Image and Kernel (A3)

Section 3.3 Image and Kernel (A3)

Definition 3.3.1.

Let \(T: V \rightarrow W\) be a linear transformation. 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*}

  \begin{tikzpicture}[x=0.15in,y=0.15in]
    \begin{scope}[shift={(0,0)}]
      \draw (0,0) node[anchor=north west] {\(\ker T\)} -- (3,0);
      \draw (0,0) -- (0,3);
      \draw (0,0)  -- (-2,-1);
      \draw[thick,latex-latex,blue] (-3,-3) -- (3,3);
    \end{scope}
    \draw[dashed,-latex] (3.5,3) to [bend left=30] (7,3);
    \begin{scope}[shift={(9,0.5)}]
      \draw (-2,0) -- (2,0);
      \draw (0,-2) -- (0,2);
      \fill[blue] (0,0) circle (0.2)
            node[anchor=south east] {\(\vec{0}\)};
    \end{scope}
  \end{tikzpicture}

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 \(\ker T\text{,}\) the set of all vectors that transform into \(\vec 0\text{?}\)

\(\displaystyle \setBuilder{\left[\begin{array}{c}a \\ a\end{array}\right]}{a\in\IR}\)
\(\displaystyle \setList{\left[\begin{array}{c}0\\0\end{array}\right]}\)
\(\displaystyle \IR^2=\setBuilder{\left[\begin{array}{c}x \\ y\end{array}\right]}{x,y\in\IR}\)

Activity 3.3.2.

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

\(\displaystyle \setBuilder{\left[\begin{array}{c}0 \\ 0\\ a\end{array}\right]}{a\in\IR}\)
\(\displaystyle \setBuilder{\left[\begin{array}{c}a \\ a\\ 0\end{array}\right]}{a\in\IR}\)
\(\displaystyle \setList{\left[\begin{array}{c}0\\0\\0\end{array}\right]}\)
\(\displaystyle \IR^3=\setBuilder{\left[\begin{array}{c}x \\ y\\z\end{array}\right]}{x,y,z\in\IR}\)

Activity 3.3.3.

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 3.3.4.

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 \(T\text{.}\)

Definition 3.3.2.

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

\begin{tikzpicture}[x=0.15in,y=0.15in]
  \begin{scope}[shift={(0,0)}]
    \draw (0,0) -- (3,0);
    \draw (0,0) -- (0,3);
    \draw (0,0) -- (-2,-1);
    \draw[thick,-latex,blue] (0,0) -- (2,1);
    \draw[thick,-latex,blue] (0,0) -- (1,2);
    \draw[thick,-latex,blue] (0,0) -- (0,2);
    \draw[thick,-latex,blue] (0,0) -- (1,-1);
    \draw[thick,-latex,blue] (0,0) -- (-2,3);
    \draw[thick,-latex,blue] (0,0) -- (-3,-2);
  \end{scope}
  \draw[dashed,-latex] (3,3) to [bend left=30] (7,3);
  \begin{scope}[shift={(9,1)}]
    \draw (-2,0) -- (2,0);
    \draw (0,-2) -- (0,2);
    \draw[thick,-latex,blue] (0,0) -- (0.5,2);
    \draw[thick,-latex,blue] (0,0) -- (2,1);
    \draw[thick,-latex,blue] (0,0) -- (-1.5,1);
    \draw[thick,-latex,blue] (0,0) -- (0,-1.5);
    \draw[thick,-latex,blue] (0,0) -- (2,-2);
    \fill[color=blue, opacity=0.5] (-2,-2) rectangle (2,2);
  \end{scope}
\end{tikzpicture}
\hspace{3em}
\begin{tikzpicture}[x=0.15in,y=0.15in]
  \begin{scope}[shift={(0,1)}]
    \draw (-2,0) -- (2,0);
    \draw (0,-2) -- (0,2);
    \draw[thick,-latex,blue] (0,0) -- (0.5,2);
    \draw[thick,-latex,blue] (0,0) -- (2,1);
    \draw[thick,-latex,blue] (0,0) -- (-1.5,1);
    \draw[thick,-latex,blue] (0,0) -- (0,-1.5);
    \draw[thick,-latex,blue] (0,0) -- (2,-2);
  \end{scope}
  \draw[dashed,-latex] (3,3) to [bend left=30] (7,3);
  \begin{scope}[shift={(9,0)}]
    \draw (0,0) -- (3,0);
    \draw (0,0) -- (0,3);
    \draw (0,0) -- (-2,-1);
    \draw[thick,-latex,blue] (0,0) -- (0.5,1.5);
    \draw[thick,-latex,blue] (0,0) -- (2,1);
    \draw[thick,-latex,blue] (0,0) -- (-2.5,1);
    \draw[thick,-latex,blue] (0,0) -- (-0.5,-1.5);
    \draw[thick,-latex,blue] (0,0) -- (2.5,-0.5);
    \fill[color=blue, opacity=0.5] (-2,-2) -- (3,-1) -- (2,2) -- (-3,1) -- (-2,-2);
  \end{scope}
\end{tikzpicture}

Activity 3.3.5.

Let \(T: \IR^2 \rightarrow \IR^3\) be given by

Which of these subspaces of \(\IR^3\) describes \(\Im T\text{,}\) the set of all vectors that are the result of using \(T\) to transform \(\IR^2\) vectors?

\(\displaystyle \setBuilder{\left[\begin{array}{c}0 \\ 0\\ a\end{array}\right]}{a\in\IR}\)
\(\displaystyle \setBuilder{\left[\begin{array}{c}a \\ b\\ 0\end{array}\right]}{a,b\in\IR}\)
\(\displaystyle \setList{\left[\begin{array}{c}0\\0\\0\end{array}\right]}\)
\(\displaystyle \IR^3=\setBuilder{\left[\begin{array}{c}x \\ y\\z\end{array}\right]}{x,y,z\in\IR}\)

Activity 3.3.6.

Let \(T: \IR^3 \rightarrow \IR^2\) be given by

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?

\(\displaystyle \setBuilder{\left[\begin{array}{c}a \\ a\end{array}\right]}{a\in\IR}\)
\(\displaystyle \setList{\left[\begin{array}{c}0\\0\end{array}\right]}\)
\(\displaystyle \IR^2=\setBuilder{\left[\begin{array}{c}x \\ y\end{array}\right]}{x,y\in\IR}\)

Activity 3.3.7.

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 \(T(\vec v)=T(x_1\vec e_1+x_2\vec e_2+x_3\vec e_3+x_4\vec e_4)\text{,}\) 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] } \end{equation*}

spans \(\Im T\)
is a linearly independent subset of \(\Im T\)
is a basis for \(\Im T\)

Observation 3.3.3.

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 3.3.4.

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 3.3.8.

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 3.3.9.

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 3.3.10.

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 3.3.5.

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 3.3.11.

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 \(T\text{.}\)

Exercises 3.3.1 Exercises

1.

Let \(T:\mathbb{R}^ 3 \to \mathbb{R}^ 4 \) be the linear transformation given by

\begin{equation*} T\left( \left[\begin{array}{c} x \\ y \\ z \end{array}\right] \right) = \left[\begin{array}{c} x + 3 \, y - z \\ -3 \, x - 8 \, y + 2 \, z \\ -4 \, x - 8 \, y \\ -x - 7 \, y + 5 \, z \end{array}\right] . \end{equation*}

Explain how to find the image of \(T\) and the kernel of \(T\text{.}\)
Explain how to find a basis of the image of \(T\) and a basis of the kernel of \(T\text{.}\)
Explain how to find the rank and nullity of \(T\text{,}\) and why the rank-nullity theorem holds for \(T\text{.}\)

Answer

\begin{equation*} \operatorname{RREF} \left[\begin{array}{ccc} 1 & 3 & -1 \\ -3 & -8 & 2 \\ -4 & -8 & 0 \\ -1 & -7 & 5 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & 2 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \end{equation*}

\begin{equation*} \operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 1 \\ -3 \\ -4 \\ -1 \end{array}\right] , \left[\begin{array}{c} 3 \\ -8 \\ -8 \\ -7 \end{array}\right] \right\} \end{equation*}

\begin{equation*} \operatorname{ker}\ T = \left\{ \left[\begin{array}{c} -2 \, a \\ a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \end{equation*}
A basis of \(\operatorname{Im}\ T\) is \(\left\{ \left[\begin{array}{c} 1 \\ -3 \\ -4 \\ -1 \end{array}\right] , \left[\begin{array}{c} 3 \\ -8 \\ -8 \\ -7 \end{array}\right] \right\} \text{.}\) A basis of \(\operatorname{ker}\ T\) is \(\left\{ \left[\begin{array}{c} -2 \\ 1 \\ 1 \end{array}\right] \right\} \)
The rank of \(T\) is \(2 \text{,}\) the nullity of \(T\) is \(1 \text{,}\) and the dimension of the domain of \(T\) is \(3 \text{.}\) The rank-nullity theorem asserts that \(2 + 1 = 3 \text{,}\) which we see to be true.

2.

Let \(T:\mathbb{R}^ 3 \to \mathbb{R}^ 4 \) be the linear transformation given by

\begin{equation*} T\left( \left[\begin{array}{c} x \\ y \\ z \end{array}\right] \right) = \left[\begin{array}{c} 3 \, x + 2 \, y - 3 \, z \\ -3 \, x - 2 \, y + 3 \, z \\ 2 \, x + y - z \\ x - y + 4 \, z \end{array}\right] . \end{equation*}

Explain how to find the image of \(T\) and the kernel of \(T\text{.}\)
Explain how to find a basis of the image of \(T\) and a basis of the kernel of \(T\text{.}\)
Explain how to find the rank and nullity of \(T\text{,}\) and why the rank-nullity theorem holds for \(T\text{.}\)

Answer

\begin{equation*} \operatorname{RREF} \left[\begin{array}{ccc} 3 & 2 & -3 \\ -3 & -2 & 3 \\ 2 & 1 & -1 \\ 1 & -1 & 4 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & 1 \\ 0 & 1 & -3 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \end{equation*}

\begin{equation*} \operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 3 \\ -3 \\ 2 \\ 1 \end{array}\right] , \left[\begin{array}{c} 2 \\ -2 \\ 1 \\ -1 \end{array}\right] \right\} \end{equation*}

\begin{equation*} \operatorname{ker}\ T = \left\{ \left[\begin{array}{c} -a \\ 3 \, a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \end{equation*}
A basis of \(\operatorname{Im}\ T\) is \(\left\{ \left[\begin{array}{c} 3 \\ -3 \\ 2 \\ 1 \end{array}\right] , \left[\begin{array}{c} 2 \\ -2 \\ 1 \\ -1 \end{array}\right] \right\} \text{.}\) A basis of \(\operatorname{ker}\ T\) is \(\left\{ \left[\begin{array}{c} -1 \\ 3 \\ 1 \end{array}\right] \right\} \)
The rank of \(T\) is \(2 \text{,}\) the nullity of \(T\) is \(1 \text{,}\) and the dimension of the domain of \(T\) is \(3 \text{.}\) The rank-nullity theorem asserts that \(2 + 1 = 3 \text{,}\) which we see to be true.

3.

Let \(T:\mathbb{R}^ 3 \to \mathbb{R}^ 4 \) be the linear transformation given by

\begin{equation*} T\left( \left[\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \end{array}\right] \right) = \left[\begin{array}{c} x_{1} - 2 \, x_{2} \\ x_{2} \\ x_{1} + 2 \, x_{2} \\ 2 \, x_{1} - 7 \, x_{2} \end{array}\right] . \end{equation*}

Explain how to find the image of \(T\) and the kernel of \(T\text{.}\)
Explain how to find a basis of the image of \(T\) and a basis of the kernel of \(T\text{.}\)
Explain how to find the rank and nullity of \(T\text{,}\) and why the rank-nullity theorem holds for \(T\text{.}\)

Answer

\begin{equation*} \operatorname{RREF} \left[\begin{array}{ccc} 1 & -2 & 0 \\ 0 & 1 & 0 \\ 1 & 2 & 0 \\ 2 & -7 & 0 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \end{equation*}

\begin{equation*} \operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 1 \\ 0 \\ 1 \\ 2 \end{array}\right] , \left[\begin{array}{c} -2 \\ 1 \\ 2 \\ -7 \end{array}\right] \right\} \end{equation*}

\begin{equation*} \operatorname{ker}\ T = \left\{ \left[\begin{array}{c} 0 \\ 0 \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \end{equation*}
A basis of \(\operatorname{Im}\ T\) is \(\left\{ \left[\begin{array}{c} 1 \\ 0 \\ 1 \\ 2 \end{array}\right] , \left[\begin{array}{c} -2 \\ 1 \\ 2 \\ -7 \end{array}\right] \right\} \text{.}\) A basis of \(\operatorname{ker}\ T\) is \(\left\{ \left[\begin{array}{c} 0 \\ 0 \\ 1 \end{array}\right] \right\} \)
The rank of \(T\) is \(2 \text{,}\) the nullity of \(T\) is \(1 \text{,}\) and the dimension of the domain of \(T\) is \(3 \text{.}\) The rank-nullity theorem asserts that \(2 + 1 = 3 \text{,}\) which we see to be true.

4.

Let \(T:\mathbb{R}^ 3 \to \mathbb{R}^ 4 \) be the linear transformation given by

\begin{equation*} T\left( \left[\begin{array}{c} x \\ y \\ z \end{array}\right] \right) = \left[\begin{array}{c} x - 2 \, y + 5 \, z \\ -x + 3 \, y - 7 \, z \\ -3 \, x + 2 \, y - 7 \, z \\ 4 \, y - 8 \, z \end{array}\right] . \end{equation*}

Explain how to find the image of \(T\) and the kernel of \(T\text{.}\)
Explain how to find a basis of the image of \(T\) and a basis of the kernel of \(T\text{.}\)
Explain how to find the rank and nullity of \(T\text{,}\) and why the rank-nullity theorem holds for \(T\text{.}\)

Answer

\begin{equation*} \operatorname{RREF} \left[\begin{array}{ccc} 1 & -2 & 5 \\ -1 & 3 & -7 \\ -3 & 2 & -7 \\ 0 & 4 & -8 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & 1 \\ 0 & 1 & -2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \end{equation*}

\begin{equation*} \operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 1 \\ -1 \\ -3 \\ 0 \end{array}\right] , \left[\begin{array}{c} -2 \\ 3 \\ 2 \\ 4 \end{array}\right] \right\} \end{equation*}

\begin{equation*} \operatorname{ker}\ T = \left\{ \left[\begin{array}{c} -a \\ 2 \, a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \end{equation*}
A basis of \(\operatorname{Im}\ T\) is \(\left\{ \left[\begin{array}{c} 1 \\ -1 \\ -3 \\ 0 \end{array}\right] , \left[\begin{array}{c} -2 \\ 3 \\ 2 \\ 4 \end{array}\right] \right\} \text{.}\) A basis of \(\operatorname{ker}\ T\) is \(\left\{ \left[\begin{array}{c} -1 \\ 2 \\ 1 \end{array}\right] \right\} \)
The rank of \(T\) is \(2 \text{,}\) the nullity of \(T\) is \(1 \text{,}\) and the dimension of the domain of \(T\) is \(3 \text{.}\) The rank-nullity theorem asserts that \(2 + 1 = 3 \text{,}\) which we see to be true.

5.

Let \(T:\mathbb{R}^ 3 \to \mathbb{R}^ 4 \) be the linear transformation given by

\begin{equation*} T\left( \left[\begin{array}{c} x \\ y \\ z \end{array}\right] \right) = \left[\begin{array}{c} x - z \\ -2 \, x + y + z \\ -5 \, x + 3 \, y + 2 \, z \\ -2 \, x - 2 \, y + 4 \, z \end{array}\right] . \end{equation*}

Explain how to find the image of \(T\) and the kernel of \(T\text{.}\)
Explain how to find a basis of the image of \(T\) and a basis of the kernel of \(T\text{.}\)
Explain how to find the rank and nullity of \(T\text{,}\) and why the rank-nullity theorem holds for \(T\text{.}\)

Answer

\begin{equation*} \operatorname{RREF} \left[\begin{array}{ccc} 1 & 0 & -1 \\ -2 & 1 & 1 \\ -5 & 3 & 2 \\ -2 & -2 & 4 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & -1 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \end{equation*}

\begin{equation*} \operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 1 \\ -2 \\ -5 \\ -2 \end{array}\right] , \left[\begin{array}{c} 0 \\ 1 \\ 3 \\ -2 \end{array}\right] \right\} \end{equation*}

\begin{equation*} \operatorname{ker}\ T = \left\{ \left[\begin{array}{c} a \\ a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \end{equation*}
A basis of \(\operatorname{Im}\ T\) is \(\left\{ \left[\begin{array}{c} 1 \\ -2 \\ -5 \\ -2 \end{array}\right] , \left[\begin{array}{c} 0 \\ 1 \\ 3 \\ -2 \end{array}\right] \right\} \text{.}\) A basis of \(\operatorname{ker}\ T\) is \(\left\{ \left[\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right] \right\} \)
The rank of \(T\) is \(2 \text{,}\) the nullity of \(T\) is \(1 \text{,}\) and the dimension of the domain of \(T\) is \(3 \text{.}\) The rank-nullity theorem asserts that \(2 + 1 = 3 \text{,}\) which we see to be true.

Additional exercises available at checkit.clontz.org.