Chapter 9: Linear Mappings

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We will start out this chapter by rehearsing what a mapping or a function for regular real values (scalars), and then move to vector mappings.

This may look abstract, but in fact you have already been dealing with mappings, but under the name functions. Another way to state the same thing is

\begin{equation} y = F(x). \end{equation}

(9.2)

The form

\begin{equation} F: x \rightarrow y, x \in N. \end{equation}

(9.3)

is also used. For example the function $y = x^2$, shown in Figure 9.3, is a rule that, for every item in the set of real numbers $\mathbb{R}$, provides another item from the set of real numbers $\mathbb{R}$. Thus, in this example, both $N$ and $M$ equals $\mathbb{R}$. In linear algebra, the term mapping is traditionally used instead of function, but the meaning is the same, i.e., you start out with an item $x$, and you obtain an item $y$. We say that the value $x$ maps to the value $y$. Note that every $x$ only maps to one value $y$. This means that the curve in Figure 9.4 cannot be the graph of a function $y =f(x)$ since, for instance, the value $x=1$ is associated with two $y$-values, namely $+1$ and $-1$. The set $N$ is called the domain of the mapping, whereas $M$ is called the codomain. The subset of the codomain that can be reached by the mapping is called the range of the codomain, or alternatively, the image of the codomain, and we denote it $V_F$. This is summarized in the following definition.

The vertical bar should be read as "such that" or "with the property of". In this example, the expression can be read out as "$V_F$ is the set of all elements $F(x)$ such that $x$ belongs to the set $N$". For the example $y=x^2$, the range equals the set of positive real numbers including zero, $V_F = \mathbb{R}_{\geq 0}$. Therefore, in this case, we only reach a subset of the codomain, i.e., $V_F$ is a subset of $M$.

In linear algebra, the inputs and outputs of a function are vectors instead of scalar. Assume we have a coordinate system $\vc{e}_1,\vc{e}_2$ and that $\begin{pmatrix}x_1 \\ x_2 \\\end{pmatrix}$ is the coordinate for the vector $\vc{x}$. We can now have a function $\vc{y} = F( \vc{x} )$ that maps every $\vc{x}$ to a new vector $\vc{y} = \begin{pmatrix}y_1 \\ y_2 \\\end{pmatrix}$ according to, for instance,

\begin{equation} \begin{cases} y_1 = x_1 \\ y_2 = 0 \end{cases} \end{equation}

(9.5)

It is not possible to draw a simple graph for this mapping, since four dimensions would be needed for that (two for the input and two for the output). However, it is often possible to get an intuitive understanding of the mapping by drawing both the input and the output in the same diagram. Interactive Illustration 9.5 shows this for the mapping mentioned above. Note that you can move the red input arrow $\vc{x}$ and see how the blue output arrow $\vc{y}$ moves. As can be seen, the effect of the mapping is to project any input vector on to the $\vc{e}_1$-axis. Any vector in the plane can be used as input, hence the domain is $\mathbb{R}^2$. The codomain is also $\mathbb{R}^2$, since the output is a vector of two dimensions, but the range or image is the $\vc{e}_1$-axis. The range is marked with green in the second step of the figure.

A slightly more interesting example of a mapping is the following

\begin{equation} \begin{cases} y_1 = \cos(\frac{\pi}{3}) x_1 - \sin(\frac{\pi}{3}) x_2, \\ y_2 = \sin(\frac{\pi}{3}) x_1 + \cos(\frac{\pi}{3}) x_2. \end{cases} \end{equation}

(9.6)

As can be seen, the factors before the $x_1$s and $x_2$s resemble a rotation matrix (see Definition 6.10) by $\pi/3$ radians. This mapping is illustrated in the Interactive Illustration 9.6, where again the input vector is marked with red and the output vector is marked with blue. We have now learned that vector mappings are simply functions that take a vector as input and outputs another vector. Next we will explore when such mappings can be written using matrices.

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As can be seen by playing around with Interactive Ilustration 9.6, the output vector is a rotated copy of the input vector with the rotation angle $\frac{\pi}{3}$. As a matter of fact, we can write Equation (9.6) in matrix form as

\begin{equation} \begin{pmatrix} y_1 \\ y_2 \end{pmatrix} = \left(\begin{array}{rr} \cos \frac{\pi}{3} & -\sin \frac{\pi}{3} \\ \sin \frac{\pi}{3} & \cos \frac{\pi}{3} \end{array}\right) \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} \end{equation}

(9.7)

or, shorter,

\begin{equation} \vc{y} = \mx{A} \vc{x}. \end{equation}

(9.8)

It is now easy to see that the matrix $\mx{A}$ is just a two-dimensional rotation matrix as defined in Definition 6.10 in Chapter 6. When a mapping can be written in matrix form, i.e., in the form $\vc{y} = \mx{A} \vc{x}$, we call $\mx{A}$ the transformation matrix.

The example in Interactive Illustration 9.3 can also be written in matrix form,

\begin{equation} \begin{pmatrix} y_1 \\ y_2 \end{pmatrix} = \left(\begin{array}{rr} 1 & 0 \\ 0 & 0 \end{array}\right) \begin{pmatrix} x_1 \\ x_2 \end{pmatrix}, \end{equation}

(9.9)

where the transformation matrix in this case equals $\left(\begin{array}{rr} 1 & 0 \\ 0 & 0 \end{array}\right)$. That raises the question whether all vector mappings be written on the form $\vc{y} = \mx{A} \vc{x}$ for some $\mx{A}$ with constant coefficients? The answer is no. As an example, the mapping

\begin{equation} \begin{cases} y_1 = x_1 x_2 + x_2\\ y_2 = x_1 + e^{x_2} \end{cases} \end{equation}

(9.10)

cannot be written as $\vc{y} = \mx{A}\vc{x}$. It is of course possible to write $\begin{pmatrix} y_1 \\ y_2 \end{pmatrix} = \left(\begin{array}{rr} x_2 & 1 \\ 1 & \frac{e^{x_2}}{x_2} \end{array}\right) \begin{pmatrix} x_1 \\ x_2 \end{pmatrix}$, but that violates the rule that $\mx{A}$ should consist of constant coefficients, i.e, independent of $\vc{x}$. To investigate which mappings can be written in this form, we first introduce the concept of a linear mapping.

The intuitive description of the first condition is that it shouldn't matter if you take the function of the sum or the sum of the functions, you get the same result. The second condition states that amplifying the input by a certain amount gives the same result as amplifying the output by the same amount.

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It is often the case that you want to find the matrix $\mx{A}$ for a particular linear mapping. Then the following theorem can be of great help.

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We can now use this fact to easily find the matrix $\mx{A}$ for a linear mapping by simply observing what happens to the basis function.

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Assume we have a linear mapping $\vc{y} = F(\vc{x})$, and that the output vector $\vc{y}$ can be fed into another linear mapping $\vc{z}= G(\vc{y})$. For this to work we need the domain of $G$ to be equal the codomain of $F$. As an example, if $G(\vc{y})$ accepts two-dimensional vectors, i.e., the domain of $G$ is $\mathbb{R}^2$, then the output of $F$, must be a two-dimensional vector, i.e., the codomain of $F(\vc{x})$ must also be $\mathbb{R}^2$. We then call the mapping

\begin{equation} \vc{z} = G(F(\vc{x})) \end{equation}

(9.26)

a composite mapping.

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For some mappings, several inputs $\vc{x}$ can map to the same value $F(\vc{x})$. As an example, in Interactive Illustration 9.5, where we simply project a point onto the $x$-axis by setting the $y$-coordinate to zero, both $\vc{x}_1 = (1,5)$ and $\vc{x}_2 = (1,2)$ will map to the same point $(1,0) = F(\vc{x}_1) =F(\vc{x}_2)$. However, for some mappings, the result will be unique, so if we start out with two different $\vc{x}_1 \neq \vc{x}_2$, we know that $F(\vc{x}_1) \neq F(\vc{x}_2)$. We call such mappings injective.

Another way to state this is to say that an injective mapping has the property that if two images $F(\vc{x}_1)$ and $F(\vc{x}_2)$ are equal, then $\vc{x_1}$ must equal $\vc{x_2}$.

For some mappings $\vc{y} = F(\vc{x})$, we can reach every point $\vc{y}$ in the codomain by selecting a suitable $\vc{x}$. When the range of the mapping covers the codomain like this, we call the mapping surjective. A mapping that is both injective and surjective is called bijective.

We will now go through a few examples of mappings and investigate whether they are injective, surjective, and bijective.

As it turns out, linear mappings that are bijective have many useful properties. As an example, it is simple to find the inverse mapping, as we will see in the following theorem.

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From this proof, the following theorem follows.

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In Interactive Illustration 9.8, we show how linear mappings can be used to generate a shadow effect.

Now that we have this shadow example, we can investigate whether the mapping is bijective.