Skip to main content

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 \(\vec 0\text{?}\)

  1. \(\displaystyle \setBuilder{\left[\begin{array}{c}a \\ a\end{array}\right]}{a\in\IR}\)

  2. \(\displaystyle \setList{\left[\begin{array}{c}0\\0\end{array}\right]}\)

  3. \(\displaystyle \IR^2=\setBuilder{\left[\begin{array}{c}x \\ y\end{array}\right]}{x,y\in\IR}\)

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*}
Figure 27. The kernel of a linear transformation

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 \(\vec 0\text{?}\)

  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. \(\displaystyle \IR^3=\setBuilder{\left[\begin{array}{c}x \\ y\\z\end{array}\right]}{x,y,z\in\IR}\)

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?

  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. \(\displaystyle \IR^3=\setBuilder{\left[\begin{array}{c}x \\ y\\z\end{array}\right]}{x,y,z\in\IR}\)

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 \(\IR^3\text{.}\)

Figure 28. The image of a linear transformation

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?

  1. \(\displaystyle \setBuilder{\left[\begin{array}{c}a \\ a\end{array}\right]}{a\in\IR}\)

  2. \(\displaystyle \setList{\left[\begin{array}{c}0\\0\end{array}\right]}\)

  3. \(\displaystyle \IR^2=\setBuilder{\left[\begin{array}{c}x \\ y\end{array}\right]}{x,y\in\IR}\)

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*}

  1. The set of vectors spans \(\Im T\) but is not linearly independent.

  2. The set of vectors is a linearly independent subset of \(\Im T\) but does not span \(\Im T\text{.}\)

  3. 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*}

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 \(T\text{?}\)

  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 3.3.14.

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 \(T\text{?}\)

  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 3.3.15.

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 \(T: \IR^3 \rightarrow \IR^4\) 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*}

Verify that the rank-nullity theorem holds for \(T\text{.}\)

Subsection 3.3.2 Videos

Figure 29. Video: The kernel and image of a linear transformation
Figure 30. Video: Finding a basis of the image of a linear transformation
Figure 31. Video: Finding a basis of the kernel of a linear transformation
Figure 32. Video: The rank-nullity theorem

Subsection 3.3.3 Slideshow

Slideshow of activities available at https://teambasedinquirylearning.github.io/linear-algebra/2022/AT3.slides.html.

Exercises 3.3.4 Exercises

Exercises available at https://teambasedinquirylearning.github.io/linear-algebra/2022/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.1.16.

You have attempted of activities on this page.