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 Class Activities
Activity 3.3.1.
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 3.3.2.
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 3.3.3.
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 3.3.4.
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 3.3.5.
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 3.3.6.
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 3.3.7.
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 3.3.8.
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 3.3.9.
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*}
Since for a vector \(\vec v =\left[\begin{array}{c}x_1 \\ x_2 \\ x_3 \\ x_4 \end{array}\right] \text{,}\) \(T(\vec v)=T(x_1\vec e_1+x_2\vec e_2+x_3\vec e_3+x_4\vec e_4)\text{,}\) which of the following best describes the set of vectors
\begin{equation*}
\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]
}\text{?}
\end{equation*}
- The set of vectors spans \(\Im T\) but is not linearly independent.
- The set of vectors is a linearly independent subset of \(\Im T\) but does not span \(\Im T\text{.}\)
- The set of vectors is linearly independent and spans \(\Im T\text{;}\) that is, the set of vectors is a basis for \(\Im T\text{.}\)
Observation 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]
.
\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 3.3.11.
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.
- 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.
Activity 3.3.12.
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 3.3.13.
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 3.3.14.
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 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 3.3.15.
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 3.3.16.
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 3.3.2 Videos
Subsection 3.3.3 Slideshow
Slideshow of activities available at
https://teambasedinquirylearning.github.io/linear-algebra/2023/AT3.slides.html.Exercises 3.3.4 Exercises
Exercises available at
https://teambasedinquirylearning.github.io/linear-algebra/2023/exercises/#/bank/AT3/.Subsection 3.3.5 Mathematical Writing Explorations
Exploration 3.3.17.
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 3.3.18.
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 3.3.6 Sample Problem and Solution
Sample problem Example B.1.14.
