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Linear Algebra for Team-Based Inquiry Learning: PREVIEW Edition — Instructor Version

Steven Clontz, Drew Lewis

Section 3.3 Image and Kernel (AT3)

Subsection 3.3.1 Warm Up

Activity 3.3.1.

Consider the matrix \(A=\left[\begin{array}{cccc} 3 & 4 & 7 & 1\\ -1 & 1 & 0 & 2 \\ 2 & 1 & 3 & -1 \end{array}\right].\)
(a)
The matrix \(A\) is the standard matrix of a linear transformation \(T\text{.}\) What is the domain and the codomain of the transformation \(T\text{?}\)
(b)
Describe how \(T\) transforms the standard basis vectors of the domain that you found above.

Subsection 3.3.2 Class Activities

Activity 3.3.2.

Let \(T: \IR^2 \rightarrow \IR^3\) be given by
\begin{equation*} T\left(\left[\begin{array}{c}x \\ y \end{array}\right] \right) = \left[\begin{array}{c} x \\ y \\ 0 \end{array}\right] \hspace{3em} \text{with standard matrix } \left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \\ 0 & 0 \end{array}\right] \end{equation*}
Which of these sets contain all vectors \(\left[\begin{array}{c}x \\ y \end{array}\right]\) that transform into \(T\left(\left[\begin{array}{c}x \\ y \end{array}\right] \right) = \left[\begin{array}{c} x \\ y \\ 0 \end{array}\right]= \left[\begin{array}{c} 0 \\ 0 \\ 0 \end{array}\right]\text{?}\)
  1. \(\displaystyle \setBuilder{\left[\begin{array}{c}a \\ a\end{array}\right]}{a\in\IR}\)
  2. \(\displaystyle \setBuilder{\left[\begin{array}{c}a \\ 0\end{array}\right]}{a\in\IR}\)
  3. \(\displaystyle \setList{\left[\begin{array}{c}0\\0\end{array}\right]}\)
  4. \(\displaystyle \setBuilder{\left[\begin{array}{c}a \\ b\end{array}\right]}{a,b\in\IR}\)

Definition 3.3.3.

Let \(T: V \rightarrow W\) be a linear transformation, and let \(\vec{0}\) be the zero vector of \(W\text{.}\) The kernel of \(T\) (also known as the null space of \(T\)) is an important subset of \(V\) defined by
\begin{equation*} \ker T = \left\{ \vec{x} \in V\ \big|\ T(\vec{x})=\vec{0}\right\} \end{equation*}
Figure 24. The kernel of a linear transformation

Activity 3.3.4.

Let \(T: \IR^3 \rightarrow \IR^2\) be given by
\begin{equation*} T\left(\left[\begin{array}{c}x \\ y\\z \end{array}\right] \right) = \left[\begin{array}{c} x \\ y \end{array}\right] \hspace{3em} \text{with standard matrix } \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \end{array}\right] \end{equation*}
Which of these subsets of \(\IR^3\) is equal to \(\ker T=\left\{ \vec{x} \in \mathbb R^3\ \big|\ T(\vec{x})=\vec{0}\right\}\text{,}\) the set of all vectors \(\left[\begin{array}{c}x \\ y \\ z\end{array}\right]\) that transform into \(T\left(\left[\begin{array}{c}x \\ y \\ z \end{array}\right] \right) = \left[\begin{array}{c} x \\ y\end{array}\right]= \left[\begin{array}{c} 0 \\ 0 \end{array}\right]\text{?}\)
  1. \(\displaystyle \setBuilder{\left[\begin{array}{c}0 \\ 0\\ a\end{array}\right]}{a\in\IR}\)
  2. \(\displaystyle \setBuilder{\left[\begin{array}{c}a \\ a\\ 0\end{array}\right]}{a\in\IR}\)
  3. \(\displaystyle \setList{\left[\begin{array}{c}0\\0\\0\end{array}\right]}\)
  4. \(\displaystyle \setBuilder{\left[\begin{array}{c}a \\ b\\c\end{array}\right]}{a,b,c\in\IR}\)

Activity 3.3.5.

Let \(T: \IR^3 \rightarrow \IR^2\) be the linear transformation given by the standard matrix
\begin{equation*} A=\left[\begin{array}{ccc} 3 & 4 & -1 \\ 1 & 2 & 1 \end{array}\right] =\left[\begin{array}{ccc} T(\vec e_1) & T(\vec e_2) & T(\vec e_3)\end{array}\right]. \end{equation*}
(a)
Which of these is the most appropriate method to determine whether a \(\mathbb R^3\) vector \(\vec{x}=\left[\begin{array}{c}x_1\\x_2\\x_3\end{array}\right]=x_1\vec{e}_1+x_2\vec{e}_2+x_3\vec{e}_3\text{,}\) belongs to \(\ker T=\left\{ \vec{x} \in \mathbb R^3\ \big|\ T(\vec{x})=\vec{0}\right\}\text{?}\)
  1. Determine if the set of vectors \(\left\{T(\vec e_1), T(\vec e_2), T(\vec e_3), T(\vec x)\right\}\) spans \(\IR^3\text{.}\)
  2. Determine if the set of vectors \(\left\{T(\vec e_1), T(\vec e_2), T(\vec e_3), T(\vec x)\right\}\) is linearly dependent.
  3. Determine if \(\vec x=\left[\begin{array}{c} x_1 \\ x_2\\ x_3\end{array}\right]\) belongs to the solution set of the vector equation
    \begin{equation*} T(\vec x)=T\left(\left[\begin{array}{c} x_1 \\ x_2\\ x_3\end{array}\right]\right)= x_1T(\vec e_1)+x_2T(\vec e_2)+x_3T(\vec e_3)= \left[\begin{array}{c} 0 \\ 0\end{array}\right]\text{.} \end{equation*}
  4. Determine if the equation
    \begin{equation*} T(\vec 0)=T\left(\left[\begin{array}{c} 0 \\ 0\\ 0\end{array}\right]\right)= 0T(\vec e_1)+0T(\vec e_2)+0T(\vec e_3)= T(\vec x) \end{equation*}
    is consistent.
Answer.
B. This is exactly the same as finding the solution space for the homoegeneous vector equation \(T(\vec x)=\vec 0\text{.}\)
(b)
Use this method to find the kernel of \(T\text{.}\)
Answer.
\(\ker T=\left\{\left[\begin{array}{c}3a\\-2a\\a\end{array}\right]\middle| a\in\IR\right\}\)

Observation 3.3.6.

The kernel of a transformation \(T\) is exactly the solution space of the homogeneous equation \(T(\vec{x})=\vec{0}\text{.}\) If its standard matrix is \(A\text{,}\) then we may write \(A\vec x=\vec 0\) and use \(\RREF[A\,|\,\vec 0]\) to find this kernel.
In particular, the kernel is a subspace of the transformation’s domain, and has a basis which may be found as in Fact 2.7.5:
\begin{equation*} \ker T=\left\{\left[\begin{array}{c}3a\\-2a\\a\end{array}\right]\middle| a\in\IR\right\} \hspace{2em} \text{Basis for }\ker T=\left\{\left[\begin{array}{c}3\\-2\\1\end{array}\right]\right\}. \end{equation*}

Activity 3.3.7.

Let \(T: \IR^4 \rightarrow \IR^3\) be the linear transformation given by
\begin{equation*} T\left(\left[\begin{array}{c} x \\ y \\ z \\ w \end{array}\right] \right) = \left[\begin{array}{c} 2x+4y+2z-4w \\ -2x-4y+z+w \\ 3x+6y-z-4w\end{array}\right]. \end{equation*}
Find a basis for the kernel of \(T\text{.}\)

Activity 3.3.8.

Let \(T: \IR^2 \rightarrow \IR^3\) be given by
\begin{equation*} T\left(\left[\begin{array}{c}x \\ y \end{array}\right] \right) = \left[\begin{array}{c} x \\ y \\ 0 \end{array}\right] \hspace{3em} \text{with standard matrix } \left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \\ 0 & 0 \end{array}\right] \end{equation*}
Which of these subspaces of \(\IR^3\) describes the set of all vectors that are the result of using \(T\) to transform \(\IR^2\) vectors?
  1. \(\displaystyle \setBuilder{\left[\begin{array}{c}0 \\ 0\\ a\end{array}\right]}{a\in\IR}\)
  2. \(\displaystyle \setBuilder{\left[\begin{array}{c}a \\ b\\ 0\end{array}\right]}{a,b\in\IR}\)
  3. \(\displaystyle \setList{\left[\begin{array}{c}0\\0\\0\end{array}\right]}\)
  4. \(\displaystyle \setBuilder{\left[\begin{array}{c}a \\ b\\c\end{array}\right]}{a,b,c\in\IR}\)

Definition 3.3.9.

Let \(T: V \rightarrow W\) be a linear transformation. The image of \(T\) is an important subset of \(W\) defined by
\begin{equation*} \Im T = \left\{ T(\vec v) \in W\ \big| \vec v\in V \right\} \end{equation*}
In the examples below, the left example’s image is all of \(\IR^2\text{,}\) but the right example’s image is a planar subspace of \(\IR^3\text{.}\)
Figure 25. The image of a linear transformation

Activity 3.3.10.

Let \(T: \IR^3 \rightarrow \IR^2\) be given by
\begin{equation*} T\left(\left[\begin{array}{c}x \\ y\\z \end{array}\right] \right) = \left[\begin{array}{c} x \\ y \end{array}\right] \hspace{3em} \text{with standard matrix } \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \end{array}\right] \end{equation*}
Which of these subsets of \(\IR^2\) describes \(\Im T = \left\{ T(\vec v) \in \IR^2\ \big| \vec v\in \IR^3 \right\}\text{,}\) the set of all vectors that are the result of using \(T\) to transform \(\IR^3\) vectors?
  1. \(\displaystyle \setBuilder{\left[\begin{array}{c}a \\ a\end{array}\right]}{a\in\IR}\)
  2. \(\displaystyle \setBuilder{\left[\begin{array}{c}a \\ 0\end{array}\right]}{a\in\IR}\)
  3. \(\displaystyle \setList{\left[\begin{array}{c}0\\0\end{array}\right]}\)
  4. \(\displaystyle \setBuilder{\left[\begin{array}{c}a \\ b\end{array}\right]}{a,b\in\IR}\)

Activity 3.3.11.

Let \(T: \IR^4 \rightarrow \IR^3\) be the linear transformation given by the standard matrix
\begin{equation*} A = \left[\begin{array}{cccc} 3 & 4 & 7 & 1\\ -1 & 1 & 0 & 2 \\ 2 & 1 & 3 & -1 \end{array}\right] = \left[\begin{array}{cccc}T(\vec e_1)&T(\vec e_2)&T(\vec e_3)&T(\vec e_4)\end{array}\right] . \end{equation*}
Consider the question: Which vectors \(\vec{w}\) in \(\IR^3\) belong to \(\Im T=\left\{ T(\vec v) \in \IR^3\ \big| \vec v\in \IR^4 \right\}\text{?}\)
(a)
Recall that
\begin{equation*} T\left(\left[\begin{array}{c} x_1 \\ x_2 \\ x_3 \\ x_4 \end{array}\right] \right) = x_1T(\vec e_1)+x_2T(\vec e_2)+x_3T(\vec e_3)+x_4T(\vec e_4)\text{.} \end{equation*}
Complete the following vector equation which must be consistent in order for \(\left[\begin{array}{c} 12 \\ 3 \\ 3 \end{array}\right]\) to belong to \(\Im T\text{:}\)
\begin{equation*} x_1 \left[\begin{array}{c}\unknown\\\unknown\\\unknown\end{array}\right]+ x_2 \left[\begin{array}{c}\unknown\\\unknown\\\unknown\end{array}\right]+ x_3\left[\begin{array}{c}\unknown\\\unknown\\\unknown\end{array}\right]+ x_4\left[\begin{array}{c}\unknown\\\unknown\\\unknown\end{array}\right]= \left[\begin{array}{c} 12 \\ 3 \\ 3 \end{array}\right] \end{equation*}
(b)
Determine if \(\left[\begin{array}{c} 12 \\ 3 \\ 3 \end{array}\right]\) belongs to \(\Im T\text{.}\)
(c)
Write down the vector equation which must be consistent in order for \(\left[\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right]\) to belong to \(\Im T\text{.}\)
(d)
Determine if \(\left[\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right]\) belongs to \(\Im T\text{.}\)
(e)
An arbitrary vector \(\left[\begin{array}{c}\unknown\\\unknown\\\unknown\end{array}\right]\) belongs to \(\Im T\) provided the equation
\begin{equation*} x_1 T(\vec{e}_1)+x_2 T(\vec{e}_2)+x_3T(\vec{e}_3)+x_4T(\vec{e}_4)= \left[\begin{array}{c}\unknown\\\unknown\\\unknown\end{array}\right] \end{equation*}
has...
  1. no solutions.
  2. at most one solution.
  3. exactly one solution.
  4. at least one solution.
(f)
Based on this, what do we know?
  1. The set \(\left\{T(\vec{e}_1),T(\vec{e}_2),T(\vec{e}_3),T(\vec{e}_4)\right\}\) spans \(\Im T\) whenever \(\left\{T(\vec{e}_1),T(\vec{e}_2),T(\vec{e}_3),T(\vec{e}_4)\right\}\) is linearly independent.
  2. The set \(\Im T\) equals the codomain whenever \(\left\{T(\vec{e}_1),T(\vec{e}_2),T(\vec{e}_3),T(\vec{e}_4)\right\}\) is linearly independent.
  3. The set \(\Im T\) is simply \(\{\vec 0\}\) whenever \(\left\{T(\vec{e}_1),T(\vec{e}_2),T(\vec{e}_3),T(\vec{e}_4)\right\}\) is linearly dependent.
  4. The set \(\Im T\) and \(\vspan\left\{T(\vec{e}_1),T(\vec{e}_2),T(\vec{e}_3),T(\vec{e}_4)\right\}\) are equal to each other.

Observation 3.3.12.

Let \(T: \IR^n \rightarrow \IR^m\) be the linear transformation given by the standard matrix
\begin{equation*} A = \left[\begin{array}{ccc} T(\vec e_1) & \cdots & T(\vec e_n) \end{array}\right] . \end{equation*}
Then we have
\begin{equation*} \Im T=\vspan\setList{ T(\vec e_1), \cdots, T(\vec e_n) } \end{equation*}
and in particular, see that \(\Im T\) is a subspace of the codomain, and thus has a kernel which may be found as in Observation 2.6.5: given
\begin{equation*} \RREF \left[\begin{array}{cccc} 3 & 4 & 7 & 1\\ -1 & 1 & 0 & 2 \\ 2 & 1 & 3 & -1 \end{array}\right] = \left[\begin{array}{cccc} \markedPivot{1} & 0 & 1 & -1\\ 0 & \markedPivot{1} & 1 & 1 \\ 0 & 0 & 0 & 0 \end{array}\right] \end{equation*}
we see that we have
\begin{equation*} \Im T=\vspan\left\{ \left[\begin{array}{c} 3 \\ -1 \\ 2 \end{array}\right], \left[\begin{array}{c} 4 \\ 1 \\ 1 \end{array}\right], \left[\begin{array}{c} 7\\ 0 \\ 3 \end{array}\right], \left[\begin{array}{c} 1\\ 2 \\ -1 \end{array}\right], \right\} \end{equation*}
and
\begin{equation*} \text{Basis for }\Im T= \setList{ \left[\begin{array}{c}3\\-1\\2\end{array}\right], \left[\begin{array}{c}4\\1\\1\end{array}\right] }. \end{equation*}
This justifies why the image of a transformation is also called the column space of its standard matrix.

Activity 3.3.14.

Let \(T: \IR^3 \rightarrow \IR^4\) be the linear transformation given by the standard matrix
\begin{equation*} A = \left[\begin{array}{ccc} 1 & -3 & 2\\ 2 & -6 & 0 \\ 0 & 0 & 1 \\ -1 & 3 & 1 \end{array}\right] . \end{equation*}
Find a basis for the kernel and a basis for the image of \(T\text{.}\)

Activity 3.3.15.

Let \(T: \IR^n \rightarrow \IR^m\) be a linear transformation with standard matrix \(A\text{.}\) Which of the following is equal to the dimension of the kernel of \(T\text{?}\)
  1. The number of pivot columns
  2. The number of non-pivot columns
  3. The number of pivot rows
  4. The number of non-pivot rows

Activity 3.3.16.

Let \(T: \IR^n \rightarrow \IR^m\) be a linear transformation with standard matrix \(A\text{.}\) Which of the following is equal to the dimension of the image of \(T\text{?}\)
  1. The number of pivot columns
  2. The number of non-pivot columns
  3. The number of pivot rows
  4. The number of non-pivot rows

Observation 3.3.17.

Combining these with the observation that the number of columns is the dimension of the domain of \(T\text{,}\) we have the rank-nullity theorem:
The dimension of the domain of \(T\) equals \(\dim(\ker T)+\dim(\Im T)\text{.}\)
The dimension of the image is called the rank of \(T\) (or \(A\)) and the dimension of the kernel is called the nullity.

Activity 3.3.18.

Let \(T:\mathbb{R}^4 \to \mathbb{R}^3\) be the linear transformation given by
\begin{equation*} T\left( \left[\begin{array}{c} x \\ y \\ z \\ {w} \end{array}\right] \right) = \left[\begin{array}{c} x - y + 5 \, z + 3 \, {w} \\ -x - 4 \, z - 2 \, {w} \\ y - 2 \, z - {w} \end{array}\right]. \end{equation*}
(a)
Explain and demonstrate how to find the image of \(T\) and a basis for that image.
(b)
Explain and demonstrate how to find the kernel of \(T\) and a basis for that kernel.
(c)
Explain and demonstrate how to find the rank and nullity of \(T\text{,}\) and why the rank-nullity theorem holds for \(T\text{.}\)

Subsection 3.3.3 Individual Practice

Activity 3.3.19.

In this section, we’ve introduced two important subspaces that are associated with a linear transformation \(T\colon V\to W\text{,}\) namely: \(\Im T\text{,}\) the image of \(T\text{,}\) and \(\ker T\text{,}\) the kernel of \(T\text{.}\) The following sequence is designed to help you internalize these definitions. Try to complete them without referring to your Activity Book, and then check your answers.
(a)
One of \(\ker T\) and \(\Im T\) is a subspace of the domain and the other is a subspace of the codomain. Which is which?
(b)
Write down the precise definitions of these subspaces.
(c)
How would you describe these definitions to a layperson?
(d)
What picture, or other study strategy would be helpful to you in conceptualizing how these definitions fit together?

Activity 3.3.20.

We can use our notation of span in relation to a matrix, not just in relation to a set of vectors. Given a matrix \(M\)
  • the span of the set of all columns is the column space
  • the span of the set of all rows is the row space
\begin{equation*} \mbox{Let } M = \left[\begin{array}{ccc|c}1&-1&0\\2&2&4\\-1&0&-1\end{array}\right] \end{equation*}
(a)
Is \(\left[\begin{array}{c}2\\1\\3\end{array}\right]\) in the column space of \(M\text{?}\) Is it in the row space of \(M\text{?}\)
  1. Yes.
  2. No.
(b)
Is \(\left[\begin{array}{c}1\\10\\-3\end{array}\right]\) in the column space of \(M\text{?}\) Is it in the row space of \(M\text{?}\)
  1. Yes.
  2. No.
(c)
\begin{equation*} \mbox{Let } N = \left[\begin{array}{ccc|c}1&-1&1\\2&2&-3\\-1&0&-1\end{array}\right] \end{equation*}
Are the row space and column space of \(N\) both equal to \(\mathbb{R}^3\text{?}\)
  1. Yes.
  2. No.

Subsection 3.3.4 Videos

Figure 26. Video: The kernel and image of a linear transformation. Note that there is a typo: if you’re following along, you should find that \(T\left(\left[\begin{array}{c}2\\1\\3\\0\end{array}\right]\right)=\left[\begin{array}{c}14\\-5\\9\end{array}\right]\text{.}\)
Figure 27. Video: Finding a basis of the image of a linear transformation
Figure 28. Video: Finding a basis of the kernel of a linear transformation
Figure 29. Video: The rank-nullity theorem

Subsection 3.3.5 Exercises

Subsection 3.3.6 Mathematical Writing Explorations

Exploration 3.3.21.

Assume \(f:V \rightarrow W\) is a linear map. Let \(\{\vec{v_1},\vec{v_2},\ldots,\vec{v_n}\}\) be a set of vectors in \(V\text{,}\) and set \(\vec{w_i} = f(\vec{v_i})\text{.}\)
  • If the set \(\{\vec{w_1},\vec{w_2},\ldots,\vec{w_n}\}\) is linearly independent, must the set \(\{\vec{v_1},\vec{v_2},\ldots,\vec{v_n}\}\) also be linearly independent?
  • If the set \(\{\vec{v_1},\vec{v_2},\ldots,\vec{v_n}\}\) is linearly independent, must the set \(\{\vec{w_1},\vec{w_2},\ldots,\vec{w_n}\}\) also be linearly independent?
  • If the set \(\{\vec{w_1},\vec{w_2},\ldots,\vec{w_n}\}\) spans \(W\text{,}\) must the set \(\{\vec{v_1},\vec{v_2},\ldots,\vec{v_n}\}\) also span \(V\text{?}\)
  • If the set \(\{\vec{v_1},\vec{v_2},\ldots,\vec{v_n}\}\) spans \(V\text{,}\) must the set \(\{\vec{w_1},\vec{w_2},\ldots,\vec{w_n}\}\) also span \(W\text{?}\)
  • In light of this, is the image of the basis of a vector space always a basis for the codomain?

Exploration 3.3.22.

Prove the Rank-Nullity Theorem. Use the steps below to help you.
  • The theorem states that, given a linear map \(h:V \rightarrow W\text{,}\) with \(V\) and \(W\) vector spaces, the rank of \(h\text{,}\) plus the nullity of \(h\text{,}\) equals the dimension of the domain \(V\text{.}\) Assume that the dimension of \(V\) is \(n\text{.}\)
  • For simplicity, denote the rank of \(h\) by \(\mathcal{R}(h)\text{,}\) and the nullity by \(\mathcal{N}(h)\text{.}\)
  • Recall that \(\mathcal{R}(h)\) is the dimension of the range space of \(h\text{.}\) State the precise definition.
  • Recall that \(\mathcal{N}(h)\) is the dimension of the null space of \(h\text{.}\) State the precise definition.
  • Begin with a basis for the null space, denoted \(B_N = \{\vec{\beta_1}, \vec{\beta_2}, \ldots, \vec{\beta_k}\}\text{.}\) Show how this can be extended to a basis \(B_V\) for \(V\text{,}\) with \(B_V = \{\vec{\beta_1}, \vec{\beta_2}, \ldots, \vec{\beta_k}, \vec{\beta_{k+1}}, \vec{\beta_{k+2}}, \ldots, \vec{\beta_n}\}.\) In this portion, you should assume \(k \leq n\text{,}\) and construct additional vectors which are not linear combinations of vectors in \(B_N\text{.}\) Prove that you can always do this until you have \(n\) total linearly independent vectors.
  • Show that \(B_R = \{h(\vec{\beta_{k+1}}), h(\vec{\beta_{k+2}}), \ldots, h(\vec{\beta_n})\}\) is a basis for the range space. Start by showing that it is linearly independent, and be sure you prove that each element of the range space can be written as a linear combination of \(B_R\text{.}\)
  • Show that \(B_R\) spans the range space.
  • State your conclusion.

Subsection 3.3.7 Sample Problem and Solution

Sample problem Example B.1.14.
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