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Coordinated Linear Algebra

Anna Davis, Paul Zachlin, Allan Donsig, Levi Heath, Mikkel Munkholm

Section 6.2 Linear Transformations

We start by reviewing the definition of a function.

Definition 6.2.1.

Let \(V\) and \(W\) be sets. A function \(f\) from \(V\) into \(W\text{,}\) denoted by
\begin{equation*} f:V\rightarrow W \end{equation*}
assigns to each element \(x\) of \(V\text{,}\) an element \(y=f(x)\) of \(W\text{.}\) Moreover, we use the the following terminology.
The set \(V\) is called the domain of \(f\text{,}\) and the set \(W\) is called the codomain.
If \(y=f(x)\text{,}\) we say that \(x\) maps to \(y\text{,}\) and \(y\) is the image of \(x\text{.}\)
The collection of images of all points of \(V\) is called the image of \(V\) under \(f\text{,}\) or the image of \(f\text{.}\) (It is also known as the range of \(f\text{.}\))
In algebra and calculus you worked with functions \(f:\R\rightarrow\R\) 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.

Exploration 6.2.1.

In this exercise we will introduce a very special type of transformation by contrasting the effects of two transformations on vectors of \(\R^2\text{.}\) We will see that some transformations have ``nice" properties, while others do not. Define \(T_1\) and \(T_2\) as follows:
\begin{equation*} T_1:\R^2\rightarrow\R^2; \quad T_1\left(\begin{bmatrix} x\\ y \end{bmatrix}\right)=\begin{bmatrix} x-y\\ x \end{bmatrix}, \end{equation*}
\begin{equation*} T_2:\R^2\rightarrow\R^2; \quad T_2\left(\begin{bmatrix} x\\ y \end{bmatrix}\right)=\begin{bmatrix} -x+y+1\\ y-2 \end{bmatrix}. \end{equation*}
Each of these transformations takes a vector in \(\R^2\text{,}\) and maps it to another vector in \(\R^2\text{.}\) To see if you understand how these transformations are defined, see if you can determine what these transformations do to the vector \([4,3]\text{.}\)

Problem 6.2.2.

Compute the following two images:
\begin{equation*} T_1\left(\begin{bmatrix} 4\\ 3 \end{bmatrix}\right) \quad \text{and} \quad T_2\left(\begin{bmatrix} 4\\ 3 \end{bmatrix}\right) \end{equation*}
Answer.
\begin{equation*} T_1\left(\begin{bmatrix} 4\\ 3 \end{bmatrix}\right)=\begin{bmatrix} 1\\ 4 \end{bmatrix} \quad \text{and} \quad T_2\left(\begin{bmatrix} 4\\ 3 \end{bmatrix}\right)=\begin{bmatrix} 0\\ 1\end{bmatrix}. \end{equation*}
Now, let’s take the vector \([4,3]\) and multiply it by a scalar, say \(7\text{.}\)
\begin{equation*} 7\begin{bmatrix} 4\\ 3 \end{bmatrix} = \begin{bmatrix} 28\\ 21 \end{bmatrix}. \end{equation*}
Now let’s compare how \(T_1\) and \(T_2\) ``handle" this product. Starting with \(T_1\text{,}\) we compute:
\begin{equation*} 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}. \end{equation*}
Observe that multiplying the original vector by \(7\text{,}\) then applying \(T_1\text{,}\) has the same effect as applying \(T_1\) to the original vector, then multiplying the image by \(7\text{.}\) In other words,
\begin{equation*} 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). \end{equation*}
Diagrammatically, this can be represented as follows.
T with domain and codomain pictured
You should try to verify that this property does not hold for transformation \(T_2\text{.}\) In other words,
\begin{equation*} T_2\left(7\begin{bmatrix} 4\\ 3 \end{bmatrix}\right)\neq 7T_2\left(\begin{bmatrix} 4\\ 3 \end{bmatrix}\right). \end{equation*}
There is nothing special about the number \(7\text{,}\) and it is not hard to prove that for any scalar \(k\) and vector \(\mathbf{u}\) of \(\R^2\text{,}\) \(T_1\) satisfies
\begin{equation} kT_1(\mathbf{u})= T_1(k\mathbf{u}).\tag{6.2.1} \end{equation}
It turns out that \(T_1\) satisfies another important property. For all vectors \(\mathbf{u}\) and \(\mathbf{v}\) of \(\R^2\) we have:
\begin{equation} T_1(\mathbf{u}+\mathbf{v}) = T_1(\mathbf{u})+T_1(\mathbf{v})\tag{6.2.2} \end{equation}
We leave it to the reader to illustrate this property with a specific example (see Exercise 6.2.7.1). We will show that \(T_1\) satisfies (6.2.2) in general. Let
\begin{equation*} \mathbf{u}=\begin{bmatrix} u_1\\ u_2 \end{bmatrix} \quad \text{and} \mathbf{v}=\begin{bmatrix} v_1\\ v_2 \end{bmatrix}. \end{equation*}
then
\begin{align*} T_1(\mathbf{u}+\mathbf{v})\amp =T_1\left(\begin{bmatrix} u_1\\ u_2 \end{bmatrix}+\begin{bmatrix} v_1\\ v_2 \end{bmatrix}\right) \\ \amp =T_1\left(\begin{bmatrix} u_1+v_1\\ u_2+v_2 \end{bmatrix}\right) \\ \amp =\begin{bmatrix} u_1+v_1-u_2-v_2\\ u_1+v_1 \end{bmatrix} \\ \amp =\begin{bmatrix} u_1-u_2\\ u_1 \end{bmatrix}+\begin{bmatrix} v_1-v_2\\ v_1 \end{bmatrix} \\ \amp=T_1\left(\begin{bmatrix} u_1\\ u_2 \end{bmatrix}\right)+T_1\left(\begin{bmatrix} v_1\\ v_2 \end{bmatrix}\right) \\ \amp =T_1(\mathbf{u})+T_1(\mathbf{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 \(\mathbf{u}\) and \(\mathbf{v}\) such that
\begin{equation} T_2(\mathbf{u}+\mathbf{v}) \neq T_2(\mathbf{u})+T_2(\mathbf{v}).\tag{6.2.3} \end{equation}
See Exercise 6.2.7.2 for more on this.
Transformations satisfying (6.2.1) and (6.2.2), like \(T_1\text{,}\) are called linear transformations. Transformations like \(T_2\) are not linear. You have encountered several linear transformations in the form of matrix transformations previously.

Definition 6.2.3.

A transformation \(T:\R^n\rightarrow \R^m\) is called a linear transformation if the following are true for all vectors \(\mathbf{u}\) and \(\mathbf{v}\) in \(\R^n\text{,}\) and scalars \(k\text{.}\)
\begin{equation} T(k\mathbf{u})= kT(\mathbf{u})\tag{6.2.4} \end{equation}
\begin{equation} T(\mathbf{u}+\mathbf{v})= T(\mathbf{u})+T(\mathbf{v})\tag{6.2.5} \end{equation}
Equations (6.2.4) and (6.2.5) of the above definition can be illustrated diagrammatically as follows.
Linearity drawn as function diagram
Continuation of above

Remark 6.2.4.

The properties (6.2.4) and (6.2.5) are often combined into a single property, namely
\begin{equation} T(k_1\mathbf{u}+k_2\mathbf{v})= k_1T(\mathbf{u})+k_2T(\mathbf{v})\tag{6.2.6} \end{equation}

Example 6.2.5.

Suppose \(T:\R^2\rightarrow \R^3\) is a linear transformation such that
\begin{equation*} T\left(\begin{bmatrix}1\\2\end{bmatrix}\right)=\begin{bmatrix}-1\\0\\3\end{bmatrix}\quad\text{and}\quad T\left(\begin{bmatrix}0\\-1\end{bmatrix}\right)=\begin{bmatrix}2\\-1\\0\end{bmatrix}. \end{equation*}
Find each of the following:
  1. \begin{equation*} 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). \end{equation*}
  2. \begin{equation*} T\left(\begin{bmatrix}1\\1\end{bmatrix}\right). \end{equation*}
Answer.
Item 1: Because \(T\) is a linear transformation, it satisfies (6.2.6). We compute:
\begin{align*} T\left(\begin{bmatrix}2\\5\end{bmatrix}\right)\amp =T\left(2\begin{bmatrix}1\\2\end{bmatrix}-\begin{bmatrix}0\\-1\end{bmatrix}\right) \\ \amp =2T\left(\begin{bmatrix}1\\2\end{bmatrix}\right)-T\left(\begin{bmatrix}0\\-1\end{bmatrix}\right) \\ \amp =2\begin{bmatrix}-1\\0\\3\end{bmatrix}-\begin{bmatrix}2\\-1\\0\end{bmatrix} \\ \amp =\begin{bmatrix}-2\\0\\6\end{bmatrix}-\begin{bmatrix}2\\-1\\0\end{bmatrix}=\begin{bmatrix}-4\\1\\6\end{bmatrix}. \end{align*}
Item 2 Observe that
\begin{equation*} \begin{bmatrix}1\\1\end{bmatrix}=\begin{bmatrix}1\\2\end{bmatrix}+\begin{bmatrix}0\\-1\end{bmatrix}. \end{equation*}
By (6.2.6) we have:
\begin{align*} T\left(\begin{bmatrix}1\\1\end{bmatrix}\right)\amp =T\left(\begin{bmatrix}1\\2\end{bmatrix}+\begin{bmatrix}0\\-1\end{bmatrix}\right) \\ \amp =T\left(\begin{bmatrix}1\\2\end{bmatrix}\right)+T\left(\begin{bmatrix}0\\-1\end{bmatrix}\right) \\ \amp =\begin{bmatrix}-1\\0\\3\end{bmatrix}+\begin{bmatrix}2\\-1\\0\end{bmatrix}=\begin{bmatrix}1\\-1\\3\end{bmatrix}. \end{align*}

Observation 6.2.6.

In Example 6.2.5 we were given the images of two vectors, \([1,2]\) and \([0,-1]\text{,}\) under a linear transformation \(T\text{.}\)
Based on this information, we were able to determine the images of two additional vectors: \([2,5]\) and \([1,1]\text{.}\) The reason we were able to determine \(T([2,5])\) and \(T([1,1])\) is because \([2,5]\) and \([1,1]\) can be written as unique linear combinations of \([1,2]\) and \([0,-1]\text{.}\)

Problem 6.2.7.

Can every vector of \(\R^2\) be written as a linear combination of \([1,2]\) and \([0,-1]\text{?}\)
Answer.
Yes.

Problem 6.2.8.

Is the information provided in Example 6.2.5 sufficient to determine the image of every vector in \(\R^2\) under \(T\text{?}\)
Answer.
Yes.

Example 6.2.9.

Suppose \(T:\R^2\rightarrow\R^2\) is a transformation such that
\begin{equation*} T\left(\begin{bmatrix} 2\\ 1 \end{bmatrix}\right)=\begin{bmatrix} 3\\ 2 \end{bmatrix},\,\,\,T\left(\begin{bmatrix} -1\\ 0 \end{bmatrix}\right)=\begin{bmatrix} 1\\ 1 \end{bmatrix},\,\,\,T\left(\begin{bmatrix} -1\\ 1 \end{bmatrix}\right)=\begin{bmatrix} 2\\ -4 \end{bmatrix}. \end{equation*}
Determine whether \(T\) is a linear transformation.
Answer.
Observe that
\begin{equation*} \begin{bmatrix} -1\\ 1 \end{bmatrix}=\begin{bmatrix} 2\\ 1 \end{bmatrix}+3\begin{bmatrix} -1\\ 0 \end{bmatrix} \end{equation*}
If \(T\) were a linear transformation, then we would have:
\begin{equation*} T\left(\begin{bmatrix} -1\\ 1 \end{bmatrix}\right)=T\left(\begin{bmatrix} 2\\ 1 \end{bmatrix}+3\begin{bmatrix} -1\\ 0 \end{bmatrix}\right)=\begin{bmatrix} 3\\ 2 \end{bmatrix}+3\begin{bmatrix} 1\\ 1 \end{bmatrix}=\begin{bmatrix} 6\\ 5 \end{bmatrix}. \end{equation*}
But according to the given,
\begin{equation*} T\left(\begin{bmatrix}-1\\1\end{bmatrix}\right)=\begin{bmatrix}2\\-4\end{bmatrix} \end{equation*}
Since \(\begin{bmatrix} 6\\ 5 \end{bmatrix}\neq \begin{bmatrix} 2\\ -4 \end{bmatrix}\) we conclude that transformation \(T\) is not linear.
In Exploration 6.2.1 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.

Proof.

To prove Item 1, let \(\mathbf{v}\) be any vector in \(\R^n\text{.}\) By linearity of \(T\text{,}\) we have:
\begin{equation*} T(\mathbf{0})=T(\mathbf{v}-\mathbf{v})=T(\mathbf{v})-T(\mathbf{v})=\mathbf{0}. \end{equation*}
Part Item 2 will become evident after the next section by combinining observations there with Exercise 6.1.5.3.

Example 6.2.11.

Use Theorem 6.2.10 to show that transformation \(T_2\) of Exploration 6.2.1 is not linear.
Answer.
Recall that \(T_2:\R^2\rightarrow\R^2\) was defined by
\begin{equation*} T_2\left(\begin{bmatrix} x\\ y \end{bmatrix}\right)=\begin{bmatrix} -x+y+1\\ y-2 \end{bmatrix} \end{equation*}
We evaluate \(T_2\) at \(\mathbf{0}\text{:}\)
\begin{equation*} T_2(\mathbf{0})=\begin{bmatrix} -0+0+1\\ 0-2 \end{bmatrix}=\begin{bmatrix}1\\-2\end{bmatrix}\neq\mathbf{0} \end{equation*}
Since \(T_2(\mathbf{0})\neq\mathbf{0}\text{,}\) \(T_2\) is not linear.

Subsection 6.2.1 Linear Transformations Induced by Matrices

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

Proof.

Let \(\mathbf{u}\) and \(\mathbf{v}\) be vectors in \(\R^n\text{,}\) and let \(k\) be a scalar. By properties of matrix multiplication we have:
\begin{equation*} T(\mathbf{u}+\mathbf{v})=A(\mathbf{u}+\mathbf{v})=A\mathbf{u}+A\mathbf{v}=T(\mathbf{u})+T(\mathbf{v}), \end{equation*}
\begin{equation*} T(k\mathbf{u})=A(k\mathbf{u})=kA\mathbf{u}=kT(\mathbf{u}). \end{equation*}
Therefore \(T\) is a linear transformation.

Example 6.2.13.

Let \(T:\R^n\rightarrow\R^m\) be a linear transformation induced by
\begin{equation*} A=\begin{bmatrix} 2\amp 0\\ 1\amp 4\\ 0\amp 1 \end{bmatrix} \end{equation*}
  1. Find \(n\) and \(m\text{.}\)
  2. Find the image of \(T\text{.}\)
Answer.
Item 1 \(A\) is a \(3\times 2\) matrix, so for the expression \(T(\mathbf{x})=A\mathbf{x}\) to make sense, \(\mathbf{x}\) has to be a \(2\times 1\) vector. Thus, the domain of \(T\) is \(\R^2\) (\(n=2\)). The product \(A\mathbf{x}\) is a \(3\times 1\) vector. The codomain of \(T\) is \(\R^3\) (\(m=3\)). Item 2 By Definition 6.2.1, the image of \(T\) consists of images of all individual vectors in \(\R^2\) under \(T\text{.}\)
Every vector \(\mathbf{v}\) in \(\R^2\) can be written as \(\mathbf{v}=a\mathbf{i}+b\mathbf{j}\) for some real numbers \(a\) and \(b\text{.}\) Consider the image of \(\mathbf{v}\text{:}\)
\begin{equation*} T(\mathbf{v})=T(a\mathbf{i}+b\mathbf{j})=aT(\mathbf{i})+bT(\mathbf{j})=a\begin{bmatrix}2\\1\\0\end{bmatrix}+b\begin{bmatrix}0\\4\\1\end{bmatrix}. \end{equation*}
This shows that the range, or the image, of \(T\) consists of all linear combinations of the columns of \(A\text{.}\) In other words, the image of \(T\) is the span of vectors \([2,1,0]\) and \([0,4,1]\text{.}\) The two vectors are not scalar multiples of each other, therefore they span a plane in \(\R^3\text{.}\)

Example 6.2.14.

Let \(T:\R^n\rightarrow\R^m\) be a linear transformation induced by
\begin{equation*} A=\begin{bmatrix} -2\amp 1\amp 3\\ 4\amp -2\amp -6 \end{bmatrix} \end{equation*}
  1. Find \(n\) and \(m\text{.}\)
  2. Find and draw the image of \(T\text{.}\)
Answer.
For part Item 1: \(n=3 \) and \(m = 2 \text{.}\)
For Item 2: To find the image of \(T\text{,}\) we will take a slightly different approach from what we did in Example 6.2.13Item 2.
Let \(\mathbf{v}=[a,b,c]\) be an arbitrary vector of \(\R^3\text{.}\) The image of \(\mathbf{v}\) is given by
\begin{align*} T(\mathbf{v})=\begin{bmatrix} -2\amp 1\amp 3\\ 4\amp -2\amp -6 \end{bmatrix}\begin{bmatrix}a\\b\\c\end{bmatrix}\amp =a\begin{bmatrix}-2\\4\end{bmatrix}+b\begin{bmatrix}1\\-2\end{bmatrix}+c\begin{bmatrix}3\\-6\end{bmatrix} \\ \amp =(a(-2)+b+c(3))\begin{bmatrix}1\\-2\end{bmatrix}. \end{align*}
This shows that the image of every vector in \(\R^3\) is a scalar multiple of \([1,-2]\text{.}\) This means that the image of \(T\) is a line in \(\R^2\text{.}\)
Line generated is graphed

Subsection 6.2.2 Linear Transformations of Subspaces of \(\R^n\)

Definition 6.2.3 defines a linear transformation as a map from \(\R^n\) into \(\R^m\text{.}\) We will now make this definition more general by allowing the domain and the codomain of the transformation to be subspaces of \(\R^n\) and \(\R^m\text{.}\) Eventually, a linear transformation will be defined as a mapping between vector spaces.

Definition 6.2.15.

Let \(V\) and \(W\) be subspaces of \(\R^n\) and \(\R^m\text{.}\) Let \(T:V\rightarrow W\) be a transformation. We call \(T\) linear transformation if for all vectors \(\mathbf{u}\) and \(\mathbf{v}\) in \(V\text{,}\) and scalars \(k\text{,}\) the following two rules hold:
\begin{equation*} T(k\mathbf{u})= kT(\mathbf{u}), \end{equation*}
\begin{equation*} T(\mathbf{u}+\mathbf{v})= T(\mathbf{u})+T(\mathbf{v}). \end{equation*}

Example 6.2.16.

Let \(V\) be a subspace of \(\R^3\) consisting of all vectors in the \(xy\)-plane. Let \(W\) be a subspace of \(\R^3\) consisting of all vectors along the \(z\)-axis. Do a quick verification that \(V\) and \(W\) are subspaces of \(\R^3\text{.}\) Define a transformation \(T:V\rightarrow W\) by
\begin{equation*} T\left(\begin{bmatrix}a\\b\\0\end{bmatrix}\right)=\begin{bmatrix}0\\0\\a+b\end{bmatrix}. \end{equation*}
Show that \(T\) is a linear transformation, and describe its action geometrically.
Answer.
Consider two arbitrary elements \([a_1, b_1, 0]\) and \([a_2, b_2, 0]\) of \(V\text{.}\)
\begin{align*} T\left(\begin{bmatrix}a_1\\b_1\\0\end{bmatrix}+ \begin{bmatrix}a_2\\b_2\\0\end{bmatrix}\right)\amp =T\left(\begin{bmatrix}a_1+a_2\\b_1+b_2\\0\end{bmatrix}\right) \\ \amp =\begin{bmatrix}0\\0\\(a_1+a_2)+(b_1+b_2)\end{bmatrix} \\ \amp =\begin{bmatrix}0\\0\\(a_1+b_1)+(a_2+b_2)\end{bmatrix} \\ \amp =\begin{bmatrix}0\\0\\a_1+b_1\end{bmatrix}+\begin{bmatrix}0\\0\\a_2+b_2\end{bmatrix} \\ \amp =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
\begin{equation*} T\left(k\begin{bmatrix}a_1\\b_1\\0\end{bmatrix}\right)=kT\left(\begin{bmatrix}a_1\\b_1\\0\end{bmatrix}\right) \end{equation*}
is similar, and we omit the details.
We have shown that \(T\) is a linear transformation. \(T\) maps all vectors in the \(xy\)-plane to the \(z\)-axis. The following diagram helps us visualize the action of \(T\) on a specific vector.
[1,2,0] acted on by T
We can investigate further. Recall that \(T\) is defined by
\begin{equation*} T\left(\begin{bmatrix}a\\b\\0\end{bmatrix}\right)=\begin{bmatrix}0\\0\\a+b\end{bmatrix} \end{equation*}
Next, we ask you to consider the diagram below and try solve the following problem.
Problem 6.2.17.
Choose the proper answer for each of the statements below.
  1. Is the image of the line \(y=-x\) under \(T\) a line, a plane, the zero vector or all of \(\R^3\text{?}\)
  2. is the image of the orange part of the domain (the front triangle) the positive \(z\)-axis, the negative \(z\)-axis, the zero vector or the entire \(z\)-axis?
  3. is the image of the purple part of the domain (the back triangle) the positive \(z\)-axis, the negative \(z\)-axis, the zero vector or the entire \(z\)-axis?
Answer.
  1. The image of the line \(y=-x\) under \(T\) is the the zero vector.
  2. The image of the orange part of the domain (the front triangle) is the positive \(z\)-axis.
  3. The image of the purple part of the domain (the back triangle) is the negative \(z\)-axis.
We conclude this section by introducing two simple but important transformations.

Definition 6.2.18.

The identity transformation on \(V\text{,}\) denoted by \(\id_V\text{,}\) is a transformation that maps each element of \(V\) to itself. In other words,
\begin{equation*} \id_V:V\rightarrow V \end{equation*}
is a transformation such that
\begin{equation*} \id_V(\mathbf{v})=\mathbf{v}\quad\text{for all}\quad \mathbf{v} \in V \end{equation*}

Definition 6.2.19.

The zero transformation, \(Z\text{,}\) maps every element of the domain to the zero vector. In other words,
\begin{equation*} Z:V\rightarrow W \end{equation*}
is a transformation such that
\begin{equation*} Z(\mathbf{v})=\mathbf{0}\quad\text{for all}\quad \mathbf{v} \in V \end{equation*}

Proof.

Proof.

Subsection 6.2.3 Standard Matrix of a Linear Transformation from \(\R^n\) to \(\R^m\)

In the preceding sections, we learned several important properties of matrix transformations of \(\R^n\) and subspaces of \(\R^n\text{.}\) Let’s summarize the main points.
The last point in the summary is so important that it is worth illustrating again.

Example 6.2.23.

Let \(T:\R^3\rightarrow \R^2\) be a linear transformation. Suppose that the only information we have about this transformation is that
\begin{equation*} T(\mathbf{i})=\begin{bmatrix}3\\-1\end{bmatrix}, \quad T(\mathbf{j})=\begin{bmatrix}0\\4\end{bmatrix} \quad \text{and} \quad T(\mathbf{k})=\begin{bmatrix}-2\\1\end{bmatrix}. \end{equation*}
Is this information sufficient to determine the image of \(\mathbf{w}=\begin{bmatrix}1\\-3\\6\end{bmatrix}\text{?}\)
Answer.
Observe that
\begin{equation*} \mathbf{w}=\mathbf{i}-3\mathbf{j}+6\mathbf{k} \end{equation*}
We find \(T(\mathbf{w})\) by using the fact that \(T\) is linear:
\begin{align*} T(\mathbf{w}) \amp =T(\mathbf{i}-3\mathbf{j}+6\mathbf{k}) \\ \amp=T(\mathbf{i})-3T(\mathbf{j})+6T(\mathbf{k}) \\ \amp=\begin{bmatrix}3\\-1\end{bmatrix}-3\begin{bmatrix}0\\4\end{bmatrix}+6\begin{bmatrix}-2\\1\end{bmatrix} \\ \amp=\begin{bmatrix}-9\\-7\end{bmatrix} \end{align*}
Because of properties of linear transformations, the information about the images of the standard unit vectors proved to be sufficient for us to determine the image of \(\mathbf{w}\text{.}\)
In Example 6.2.23, there was nothing special about the vector \(\mathbf{w}\text{.}\) Any vector \(\mathbf{x}\) of \(\R^n\) can be written as a unique linear combination of the standard unit vectors \(\mathbf{e}_1,\ldots , \mathbf{e}_n\text{.}\) Therefore, the image of any vector \(\mathbf{x}\) under a linear transformation \(T:\R^n\rightarrow \R^m\) is uniquely determined by the images of \(\mathbf{e}_1, \ldots , \mathbf{e}_n\text{.}\) Knowing \(T(\mathbf{e}_1),\ldots , T(\mathbf{e}_n)\) allows us to construct a matrix \(A\text{,}\) with \(T(\mathbf{e}_1),\ldots , T(\mathbf{e}_n)\) as columns, that induces transformation \(T\text{.}\) We formalize this idea in a theorem.

Proof.

Observe that
\begin{align*} \mathbf{x}=\begin{bmatrix}x_1\\x_2\\\vdots\\x_n\end{bmatrix} \amp=x_1\begin{bmatrix}1\\0\\\vdots\\0\end{bmatrix}+x_2\begin{bmatrix}0\\1\\\vdots\\0\end{bmatrix}+\dots+x_n\begin{bmatrix}0\\0\\\vdots\\1\end{bmatrix} \\ \amp=x_1\mathbf{e}_1+x_2\mathbf{e}_2+\dots+x_n\mathbf{e}_n. \end{align*}
Because \(T\) is linear, we have
\begin{align*} T(\mathbf{x})\amp =T(x_1\mathbf{e}_1+x_2\mathbf{e}_2+\dots+x_n\mathbf{e}_n) \\ \amp=x_1T(\mathbf{e}_1)+x_2T(\mathbf{e}_2)+\dots+x_nT(\mathbf{e}_n) \\ \amp =\begin{bmatrix} | \amp |\amp \amp |\\ T(\mathbf{e}_1) \amp T(\mathbf{e}_2)\amp \dots \amp T(\mathbf{e}_n)\\ |\amp | \amp \amp | \end{bmatrix}\begin{bmatrix}x_1\\x_2\\\vdots\\x_n\end{bmatrix}=A\mathbf{x}. \end{align*}
Thus, for every \(\mathbf{x}\) in \(\R^n\text{,}\) we have \(T(\mathbf{x})=A\mathbf{x}\text{.}\)
Theorem 6.2.12 shows that every matrix transformation is linear. Theorem 6.2.24 states that every linear transformation from \(\R^n\) into \(\R^m\) is a matrix transformation. We combine these results in a corollary.
The results of this section rely on the fact that every vector of \(\R^n\) can be written as a unique linear combination of the standard unit vectors \(\mathbf{e}_1,\mathbf{e}_2,\dots,\mathbf{e}_n\text{.}\) These vectors form the standard basis for \(\R^n\text{.}\) Later on, when we encounter transformations of arbitrary bases, we will observe that the matrix used to represent a linear transformation depends on a choice of basis. Since we are using the standard basis, it is natural to name the matrix in Theorem 6.2.24 accordingly.

Definition 6.2.26.

The matrix in Theorem 6.2.24 is known as the standard matrix of the linear transformation \(T\text{.}\)

Example 6.2.27.

The standard matrix of a linear transformation \(T:\R^3\rightarrow \R^2\) such that
\begin{equation*} T(\mathbf{i})=\begin{bmatrix}2\\-1\end{bmatrix}, \quad T(\mathbf{j})=\begin{bmatrix}-1\\3\end{bmatrix} \quad \text{and} \quad T(\mathbf{k})=\begin{bmatrix}0\\4\end{bmatrix} \end{equation*}
is
\begin{equation*} A=\begin{bmatrix}2\amp -1\amp 0\\-1\amp 3\amp 4\end{bmatrix}. \end{equation*}

Example 6.2.28.

Find the standard matrix of a linear transformation \(T:\R^2\rightarrow \R^2\) such that \(T(\mathbf{i})=2\mathbf{i}\) and \(T(\mathbf{j})=2\mathbf{j}\text{.}\)
Answer.
We use the images of \(\mathbf{i}\) and \(\mathbf{j}\) as columns of the matrix. The standard matrix of \(T\) is
\begin{equation*} \begin{bmatrix}2\amp 0\\0\amp 2\end{bmatrix}. \end{equation*}

Example 6.2.29.

Find the standard matrix of a linear transformation \(T:\R^2\rightarrow \R^4\text{,}\) where
\begin{equation*} T\left(\begin{bmatrix}3\\1\end{bmatrix}\right)=\begin{bmatrix}6\\1\\13\\-1\end{bmatrix} \quad \text{and} \quad T\left(\begin{bmatrix}-2\\0\end{bmatrix}\right)=\begin{bmatrix}-2\\0\\-8\\2\end{bmatrix}. \end{equation*}
Answer.
In this example we are not given the images of the standard basis vectors \(\mathbf{i}\) and \(\mathbf{j}\text{.}\) However, we can find the images of \(\mathbf{i}\) and \(\mathbf{j}\) by expressing \(\mathbf{i}\) and \(\mathbf{j}\) as linear combinations of \([3,1]\) and \([-2,0]\text{,}\) then apply the fact that \(T\) is linear. Let’s start with the easy one.
\begin{equation*} \mathbf{i}=-\frac{1}{2}\begin{bmatrix}-2\\0\end{bmatrix}. \end{equation*}
Therefore, by linearity of \(T\text{,}\) we have:
\begin{equation*} T(\mathbf{i})=T\left(-\frac{1}{2}\begin{bmatrix}-2\\0\end{bmatrix}\right)=-\frac{1}{2}T\left(\begin{bmatrix}-2\\0\end{bmatrix}\right)=-\frac{1}{2}\begin{bmatrix}-2\\0\\-8\\2\end{bmatrix}=\begin{bmatrix}1\\0\\4\\-1\end{bmatrix}. \end{equation*}
This gives us the first column of the standard matrix for \(T\text{.}\)
You can solve the vector equation
\begin{equation*} a\begin{bmatrix}3\\1\end{bmatrix}+b\begin{bmatrix}-2\\0\end{bmatrix}=\mathbf{j} \end{equation*}
to express \(\mathbf{j}\) as a linear combination of \([3,1]\) and \([-2,0]\) as follows:
\begin{equation*} \mathbf{j}=\begin{bmatrix}3\\1\end{bmatrix}+\frac{3}{2}\begin{bmatrix}-2\\0\end{bmatrix}. \end{equation*}
By linearity of \(T\text{,}\)
\begin{align*} T(\mathbf{j})\amp =T\left(\begin{bmatrix}3\\1\end{bmatrix}+\frac{3}{2}\begin{bmatrix}-2\\0\end{bmatrix}\right) \\ \amp =T\left(\begin{bmatrix}3\\1\end{bmatrix}\right)+\frac{3}{2}T\left(\begin{bmatrix}-2\\0\end{bmatrix}\right) \\ \amp =\begin{bmatrix}6\\1\\13\\-1\end{bmatrix}+\frac{3}{2}\begin{bmatrix}-2\\0\\-8\\2\end{bmatrix}=\begin{bmatrix}3\\1\\1\\2\end{bmatrix}. \end{align*}
This gives us the second column of the standard matrix. Putting all of the information together, we get the following standard matrix for \(T\text{:}\)
\begin{equation*} A=\begin{bmatrix}1\amp 3\\0\amp 1\\4\amp 1\\-1\amp 2\end{bmatrix}. \end{equation*}

Subsection 6.2.4 The Image

In this section we will often use \(U\text{,}\) \(V\) and \(W\) to denote subspaces of \(\R^n\text{,}\) or any other finite-dimensional vector space, such as those we study in a later chaper.

Definition 6.2.30.

Let \(V\) and \(W\) be vector spaces, and let \(T:V\rightarrow W\) be a linear transformation. The image of \(T\text{,}\) denoted by \(\mbox{im}(T)\text{,}\) is the set
\begin{equation*} \mbox{im}(T)=\{T(\mathbf{v}):\mathbf{v}\in V\}. \end{equation*}
In other words, the image of \(T\) consists of individual images of all vectors of \(V\text{.}\)

Example 6.2.31.

Consider the linear transformation \(T:\R^3\rightarrow \R^2\) with standard matrix
\begin{equation*} A=\begin{bmatrix}1\amp 2\amp 3\\2\amp 4\amp 6\end{bmatrix}. \end{equation*}
  1. Find \(\mbox{im}(T)\text{.}\)
  2. Illustrate the action of \(T\) with a sketch.
Answer.
Item 1: Let \(\mathbf{v}=\begin{bmatrix}a\\b\\c\end{bmatrix}\) then
\begin{equation*} T(\mathbf{v})=A\mathbf{v}=\begin{bmatrix}1\amp 2\amp 3\\2\amp 4\amp 6\end{bmatrix}\begin{bmatrix}a\\b\\c\end{bmatrix}=a\begin{bmatrix}1\\2\end{bmatrix}+b\begin{bmatrix}2\\4\end{bmatrix}+c\begin{bmatrix}3\\6\end{bmatrix}. \end{equation*}
Thus, every element of the image can be written as a linear combination of the columns of \(A\text{.}\) We conclude that
\begin{equation*} \mbox{im}(T)=\mbox{span}\left(\begin{bmatrix}1\\2\end{bmatrix}, \begin{bmatrix}2\\4\end{bmatrix}, \begin{bmatrix}3\\6\end{bmatrix}\right)=\mbox{col}(A). \end{equation*}
Every column of \(A\) is a scalar multiple of \([1,2]\text{.}\) Thus,
\begin{equation*} \mbox{im}(T)=\mbox{span}\left(\begin{bmatrix}1\\2\end{bmatrix}, \begin{bmatrix}2\\4\end{bmatrix}, \begin{bmatrix}3\\6\end{bmatrix}\right)=\mbox{span}\left(\begin{bmatrix}1\\2\end{bmatrix}\right). \end{equation*}
The image of \(T\) is a line in \(\R^2\) determined by the vector \([1,2]\text{.}\)
Item 2: The action of \(T\) can be illustrated with a sketch.
Image of T graphed
In Example 6.2.31 we observed that the image of the linear transformation was equal to the column space of its standard matrix. In general, it is easy to see that if \(T:\R^n\rightarrow \R^m\) is a linear transformation with standard matrix \(A\) then the following relationship holds:
\begin{equation*} \mbox{im}(T)=\mbox{col}(A). \end{equation*}
In addition, by Theorem 5.3.10, we know that
\begin{equation*} \mbox{dim}(\mbox{im}(T))=\mbox{dim}(\mbox{col}(A))=\mbox{rank}(A). \end{equation*}

Example 6.2.32.

Let \(T:\R^5\rightarrow \R^4\) be a linear transformation with standard matrix
\begin{equation*} A=\begin{bmatrix}1 \amp 2 \amp 2 \amp -1 \amp 0\\-1 \amp 3 \amp 1 \amp 0 \amp -1\\3 \amp 0 \amp 0 \amp 3 \amp 6\\ 1 \amp -1 \amp 1 \amp -2 \amp -1\end{bmatrix}. \end{equation*}
Find \(\mbox{im}(T)\) and \(\mbox{dim}(\mbox{im}(T))\text{.}\)
Answer.
As in Example 6.2.31, the image of \(T\) is given by
\begin{equation*} \mbox{im}(T)=\mbox{span}\left(\begin{bmatrix}1\\-1\\3\\1\end{bmatrix}, \begin{bmatrix}2\\3\\0\\-1\end{bmatrix}, \begin{bmatrix}2\\1\\0\\1\end{bmatrix}, \begin{bmatrix}-1\\0\\3\\-2\end{bmatrix}, \begin{bmatrix}0\\-1\\6\\-1\end{bmatrix}\right)=\mbox{col}(A). \end{equation*}
This time it is harder to detect the vectors that can be eliminated from the spanning set without affecting the span. We have to rely on the reduced row-echelon form of \(A\text{.}\)
\begin{equation*} \begin{bmatrix}1 \amp 2 \amp 2 \amp -1 \amp 0\\-1 \amp 3 \amp 1 \amp 0 \amp -1\\3 \amp 0 \amp 0 \amp 3 \amp 6\\ 1 \amp -1 \amp 1 \amp -2 \amp -1\end{bmatrix} \rightsquigarrow \begin{bmatrix} 1 \amp 0 \amp 0 \amp 1 \amp 2\\0 \amp 1 \amp 0 \amp 1 \amp 1\\0 \amp 0 \amp 1 \amp -2 \amp -2\\ 0 \amp 0 \amp 0 \amp 0 \amp 0 \end{bmatrix}. \end{equation*}
We can see that \(\mbox{rank}(A)=3\text{,}\) so \(\mbox{dim}(\mbox{im}(T))=3\text{.}\) To identify vectors that span \(\mbox{im}(T)\text{,}\) we turn to Algorithm 5.3.9. We identify the first three columns as pivot columns. These columns are linearly independent and span \(\mbox{col}(A)\text{.}\) Therefore,
\begin{equation*} \mbox{im}(T)=\mbox{col}(A)=\mbox{span}\left(\begin{bmatrix}1\\-1\\3\\1\end{bmatrix}, \begin{bmatrix}2\\3\\0\\-1\end{bmatrix}, \begin{bmatrix}2\\1\\0\\1\end{bmatrix}\right) \end{equation*}
By Theorem 5.1.9 and Definition 5.3.8, we know that for an \(m\times n\) matrix \(A\text{,}\) \(\mbox{col}(A)\) is a subspace of \(\R^m\text{.}\) However, when vector spaces other than \(\R^m\) are involved, it is not yet clear that \(\mbox{im}(T)\) is a subspace of the codomain. The following theorem resolves this issue.

Proof.

To show that \(\mbox{im}(T)\) is a subspace, we need to show that \(\mbox{im}(T)\) is closed under addition and scalar multiplication. Suppose \(\mathbf{w}_1\) and \(\mathbf{w}_2\) are in \(\mbox{im}(T)\text{.}\) Then there are vectors \(\mathbf{v}_1\) and \(\mathbf{v}_2\) in \(V\) such that \(T(\mathbf{v}_1)=\mathbf{w}_1\) and \(T(\mathbf{v}_2)=\mathbf{w}_2\text{.}\) Then
\begin{equation*} \mathbf{w}_1+\mathbf{w}_2=T(\mathbf{v}_1)+T(\mathbf{v}_2)=T(\mathbf{v}_1+\mathbf{v}_2). \end{equation*}
This shows that \(\mathbf{w}_1+\mathbf{w}_2\) is in \(\mbox{im}(T)\text{.}\) For any scalar \(a\text{,}\) we have:
\begin{equation*} a\mathbf{w}_1=aT(\mathbf{v}_1)=T(a\mathbf{v}_1). \end{equation*}
This shows that \(a\mathbf{w}_1\) is in \(\mbox{im}(T)\text{.}\)
We can now define the rank of a linear transformation.

Definition 6.2.34.

The rank of a linear transformation \(T:V\rightarrow W\text{,}\) is the dimension of the image of \(T\text{.}\)
\begin{equation*} \mbox{rank}(T)=\mbox{dim}(\mbox{im}(T)). \end{equation*}
This definition gives us the following relationship between the rank of a linear transformation \(T:\R^n\rightarrow\R^m\) and the rank of the standard matrix \(A\) associated with it.

Subsection 6.2.5 The Kernel of a Linear Transformation

Exactly as in the preceding section, we will often use \(U\text{,}\) \(V\) and \(W\) to denote subspaces of \(\R^n\text{,}\) or any other finite-dimensional vector space, such as those we study in a later chaper.

Definition 6.2.36.

Let \(V\) and \(W\) be vector spaces, and let \(T:V\rightarrow W\) be a linear transformation. The kernel of \(T\text{,}\) denoted by \(\mbox{ker}(T)\text{,}\) is the set
\begin{equation*} \mbox{ker}(T)=\{\mathbf{v}:T(\mathbf{v})=\mathbf{0}\}. \end{equation*}
In other words, the kernel of \(T\) consists of all vectors of \(V\) that map to \(\mathbf{0}\) in \(W\text{.}\)
It is important to pay attention to the locations of the kernel and the image. We already proved that \(\mbox{im}(T)\) is a subspace of the codomain. In contrast, \(\mbox{ker}(T)\) is located in the domain. (We will prove shortly that it is a subspace of the domain.)
Kernel diagram shown

Example 6.2.37.

Let \(T:\R^5\rightarrow \R^4\) be a linear transformation with standard matrix
\begin{equation*} A=\begin{bmatrix}1 \amp 2 \amp 2 \amp -1 \amp 0\\-1 \amp 3 \amp 1 \amp 0 \amp -1\\3 \amp 0 \amp 0 \amp 3 \amp 6\\ 1 \amp -1 \amp 1 \amp -2 \amp -1\end{bmatrix}. \end{equation*}
  1. Find \(\mbox{ker}(T)\text{.}\)
  2. Is \(\mbox{ker}(T)\) a subspace of \(\R^5\text{?}\) If so, find \(\mbox{dim}(\mbox{ker}(T))\text{.}\)
Answer.
Item 1 To find the kernel of \(T\text{,}\) we need to find all vectors of \(\R^5\) that map to \(\mathbf{0}\) in \(\R^4\text{.}\) This amounts to solving the equation \(A\mathbf{x}=\mathbf{0}\text{.}\) Gauss-Jordan elimination yields:
\begin{equation*} \begin{bmatrix}1 \amp 2 \amp 2 \amp -1 \amp 0\\-1 \amp 3 \amp 1 \amp 0 \amp -1\\3 \amp 0 \amp 0 \amp 3 \amp 6\\ 1 \amp -1 \amp 1 \amp -2 \amp -1\end{bmatrix} \rightsquigarrow \begin{bmatrix} 1 \amp 0 \amp 0 \amp 1 \amp 2\\0 \amp 1 \amp 0 \amp 1 \amp 1\\0 \amp 0 \amp 1 \amp -2 \amp -2\\ 0 \amp 0 \amp 0 \amp 0 \amp 0 \end{bmatrix}. \end{equation*}
Thus, the kernel of \(T\) consists of all elements of the form:
\begin{equation*} \begin{bmatrix}-1\\-1\\2\\1\\0\end{bmatrix}s+\begin{bmatrix}-2\\-1\\2\\0\\1\end{bmatrix}t. \end{equation*}
We conclude that
\begin{equation*} \mbox{ker}(T)=\mbox{span}\left(\begin{bmatrix}-1\\-1\\2\\1\\0\end{bmatrix}, \begin{bmatrix}-2\\-1\\2\\0\\1\end{bmatrix}\right). \end{equation*}
Item 2: Since \(\mbox{ker}(T)\) is the span of two vectors of \(\R^5\text{,}\) we know that \(\mbox{ker}(T)\) is a subspace of \(\R^5\text{.}\) (See Theorem 5.1.9.) Observe that the two vectors in the spanning set are linearly independent. (How can we see this without performing computations?) Therefore \(\mbox{dim}(\mbox{ker}(T))=2\text{.}\)
Recall that the null space of a matrix \(A\) is defined to be set of all solutions to the homogeneous equation \(A\mathbf{x}=\mathbf{0}\text{.}\) This means that if \(T:\R^n\rightarrow \R^m\) is a linear transformation with standard matrix \(A\) then
\begin{equation*} \mbox{ker}(T)=\mbox{null}(A). \end{equation*}
We know that \(\mbox{null}(A)\) of an \(m\times n\) matrix is a subspace of \(\R^n\text{.}\) (See Theorem 5.3.14.) We conclude this section by showing that even when vector spaces other than \(\R^n\) are involved, the kernel of a linear transformation is a subspace of the domain of the transformation.

Proof.

To show that \(\mbox{ker}(T)\) is a subspace, we need to show that \(\mbox{ker}(T)\) is closed under addition and scalar multiplication. Suppose that \(\mathbf{v}_1\) and \(\mathbf{v}_2\) are in \(\mbox{ker}(T)\text{.}\) Then,
\begin{equation*} T(\mathbf{v}_1+\mathbf{v}_2)=T(\mathbf{v}_1)+T(\mathbf{v}_2)=\mathbf{0}+\mathbf{0}=\mathbf{0}. \end{equation*}
This shows that \(\mathbf{v}_1+\mathbf{v}_2\) is in \(\mbox{ker}(T)\text{.}\) For any scalar \(a\) we have:
\begin{equation*} T(a\mathbf{v}_1)=aT(\mathbf{v}_1)=a\mathbf{0}=\mathbf{0}. \end{equation*}
This shows that \(a\mathbf{v}_1\) is in \(\mbox{ker}(T)\text{.}\)

Definition 6.2.39.

The nullity of a linear transformation \(T:V\rightarrow W\text{,}\) is the dimension of the kernel of \(T\text{.}\)
\begin{equation*} \mbox{nullity}(T)=\mbox{dim}(\mbox{ker}(T)). \end{equation*}
This definition gives us the following relationship between nullity of a linear transformation \(T:\R^n\rightarrow\R^m\) and the nullity of the standard matrix \(A\) associated with it.

Subsection 6.2.6 Rank-Nullity Theorem for Linear Transformations

In Example 6.2.32 and Example 6.2.37, we found the image and the kernel of the linear transformation \(T:\R^5\rightarrow \R^4\) with standard matrix
\begin{equation*} A=\begin{bmatrix}1 \amp 2 \amp 2 \amp -1 \amp 0\\-1 \amp 3 \amp 1 \amp 0 \amp -1\\3 \amp 0 \amp 0 \amp 3 \amp 6\\ 1 \amp -1 \amp 1 \amp -2 \amp -1\end{bmatrix}. \end{equation*}
We also found that
\begin{equation*} \mbox{rank}(T)=\mbox{dim}(\mbox{im}(T))=\mbox{dim}(\mbox{col}(A))=\mbox{rank}(A)=3 \end{equation*}
and
\begin{equation*} \mbox{nullity}(T)=\mbox{dim}(\mbox{ker}(T))=\mbox{dim}(\mbox{null}(A))=\mbox{nullity}(A)=2. \end{equation*}
Because of Rank-Nullity Theorem for matrices (Theorem Theorem 5.3.17), it is not surprising that
\begin{equation*} \mbox{rank}(T)+\mbox{nullity}(T)=3+2=5=\mbox{dim}(\R^5). \end{equation*}
The following theorem is a generalization of this result.

Proof.

By Theorem 6.2.33, \(\mbox{im}(T)\) is a subspace of \(W\text{.}\) There exists a basis for \(\mbox{im}(T)\) of the form \(\{T(\mathbf{v}_1), \ldots,T(\mathbf{v}_r)\}\text{.}\) By Theorem 6.2.38, \(\mbox{ker}(T)\) is a subspace of \(V\text{.}\) Let \(\{\mathbf{u}_1,\ldots,\mathbf{u}_s\}\) be a basis for \(\mbox{ker}(T)\text{.}\) We will show that \(\{\mathbf{u}_1,\ldots ,\mathbf{u}_s, \mathbf{v}_1,\ldots ,\mathbf{v}_r\}\) is a basis for \(V\text{.}\) For any vector \(\mathbf{v}\) in \(V\text{,}\) we have:
\begin{equation*} T(\mathbf{v})=c_1T(\mathbf{v}_1)+\ldots +c_rT(\mathbf{v}_r) \end{equation*}
for some scalars \(c_i\) \((1\leq i\leq r)\text{.}\) Thus,
\begin{equation*} T(\mathbf{v})-\big(c_1T(\mathbf{v}_1)+\ldots +c_rT(\mathbf{v}_r)\big)=\mathbf{0}. \end{equation*}
By linearity,
\begin{equation*} T((\mathbf{v}-(c_1\mathbf{v}_1+\ldots +c_r\mathbf{v}_r))=\mathbf{0}. \end{equation*}
Therefore \(\mathbf{v}-(c_1\mathbf{v}_1+\ldots +c_r\mathbf{v}_r)\) is in \(\mbox{ker}(T)\text{.}\) Hence there are scalars \(a_i\) \((1\leq i\leq s)\) such that
\begin{equation*} \mathbf{v}-(c_1\mathbf{v}_1+\ldots +c_r\mathbf{v}_r)=a_1\mathbf{u}_1+\ldots +a_s\mathbf{u}_s. \end{equation*}
Thus,
\begin{equation*} \mathbf{v}=(c_1\mathbf{v}_1+\ldots +c_r\mathbf{v}_r)+(a_1\mathbf{u}_1+\ldots +a_s\mathbf{u}_s). \end{equation*}
We conclude that
\begin{equation*} V=\mbox{span}(\mathbf{u}_1,\ldots ,\mathbf{u}_s, \mathbf{v}_1,\ldots ,\mathbf{v}_r). \end{equation*}
Now we need to show that \(\{\mathbf{u}_1,\ldots ,\mathbf{u}_s, \mathbf{v}_1,\ldots ,\mathbf{v}_r\}\) is linearly independent. Suppose
\begin{equation} c_1\mathbf{v}_1+\ldots +c_r\mathbf{v}_r+a_1\mathbf{u}_1+\ldots +a_s\mathbf{u}_s=\mathbf{0}.\tag{6.2.7} \end{equation}
Applying \(T\) to both sides, we get
\begin{equation*} T(c_1\mathbf{v}_1+\ldots +c_r\mathbf{v}_r+a_1\mathbf{u}_1+\ldots +a_s\mathbf{u}_s)=T(\mathbf{0}), \end{equation*}
\begin{equation*} c_1T(\mathbf{v}_1)+\ldots +c_rT(\mathbf{v}_r)+a_1T(\mathbf{u}_1)+\ldots +a_sT(\mathbf{u}_s)=\mathbf{0}. \end{equation*}
But \(T(\mathbf{u}_i)=\mathbf{0}\) for \(1\leq i\leq s\text{,}\) thus
\begin{equation*} c_1T(\mathbf{v}_1)+\ldots +c_rT(\mathbf{v}_r)=\mathbf{0}. \end{equation*}
Since \(\{T(\mathbf{v}_1),\ldots ,T(\mathbf{v}_r)\}\) is linearly independent, it follows that each \(c_i=0\text{.}\) But then (6.2.7) implies that \(a_1\mathbf{u}_1+\ldots +a_s\mathbf{u}_s=\mathbf{0}\text{.}\) Because \(\{\mathbf{u}_1, \ldots ,\mathbf{u}_s\}\) is linearly independent, it follows that each \(a_i=0\text{.}\) We conclude that \(\{\mathbf{u}_1,\ldots ,\mathbf{u}_s,\mathbf{v}_1,\ldots ,\mathbf{v}_r\}\) is a basis for \(V\text{.}\) Thus,
\begin{equation*} \mbox{dim}(\mbox{ker}(T))+\mbox{dim}(\mbox{im}(T))=s+r=n. \end{equation*}

Exercises 6.2.7 Exercises

1.

Show that (6.2.2) of Exploration 6.2.1 holds for vectors \(\begin{bmatrix}3\\4\end{bmatrix}\) and \(\begin{bmatrix}-2\\1\end{bmatrix}\text{.}\)

3.

Suppose \(T:\R^{10}\rightarrow\R^2\) is a linear transformation such that \(T(\mathbf{u})=[2,-1]\) and \(T(\mathbf{v})=[-5,3]\text{.}\) Find the image of \(3\mathbf{u}-\mathbf{v}\text{.}\)
Answer.
\begin{equation*} T(3\mathbf{u}-\mathbf{v})=\begin{bmatrix}11\\-6\end{bmatrix}. \end{equation*}

4.

Let \(\mathbf{u}\) be a fixed vector. Define \(T_{\mathbf{u}}:\R^2\rightarrow\R^2\text{,}\) by \(T_{\mathbf{u}}(\mathbf{x})=\mathbf{u}-\mathbf{x}\text{.}\)
  1. Describe the effect of this transformation by sketching \({\bf x}\) and \(T_{\mathbf{u}}({\bf x})\) for at least four vectors \({\bf x}\) and a fixed vector \(\mathbf{u}\) of your choice.
  2. Is \(T_{\mathbf{u}}\) a linear transformation?

5.

Define \(P_{xy}:\R^3\rightarrow\R^2\text{,}\) by
\begin{equation*} P_{xy}\left(\begin{bmatrix} x\\ y\\ z \end{bmatrix} \right)=\begin{bmatrix} x\\ y \end{bmatrix}. \end{equation*}
This transformation is called an orthogonal projection onto the \(xy\)-plane. Show that \(P_{xy}\) is a linear transformation.

6.

Suppose a linear transformation \(T:\R^3\rightarrow\R^3\) maps
\begin{equation*} \mathbf{i} \ \text{to} \ \begin{bmatrix}2\\-1\\0\end{bmatrix}, \quad \ \mathbf{j} \ \text{to} \ \begin{bmatrix}-2\\4\\1\end{bmatrix}, \quad \text{and} \quad \mathbf{k} \ \text{to} \ \begin{bmatrix}3\\0\\-5\end{bmatrix}. \end{equation*}
Find the image of \(\begin{bmatrix}1\\1\\-2\end{bmatrix}\) under \(T\text{.}\)
Answer.
\begin{equation*} T\left(\begin{bmatrix}1\\1\\-2\end{bmatrix}\right)=\begin{bmatrix}-6\\3\\11\end{bmatrix}. \end{equation*}

Exercise Group.

For each matrix \(A\) below, find the domain together with codomain of the linear transformation \(T:\R^n\rightarrow\R^m\) induced by \(A\text{;}\) then find and draw the image of \(T\) (Hint: See Example 6.2.14.)
9.
\begin{equation*} A=\begin{bmatrix}0\amp 0\\1\amp 1\\2\amp 0\end{bmatrix}. \end{equation*}
Answer.
Domain: \(\R^n\text{,}\) where \(n=2\text{.}\)
Codomain: \(\R^m\text{,}\) where \(m=3\text{.}\)
10.
\begin{equation*} A=\begin{bmatrix}3\amp -1\\-3\amp 1\end{bmatrix}. \end{equation*}

11.

Suppose that a linear transformation \(T:\R^2\rightarrow\R^3\) is such that
\begin{equation*} T(\mathbf{i})=\begin{bmatrix}-4\\2\\1\end{bmatrix} \quad \text{and} \quad T(\mathbf{j})=\begin{bmatrix}0\\-1\\5\end{bmatrix}. \end{equation*}
Find
\begin{equation*} T\Big(\begin{bmatrix}4\\-1\end{bmatrix}\Big)\text{.} \end{equation*}
Answer.
\begin{equation*} T\Big(\begin{bmatrix}4\\-1\end{bmatrix}\Big)=\begin{bmatrix}-16\\9\\-1\end{bmatrix}. \end{equation*}

12.

Suppose that a linear transformation \(T:\R^2\rightarrow\R^3\) is such that
\begin{equation*} T\Big(\begin{bmatrix}1\\-1\end{bmatrix}\Big)=\begin{bmatrix}1\\4\\-1\end{bmatrix} \quad \text{and} \quad T\Big(\begin{bmatrix}2\\0\end{bmatrix}\Big)=\begin{bmatrix}0\\6\\4\end{bmatrix}\text{.} \end{equation*}
Find the standard matrix \(A\) of \(T\text{.}\)
Answer.
\begin{equation*} A=\begin{bmatrix}0\amp -1\\3\amp -1\\2\amp 3\end{bmatrix}. \end{equation*}

Exercise Group.

Find the standard matrix \(A\) of each linear transformation \(T:\R^2\rightarrow\R^2\) described below.
13.
\(T\) doubles the \(x\) component of every vector and triples the \(y\) component.
Answer.
\begin{equation*} A=\begin{bmatrix}2\amp 0\\0\amp 3\end{bmatrix}. \end{equation*}
14.
\(T\) reverses the direction of each vector.
Answer.
\begin{equation*} A=\begin{bmatrix}-1\amp 0\\0\amp -1\end{bmatrix}. \end{equation*}
15.
\(T\) doubles the length of each vector.
Answer.
\begin{equation*} A=\begin{bmatrix}2\amp 0\\0\amp 2\end{bmatrix} \end{equation*}
16.
\(T\) projects each vector onto the \(x\)-axis. (e.g. \(T([4,5])=[4,0]\)).
Answer.
\begin{equation*} A=\begin{bmatrix}1\amp 0\\0\amp 0\end{bmatrix}. \end{equation*}
17.
\(T\) projects each vector onto the \(y\)-axis. (e.g. \(T([4,5])=[0,5]\))
Answer.
\begin{equation*} A=\begin{bmatrix}0\amp 0\\0\amp 1\end{bmatrix}. \end{equation*}

Exercise Group.

Describe the image and find the rank for each linear transformation \(T:\R^n\rightarrow \R^m\) with standard matrix \(A\) given below.
18.
    \(T:\R^5\rightarrow \R^2\text{,}\)
    \begin{equation*} A=\begin{bmatrix}3\amp 2\amp 4\amp 7\amp 1\\-1\amp -9\amp 7\amp 6\amp 8\end{bmatrix}. \end{equation*}
  • \(\mbox{im}(T)=\R^2.\)
  • \(\mbox{im}\) is a line in \(\R^2\text{.}\)
  • \(\mbox{im} (T)=\{\mathbf{0}\}\text{.}\)
  • \(\mbox{im}(T)=\R^5\text{.}\)
  • \(\mbox{im} (T)\) is a plane in \(\R^5\text{.}\)
Answer.
\(\mbox{rank}(T)=2\)
19.
    \(T:\R^2\rightarrow\R^3\text{,}\)
    \begin{equation*} A=\begin{bmatrix}1\amp 1\\1\amp 1\\1\amp 1\end{bmatrix} \end{equation*}
  • \(\mbox{im}(T)=\R^3\text{.}\)
  • \(\mbox{im} \) is a line in \(\R^2\text{.}\)
  • \(\mbox{im} (T)\) is a line in \(\R^3\text{.}\)
  • \(\mbox{im}(T)=\{\mathbf{0}\}\text{.}\)
  • \(\mbox{im} \) is a plane in \(\R^3\text{.}\)
Answer.
\(\mbox{rank}(T)=1\text{.}\)

20.

Suppose linear transformations \(T:\R^2\rightarrow \R^2\) and \(S:\R^2\rightarrow \R^2\) are such that
\begin{equation*} \mbox{im}(T)=\mbox{im}(S)=\mbox{span}\left(\begin{bmatrix}1\\-3\end{bmatrix}\right). \end{equation*}
Does this mean that \(T\) and \(S\) are the same transformation? Justify your claim.

Exercise Group.

Describe the kernel and find the nullity for each linear transformation \(T:\R^n\rightarrow \R^m\) with standard matrix \(A\) given below.
21.
    \(T:\R^3\rightarrow \R^2\text{,}\)
    \begin{equation*} A=\begin{bmatrix}2\amp 1\amp 0\\-1\amp 1\amp -3\end{bmatrix}. \end{equation*}
  • \(\mbox{ker}(T)=\R^3\text{.}\)
  • \(\mbox{ker}(T)=\{\mathbf{0}\}\text{.}\)
  • \(\mbox{ker}(T)=\R^2\text{.}\)
  • \(\mbox{ker}(T)\) is a plane in \(\R^3\text{.}\)
  • \(\mbox{ker}(T)\) is a line in \(\R^3\text{.}\)
Answer.
\(\mbox{nullity}(T)=1\)
22.
    \(T:\R^2\rightarrow \R^2\text{,}\)
    \begin{equation*} A=\begin{bmatrix}2\amp -1\\3\amp 0\end{bmatrix}. \end{equation*}
  • \(\mbox{ker}(T)=\R^2\text{.}\)
  • \(\mbox{ker}(T)=\{\mathbf{0}\}\text{.}\)
  • \(\mbox{ker}(T)\) is a line in \(\R^2\text{.}\)
Answer.
\(\mbox{nullity}(T)=0\)
23.
    \(T:\R^3\rightarrow \R^5\text{,}\)
    \begin{equation*} A=\begin{bmatrix}1\amp 2\amp -1\\1\amp 2\amp -1\\1\amp 2\amp -1\\1\amp 2\amp -1\\1\amp 2\amp -1\end{bmatrix}. \end{equation*}
  • \(\mbox{ker}(T)\) is a plane in \(\R^3\text{.}\)
  • \(\mbox{ker}(T)\) is a line in \(\R^3\text{.}\)
  • \(\mbox{ker} (T)\) is a line in \(\R^5\text{.}\)
  • \(\mbox{ker}(T)=\R^3\text{.}\)
  • \(\mbox{ker}(T)=\{\mathbf{0}\}\text{.}\)
Answer.
\(\mbox{nullity}(T)=2\text{.}\)

24.

Suppose a linear transformation \(T:\R^3\rightarrow \R^3\) is such that \(\mbox{im}(T)\) is a plane in \(\R^3\text{.}\) What is the rank and nulity of \(T\text{?}\)
Answer.
\begin{equation*} \mbox{rank}(T)=2 \end{equation*}
\begin{equation*} \mbox{nullity}(T)=1 \end{equation*}

25.

Suppose a linear transformation \(T:\R^5\rightarrow \R^5\) is such that \(T(\mathbf{v})=\mathbf{0}\) for all \(\mathbf{v}\) in \(\R^5\text{.}\) What is the rank and nulity of \(T\text{?}\)
Answer.
\begin{equation*} \mbox{rank}(T)=0 \end{equation*}
\begin{equation*} \mbox{nullity}(T)=5 \end{equation*}

26.

Let \(T:\R^6\rightarrow \R^4\) be a linear transformation with standard matrix
\begin{equation*} A=\begin{bmatrix}2\amp -1\amp 1\amp -2\amp 1\amp 1\\1\amp 2\amp 3\amp 6\amp -4\amp 1\\0\amp 2\amp 2\amp 4\amp -2\amp -1\\1\amp 3\amp 2\amp 6\amp -3\amp 2\end{bmatrix} \end{equation*}
Find \(\mbox{im}(T)\) and \(\mbox{ker}(T)\) if the reduced row-echelon form of \(A\) is
\begin{equation*} \text{rref}(A)=\begin{bmatrix}1\amp 0\amp 0\amp 1\amp -1\amp 0\\0\amp 1\amp 0\amp 1\amp 0\amp 0\\0\amp 0\amp 1\amp 1\amp -1\amp 0\\0\amp 0\amp 0\amp 0\amp 0\amp 1\end{bmatrix} \end{equation*}

27.

Let
\begin{equation*} V=\mbox{span}\left(\begin{bmatrix}1\\1\end{bmatrix}\right) \end{equation*}
and let \(T:V\rightarrow \R^2\) be a linear transformation defined by \(T(\mathbf{v})=2\mathbf{v}\text{.}\) Find \(\mbox{im}(T)\) and \(\mbox{ker}(T)\text{.}\)

28.

Suppose a linear transformation \(T\) is induced by a \(4\times 6\) matrix \(A\text{.}\) Let \(S\) be a linear transformation induced by \(A^T\text{.}\) Find \(\mbox{nullity}(S)\text{,}\) if \(\mbox{nullity}(T)=3\text{.}\) Prove your claim.
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