Learning Outcomes
- Compute a basis for the kernel and a basis for the image of a linear map, and verify that the rank-nullity theorem holds for a given linear map.
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 subspaces of \(\IR^2\) describes the set of all vectors that transform into \(\vec 0\text{?}\)
- \(\displaystyle \setBuilder{\left[\begin{array}{c}a \\ a\end{array}\right]}{a\in\IR}\)
- \(\displaystyle \setBuilder{\left[\begin{array}{c}a \\ 0\end{array}\right]}{a\in\IR}\)
- \(\displaystyle \setList{\left[\begin{array}{c}0\\0\end{array}\right]}\)
- \(\displaystyle \setBuilder{\left[\begin{array}{c}a \\ b\end{array}\right]}{a,b\in\IR}\)
Definition 3.3.3.
Let \(T: V \rightarrow W\) be a linear transformation, and let \(\vec{z}\) be the additive identity (the “zero vector”) of \(W\text{.}\) The kernel of \(T\) is an important subspace of \(V\) defined by
\begin{equation*}
\ker T = \left\{ \vec{v} \in V\ \big|\ T(\vec{v})=\vec{z}\right\}
\end{equation*}
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 subspaces of \(\IR^3\) describes \(\ker T\text{,}\) the set of all vectors that transform into \(\vec 0\text{?}\)
- \(\displaystyle \setBuilder{\left[\begin{array}{c}0 \\ 0\\ a\end{array}\right]}{a\in\IR}\)
- \(\displaystyle \setBuilder{\left[\begin{array}{c}a \\ a\\ 0\end{array}\right]}{a\in\IR}\)
- \(\displaystyle \setList{\left[\begin{array}{c}0\\0\\0\end{array}\right]}\)
- \(\displaystyle \setBuilder{\left[\begin{array}{c}a \\ b\\c\end{array}\right]}{a,b,c\in\IR}\)
Activity 3.3.5.
Let \(T: \IR^3 \rightarrow \IR^2\) be the linear transformation given by the standard matrix
\begin{equation*}
T\left( \left[\begin{array}{c} x \\ y \\ z \end{array}\right]\right) = \left[\begin{array}{c} 3x+4y-z \\ x+2y+z \end{array}\right]
\end{equation*}
(a)
Set \(T\left(\left[\begin{array}{c}x\\y\\z\end{array}\right]\right)
=
\left[\begin{array}{c}0\\0\end{array}\right]\) to find a linear system of equations whose solution set is the kernel.
(b)
Use \(\RREF(A)\) to solve this homogeneous system of equations and find a basis for the kernel of \(T\text{.}\)
Activity 3.3.6.
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 3.3.7.
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?
- \(\displaystyle \setBuilder{\left[\begin{array}{c}0 \\ 0\\ a\end{array}\right]}{a\in\IR}\)
- \(\displaystyle \setBuilder{\left[\begin{array}{c}a \\ b\\ 0\end{array}\right]}{a,b\in\IR}\)
- \(\displaystyle \setList{\left[\begin{array}{c}0\\0\\0\end{array}\right]}\)
- \(\displaystyle \setBuilder{\left[\begin{array}{c}a \\ b\\c\end{array}\right]}{a,b,c\in\IR}\)
Definition 3.3.8.
Let \(T: V \rightarrow W\) be a linear transformation. The image of \(T\) is an important subspace of \(W\) defined by
\begin{equation*}
\Im T = \left\{ \vec{w} \in W\ \big|\ \text{there is some }\vec v\in V \text{ with } T(\vec{v})=\vec{w}\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 \(\IR^3\text{.}\)
Activity 3.3.9.
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 subspaces of \(\IR^2\) describes \(\Im T\text{,}\) the set of all vectors that are the result of using \(T\) to transform \(\IR^3\) vectors?
- \(\displaystyle \setBuilder{\left[\begin{array}{c}a \\ a\end{array}\right]}{a\in\IR}\)
- \(\displaystyle \setBuilder{\left[\begin{array}{c}a \\ 0\end{array}\right]}{a\in\IR}\)
- \(\displaystyle \setList{\left[\begin{array}{c}0\\0\end{array}\right]}\)
- \(\displaystyle \setBuilder{\left[\begin{array}{c}a \\ b\end{array}\right]}{a,b\in\IR}\)
Activity 3.3.10.
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 \(\Im T\text{?}\)
(a)
Determine if \(\left[\begin{array}{c} 12 \\ 3 \\ 3 \end{array}\right]\) belongs to \(\Im T\text{.}\)
(b)
Determine if \(\left[\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right]\) belongs to \(\Im T\text{.}\)
(c)
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)=\vec{w}
\end{equation*}
has...
- no solutions.
- exactly one solution.
- at least one solution.
- infinitely-many solutions.
(d)
Based on this, how do \(\Im T\) and \(\vspan\left\{T(\vec{e}_1),T(\vec{e}_2),T(\vec{e}_3),T(\vec{e}_4)\right\}\) relate to each other?
- The set \(\Im T\) contains \(\vspan\left\{T(\vec{e}_1),T(\vec{e}_2),T(\vec{e}_3),T(\vec{e}_4)\right\}\) but is not equal to it.
- The set \(\vspan\left\{T(\vec{e}_1),T(\vec{e}_2),T(\vec{e}_3),T(\vec{e}_4)\right\}\) contains \(\Im T\) but is not equal to it.
- 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.
- There is no relation between these two sets.
Observation 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]
.
\end{equation*}
Since the set \(\setList{
\left[\begin{array}{c}3\\-1\\2\end{array}\right],
\left[\begin{array}{c}4\\1\\1\end{array}\right],
\left[\begin{array}{c}7\\0\\3\end{array}\right],
\left[\begin{array}{c}1\\2\\-1\end{array}\right]
}\) spans \(\Im T\text{,}\) we can obtain a basis for \(\Im T\) by finding \(\RREF A
=
\left[\begin{array}{cccc} 1 & 0 & 1 & -1\\ 0 & 1 & 1 & 1 \\ 0 & 0 & 0 & 0 \end{array}\right]\) and only using the vectors corresponding to pivot columns:
\begin{equation*}
\setList{
\left[\begin{array}{c}3\\-1\\2\end{array}\right],
\left[\begin{array}{c}4\\1\\1\end{array}\right]
}
\end{equation*}
Fact 3.3.12.
Let \(T:\IR^n\to\IR^m\) be a linear transformation with standard matrix \(A\text{.}\)
- The kernel of \(T\) is the solution set of the homogeneous system given by the augmented matrix \(\left[\begin{array}{c|c}A&\vec 0\end{array}\right]\text{.}\) Use the coefficients of its free variables to get a basis for the kernel (as in Fact 2.7.4).
- The image of \(T\) is the span of the columns of \(A\text{.}\) Remove the vectors creating non-pivot columns in \(\RREF A\) to get a basis for the image (as in Observation 2.6.5).
Activity 3.3.13.
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 3.3.14.
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 \(T\text{?}\)
- The number of pivot columns
- The number of non-pivot columns
- The number of pivot rows
- The number of non-pivot rows
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 image of \(T\text{?}\)
- The number of pivot columns
- The number of non-pivot columns
- The number of pivot rows
- The number of non-pivot rows
Observation 3.3.16.
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 3.3.17.
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 3.3.18.
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 defintions fit together?
Subsection 3.3.4 Videos
Exercises 3.3.5 Exercises
Exercises available at
https://tbil.org/linear-algebra/preview/exercises/#/bank/AT3/.Subsection 3.3.6 Mathematical Writing Explorations
Exploration 3.3.19.
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 3.3.20.
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.
