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Mathematical Expression Editor
Introduction to Linear Transformations
We start by reviewing the definition of a function.
Let and be sets. A function from into , denoted by
assigns to each element of , a single element of .
The set is called the domain of , and the set is called the codomain.
If , we say that maps to , and is the image of .
The collection of images of all points of is called the image of under , or the image
of . (It is also known as the range of .)
In algebra and calculus you worked with functions whose domain and codomain were
each the set of all real numbers. In linear algebra, we call our functions
transformations. The domain and codomain of a transformation are vector
spaces.
In this exercise we will introduce a very special type of transformation by
contrasting the effects of two transformations on vectors of . We will see that some
transformations have “nice” properties, while others do not. Define and as
follows:
Each of these transformations takes a vector in , and maps it to another vector in .
To see if you understand how these transformations are defined, see if you can
determine what these transformations do to the vector .
Now, let’s take the vector and multiply it by a scalar, say .
.
Now let’s compare how and “handle” this product. Starting with , we
compute:
Observe that multiplying the original vector by , then applying , has the same effect
as applying to the original vector, then multiplying the image by . In other
words,
Diagrammatically, this can be represented as follows.
You should try to verify that this property does not hold for transformation . In
other words,
There is nothing special about the number , and it is not hard to prove that for any
scalar and vector of , satisfies \begin{align} \label{lin1} kT_1(\vec{u})= T_1(k\vec{u}). \end{align}
It turns out that satisfies another important property. For all vectors and of we
have: \begin{align} \label{lin2} T_1(\vec{u}+\vec{v}) = T_1(\vec{u})+T_1(\vec{v}) \end{align}
We leave it to the reader to illustrate this property with a specific example (see
Practice Problem prob:sum). We will show that satisfies (lin2) in general.
It turns out that fails to satisfy this property. Can you prove that this is
the case? Remember that to prove that a property DOES NOT hold, it
suffices to find a counter-example. See if you can find vectors and such that \begin{align} \label{t2}T_2(\vec{u}+\vec{v}) \neq T_2(\vec{u})+T_2(\vec{v}). \end{align}
Transformations satisfying (lin1) and (lin2), like , are called linear transformations.
Transformations like are not linear. You have already encountered several linear
transformations in the form of matrix transformations in sections Matrix
Transformations and Geometric Transformations of the Plane.
A transformation is called a linear transformation if the following are true for all
vectors and in , and scalars . \begin{equation} \label{eq:lintrans1} T(k\vec{u})= kT(\vec{u}) \end{equation}
\begin{equation} \label{eq:lintrans2} T(\vec{u}+\vec{v})= T(\vec{u})+T(\vec{v}) \end{equation}
Equations (eq:lintrans1) and (eq:lintrans2) of the above definition can be illustrated diagrammatically as
follows.
Properties (eq:lintrans1) and (eq:lintrans2) are often combined into a single property. \begin{equation} \label{eq:lintranscombinedprop} T(k_1\vec{u}+k_2\vec{v})= k_1T(\vec{u})+k_2T(\vec{v}) \end{equation}
Suppose is a linear transformation such that
Find each of the following:
(a)
(b)
item:lintransfirsta Because is a linear transformation, it satisfies (eq:lintranscombinedprop). We compute: \begin{align*} T\left (\begin{bmatrix}2\\5\end{bmatrix}\right )&=T\left (2\begin{bmatrix}1\\2\end{bmatrix}-\begin{bmatrix}0\\-1\end{bmatrix}\right )\\ &=2T\left (\begin{bmatrix}1\\2\end{bmatrix}\right )-T\left (\begin{bmatrix}0\\-1\end{bmatrix}\right )\\ &=2\begin{bmatrix}-1\\0\\3\end{bmatrix}-\begin{bmatrix}2\\-1\\0\end{bmatrix}\\ &=\begin{bmatrix}-2\\0\\6\end{bmatrix}-\begin{bmatrix}2\\-1\\0\end{bmatrix}=\begin{bmatrix}-4\\1\\6\end{bmatrix} \end{align*}
item:lintransfirstb Observe that . By (eq:lintranscombinedprop) we have: \begin{align*} T\left (\begin{bmatrix}1\\1\end{bmatrix}\right )&=T\left (\begin{bmatrix}1\\2\end{bmatrix}+\begin{bmatrix}0\\-1\end{bmatrix}\right )\\ &=T\left (\begin{bmatrix}1\\2\end{bmatrix}\right )+T\left (\begin{bmatrix}0\\-1\end{bmatrix}\right )\\ &=\begin{bmatrix}-1\\0\\3\end{bmatrix}+\begin{bmatrix}2\\-1\\0\end{bmatrix}=\begin{bmatrix}1\\-1\\3\end{bmatrix} \end{align*}
In Example ex:lintransfirst we were given the images of two vectors, and , under a linear
transformation . Based on this information, we were able to determine the images
of two additional vectors: and . The reason we were able to determine
and is because and can be written as unique linear combinations of and
.
Can every vector of be written as a linear combination of and ?
YesNo
Is the information provided in Example ex:lintransfirst sufficient to determine the image of every
vector in under ?
YesNo
Suppose is a transformation such that
Determine whether is a linear transformation.
Observe that
If were a linear transformation, then we would have:
But according to the given,
Since we conclude that transformation is not linear.
In Exploration init:lintransintro we introduced a transformation which turned out to be non-linear. It
took some work to show that is not linear. The following theorem would have made
our work easier.
Let be a linear transformation. Then
(a)
. In other words, linear transformations map the zero vector to the zero
vector.
(b)
maps any line in to a line (or the zero vector) in .
Proof
To prove part item:zerotozero, let be any vector in . By linearity of , we have:
Recall that
was defined by
We evaluate at :
Since , is not linear.
Linear Transformations Induced by Matrices
Recall that a transformation defined by , where is some matrix, is called a matrix
transformation (or transformation induced by ). As we had discovered in Matrix
Transformations, all matrix transformations are linear. We now formalize this result
as a theorem.
Let be an matrix. Define by . Then is a linear transformation.
Proof
Let and be vectors in , and let be a scalar. By properties of matrix
multiplication we have:
Therefore is a linear transformation.
Let be a linear transformation induced by
(a)
Find and .
(b)
Find the image of .
item:exlineartrans2a is a matrix, so for the expression to make sense, has to be a vector.
Thus, the domain of is (). The product is a vector. The codomain of is
().
item:exlineartrans2d By Definition def:function, the image of consists of images of all individual vectors in under .
Every vector in can be written as for some real numbers and . Consider the image
of
This shows that the range, or the image, of consists of all linear combinations of the
columns of . In other words, the image of is the span of vectors and . The two
vectors are not scalar multiples of each other, therefore they span a plane in
.
Let be an arbitrary vector of . The image of is given by \begin{align*} T(\vec{v})=\begin{bmatrix} -2&1&3\\ 4&-2&-6 \end{bmatrix}\begin{bmatrix}a\\b\\c\end{bmatrix}&=a\begin{bmatrix}-2\\4\end{bmatrix}+b\begin{bmatrix}1\\-2\end{bmatrix}+c\begin{bmatrix}3\\-6\end{bmatrix}\\ &=(a(-2)+b+c(3))\begin{bmatrix}1\\-2\end{bmatrix} \end{align*}
This shows that the image of every vector in is a scalar multiple of . This means that
the image of is a line in .
Linear Transformations of Subspaces of
Definition def:lin defines a linear transformation as a map from into . We will now make
this definition more general by allowing the domain and the codomain of the
transformation to be subspaces of and . Eventually, a linear transformation will be
defined as a mapping between vector spaces.
Let and be subspaces of and . A transformation is called a linear transformation
if the following are true for all vectors and in , and scalars .
Let be a subspace of consisting of all vectors in the -plane. Let be a subspace of
consisting of all vectors along the -axis. (Do a quick verification that and are
subspaces of .) Define a transformation by
Show that is a linear transformation, and describe its action geometrically.
Consider
two arbitrary elements and of . \begin{align*} T\left (\begin{bmatrix}a_1\\b_1\\0\end{bmatrix}+ \begin{bmatrix}a_2\\b_2\\0\end{bmatrix}\right )&=T\left (\begin{bmatrix}a_1+a_2\\b_1+b_2\\0\end{bmatrix}\right )\\ &=\begin{bmatrix}0\\0\\(a_1+a_2)+(b_1+b_2)\end{bmatrix}\\ &=\begin{bmatrix}0\\0\\(a_1+b_1)+(a_2+b_2)\end{bmatrix}\\ &=\begin{bmatrix}0\\0\\a_1+b_1\end{bmatrix}+\begin{bmatrix}0\\0\\a_2+b_2\end{bmatrix}\\ &=T\left (\begin{bmatrix}a_1\\b_1\\0\end{bmatrix}\right )+ T\left (\begin{bmatrix}a_2\\b_2\\0\end{bmatrix}\right ) \end{align*}
Verification of the fact that is similar, and we omit the details.
We have shown that is a linear transformation. maps all vectors in the -plane to the
-axis. The following diagram helps us visualize the action of on a specific
vector.
We can investigate further. Recall that is defined by
Complete each of the following statements referring to the diagram below.
The image of the line under is a linea planethe zero vectorall of
The image of the orange part of the domain (the front triangle) is the positive -axisthe negative -axisthe zero vectorthe entire -axis
The image of the purple part of the domain (the back triangle) is the positive -axisthe negative -axisthe zero vectorthe entire -axis
We conclude this section by introducing two simple but important transformations.
The identity transformation on , denoted by , is a transformation that maps each
element of to itself.
In other words,
is a transformation such that
The zero transformation, , maps every element of the domain to the zero
vector.
In other words,
is a transformation such that
The identity transformation is linear.
Proof
Left to the reader. (See Practice Problem prob:idtrans)
The zero transformation is linear.
Proof
Left to the reader. (See Practice Problem prob:zerotrans)
For each matrix below, find the domain and the codomain of the linear
transformation induced by ; find and draw the image of . (Hint: See Example
ex:lineartrans3.)