Introduction to Linear Transformations

We start by reviewing the definition of a function.

In algebra and calculus you worked with functions \(f:\RR \rightarrow \RR \) 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 \(\RR ^2\). We will see that some transformations have “nice" properties, while others do not. Define \(T_1\) and \(T_2\) as follows:
\[T_1:\RR ^2\rightarrow \RR ^2\]
\[T_1\left (\begin{bmatrix} x\\ y \end{bmatrix}\right )=\begin{bmatrix} x-y\\ x \end{bmatrix}\]
\[T_2:\RR ^2\rightarrow \RR ^2\]
\[T_2\left (\begin{bmatrix} x\\ y \end{bmatrix}\right )=\begin{bmatrix} -x+y+1\\ y-2 \end{bmatrix}\]

Each of these transformations takes a vector in \(\RR ^2\), and maps it to another vector in \(\RR ^2\). To see if you understand how these transformations are defined, see if you can determine what these transformations do to the vector \(\begin{bmatrix} 4\\ 3 \end{bmatrix}\).

\[ T_1\left (\begin{bmatrix} 4\\ 3 \end{bmatrix}\right )=\begin{bmatrix} \answer {1}\\ \answer {4} \end{bmatrix} \quad \text {and} \quad T_2\left (\begin{bmatrix} 4\\ 3 \end{bmatrix}\right )=\begin{bmatrix} \answer {0}\\ \answer {1}\end{bmatrix}.\]

Now, let’s take the vector \(\begin{bmatrix} 4\\ 3 \end{bmatrix}\) and multiply it by a scalar, say \(7\).

\[7\begin{bmatrix} 4\\ 3 \end{bmatrix} = \begin{bmatrix} 28\\ 21 \end{bmatrix}\]
.

Now let’s compare how \(T_1\) and \(T_2\) “handle" this product. Starting with \(T_1\), we compute:

\[ T_1\left (7\begin{bmatrix} 4\\ 3 \end{bmatrix}\right )=T_1\left (\begin{bmatrix} 28\\ 21 \end{bmatrix}\right )=\begin{bmatrix} 7\\ 28 \end{bmatrix} \]

Observe that multiplying the original vector by \(7\), then applying \(T_1\), has the same effect as applying \(T_1\) to the original vector, then multiplying the image by \(7\). In other words,

\[ T_1\left (7\begin{bmatrix} 4\\ 3 \end{bmatrix}\right )=\begin{bmatrix} 7\\ 28 \end{bmatrix}=7\begin{bmatrix} 1\\ 4 \end{bmatrix}=7T_1\left (\begin{bmatrix} 4\\ 3 \end{bmatrix}\right )\]

Diagrammatically, this can be represented as follows.

[Picture]

You should try to verify that this property does not hold for transformation \(T_2\). In other words,

\[ T_2\left (7\begin{bmatrix} 4\\ 3 \end{bmatrix}\right )\neq 7T_2\left (\begin{bmatrix} 4\\ 3 \end{bmatrix}\right )\]

There is nothing special about the number \(7\), and it is not hard to prove that for any scalar \(k\) and vector \(\vec {u}\) of \(\RR ^2\), \(T_1\) satisfies

\begin{align}\label {lin1} kT_1(\vec {u})= T_1(k\vec {u}).\end{align}

It turns out that \(T_1\) satisfies another important property. For all vectors \(\vec {u}\) and \(\vec {v}\) of \(\RR ^2\) 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 \(T_1\) satisfies (lin2) in general.

Let \(\vec {u}=\begin{bmatrix} u_1\\ u_2 \end{bmatrix}\) and \(\vec {v}=\begin{bmatrix} v_1\\ v_2 \end{bmatrix}\), then

\begin{align*} T_1(\vec {u}+\vec {v})&=T_1\left (\begin{bmatrix} u_1\\ u_2 \end{bmatrix}+\begin{bmatrix} v_1\\ v_2 \end{bmatrix}\right )=T_1\left (\begin{bmatrix} u_1+v_1\\ u_2+v_2 \end{bmatrix}\right )=\begin{bmatrix} u_1+v_1-u_2-v_2\\ u_1+v_1 \end{bmatrix}\\ &=\begin{bmatrix} u_1-u_2\\ u_1 \end{bmatrix}+\begin{bmatrix} v_1-v_2\\ v_1 \end{bmatrix}=T_1\left (\begin{bmatrix} u_1\\ u_2 \end{bmatrix}\right )+T_1\left (\begin{bmatrix} v_1\\ v_2 \end{bmatrix}\right )\\ &=T_1(\vec {u})+T_1(\vec {v}) \end{align*}

It turns out that \(T_2\) 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 \(\vec {u}\) and \(\vec {v}\) such that

\begin{align}\label {t2}T_2(\vec {u}+\vec {v}) \neq T_2(\vec {u})+T_2(\vec {v}).\end{align}

(See Practice Problem prob:prob2.)

Transformations satisfying (lin1) and (lin2), like \(T_1\), are called linear transformations. Transformations like \(T_2\) 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.

Equations (eq:lintrans1) and (eq:lintrans2) of the above definition can be illustrated diagrammatically as follows.

[Picture]

[Picture]

In Exploration init:lintransintro we introduced a transformation \(T_2\) which turned out to be non-linear. It took some work to show that \(T_2\) is not linear. The following theorem would have made our work easier.

To prove part item:zerotozero, let \(\vec {v}\) be any vector in \(\RR ^n\). By linearity of \(T\), we have:

\[T(\vec {0})=T(\vec {v}-\vec {v})=T(\vec {v})-T(\vec {v})=\vec {0}\]

Part item:linetoline will become evident after we prove Theorem th:matlin in Standard Matrix of a Linear Transformation and combine it with Practice Problem prb:linesToLines.

Linear Transformations Induced by Matrices

Recall that a transformation \(T:\RR ^n\rightarrow \RR ^m\) defined by \(T(\vec {v})=A\vec {v}\), where \(A\) is some \(m\times n\) matrix, is called a matrix transformation (or transformation induced by \(A\)). As we had discovered in Matrix Transformations, all matrix transformations are linear. We now formalize this result as a theorem.

Let \(\vec {u}\) and \(\vec {v}\) be vectors in \(\RR ^n\), and let \(k\) be a scalar. By properties of matrix multiplication we have:

\[T(\vec {u}+\vec {v})=A(\vec {u}+\vec {v})=A\vec {u}+A\vec {v}=T(\vec {u})+T(\vec {v})\]
\[T(k\vec {u})=A(k\vec {u})=kA\vec {u}=kT(\vec {u})\]
Therefore \(T\) is a linear transformation.

Linear Transformations of Subspaces of \(\RR ^n\)

Definition 2 defines a linear transformation as a map from \(\RR ^n\) into \(\RR ^m\). We will now make this definition more general by allowing the domain and the codomain of the transformation to be subspaces of \(\RR ^n\) and \(\RR ^m\). Eventually, a linear transformation will be defined as a mapping between vector spaces.

We conclude this section by introducing two simple but important transformations.

Left to the reader. (See Practice Problem )

Left to the reader. (See Practice Problem )

Practice Problems

Define \(T_1\) as follows
\[T_1:\RR ^2\rightarrow \RR ^2\]
\[T_1\left (\begin{bmatrix} x\\ y \end{bmatrix}\right )=\begin{bmatrix} x-y\\ x \end{bmatrix}\]
(See Exploration ) Use vectors \(\begin{bmatrix}3\\4\end{bmatrix}\) and \(\begin{bmatrix}-2\\1\end{bmatrix}\) to illustrate that \(T_1(\vec {u}+\vec {v}) = T_1(\vec {u})+T_1(\vec {v})\).
Define \(T_2\) as follows
\[T_2:\RR ^2\rightarrow \RR ^2\]
\[T_2\left (\begin{bmatrix} x\\ y \end{bmatrix}\right )=\begin{bmatrix} -x+y+1\\ y-2 \end{bmatrix}\]
(See Exploration )

Use a counter-example to prove that \(T_2(\vec {u}+\vec {v}) \neq T_2(\vec {u})+T_2(\vec {v})\)

Suppose \(T:\RR ^{10}\rightarrow \RR ^2\) is a linear transformation such that \(T(\vec {u})=\begin{bmatrix}2\\-1\end{bmatrix}\) and \(T(\vec {v})=\begin{bmatrix}-5\\3\end{bmatrix}\). Find the image of \(3\vec {u}-\vec {v}\).
\[T(3\vec {u}-\vec {v})=\begin{bmatrix}\answer {11}\\\answer {-6}\end{bmatrix}\]
Let \(\vec {u}\) be a fixed non-zero vector. Define \(T_{\vec {u}}:\RR ^2\rightarrow \RR ^2\), by \(T_{\vec {u}}(\vec {x})=\vec {u}-\vec {x}\).
1.
Describe the effect of this transformation by sketching \(\bf x\) and \(T_{\vec {u}}({\bf x})\) for at least four vectors \(\bf x\) and a fixed vector \(\vec {u}\) of your choice.
2.
Is \(T_{\vec {u}}\) a linear transformation?
Define \(P_{xy}:\RR ^3\rightarrow \RR ^2\), by \(P_{xy}\left (\begin{bmatrix} x\\ y\\ z \end{bmatrix} \right )=\begin{bmatrix} x\\ y\\ 0 \end{bmatrix}\). This transformation is called an orthogonal projection onto the \(xy\)-plane. Show that \(P_{xy}\) is a linear transformation.
Suppose a linear transformation \(T:\RR ^3\rightarrow \RR ^3\) maps \(\bf i\) to \(\begin{bmatrix}2\\-1\\0\end{bmatrix}\), \(\bf j\) to \(\begin{bmatrix}-2\\4\\1\end{bmatrix}\), and \(\bf k\) to \(\begin{bmatrix}3\\0\\-5\end{bmatrix}\). Find the image of \(\begin{bmatrix}1\\1\\-2\end{bmatrix}\) under \(T\).
\[T\left (\begin{bmatrix}1\\1\\-2\end{bmatrix}\right )=\begin{bmatrix}\answer {-6}\\\answer {3}\\\answer {11}\end{bmatrix}\]
Prove that the identity transformation is linear. (Theorem 27)
Prove that the zero transformation is linear. (Theorem 29)
Find the domain and the codomain of the linear transformation \(T:\RR ^n\rightarrow \RR ^m\) induced by \(A\); find and draw the image of \(T\).
\[A=\begin{bmatrix}0&0\\1&1\\2&0\end{bmatrix}\]
Domain: \(\RR ^n\), where \(n=\answer {2}\).

Codomain: \(\RR ^m\), where \(m=\answer {3}\).

Find the domain and the codomain of the linear transformation \(T:\RR ^n\rightarrow \RR ^m\) induced by \(A\); find and draw the image of \(T\).
\[A=\begin{bmatrix}3&-1\\-3&1\end{bmatrix}\]

Exercise Source

Practice Problem is adopted from Problem 5.1.3 of Ken Kuttler’s A First Course in Linear Algebra. (CC-BY)

Ken Kuttler, A First Course in Linear Algebra, Lyryx 2017, Open Edition, p. 272.