprod{\vec{a}_i}{ \vec{a}_j}$ is equal to $0$ for $i\ne j$ and equal to $1$ for $i=j$ ) then the orthogonal projection $P_U$ satisfies:

$P_U(\vec{x}) = (\dotprod{\vec{x}}{\vec{a}_1})\vec{a}_1 +\cdots + (\dotprod{\vec{x}}{\vec{a}_k})\vec{a}_k$

The linear map $L:\mathbb{R}^2\to\mathbb{R}^2$ defined by

$L\left(\rv{x,y}\right) = \rv{-6 y-4 x,6 x-y}$ is defined by a matrix $A$ .

Which matrix?

$A =$ $\matrix{-4 & -6 \\ 6 & -1 \\ }$

The fact that $L$ is determined by a matrix $A$ means $L = L_A$ . The matrix $A=\matrix{a&b\\ c & d}$ must therefore satisfy

$\matrix{a&b\\ c & d}\cv{x\\ y} =\matrix{-6 y-4 x \\ 6 x-y}$ Element by element comparison gives

$\eqs{a\cdot x+ b\cdot y &=& -6 y-4 x\\ c\cdot x+ d\cdot y &=& 6 x-y}$
It follows that $a = -4$ , $b = -6$ , $c = 6$ , and $d=-1$ , so

$A = \matrix{-4 & -6 \\ 6 & -1 \\ }$

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