Recall from
Example 6.1.4 that
\(T:\mathbb{R}^{4}\rightarrow \mathbb{R}^{4}\) is defined by
\begin{equation*}
T\left(\begin{array}{r}
x_1 \\
x_2 \\
x_3\\
x_4
\end{array}\right)=\left(\begin{array}{r}
x_1+x_2 \\
x_2+3x_3+x_4 \\
-x_1+x_3-x_4\\
2x_2+2x_3-2x_4
\end{array}\right).
\end{equation*}
Step 1: Find the kernel. By definition, \(\operatorname{ker}(T) = \{\mathbf{v} \in \mathbb{R}^4 \mid T(\mathbf{v}) = \mathbf{0}\}\text{.}\) We need to solve:
\begin{equation*}
T\left(\begin{array}{r}
x_1 \\
x_2 \\
x_3\\
x_4
\end{array}\right)=\left(\begin{array}{r}
x_1+x_2 \\
x_2+3x_3+x_4 \\
-x_1+x_3-x_4\\
2x_2+2x_3-2x_4
\end{array}\right)=\left(\begin{array}{r}
0 \\
0 \\
0\\
0
\end{array}\right).
\end{equation*}
This gives us the system of equations:
\begin{align*}
x_1 + x_2 \amp= 0\\
x_2 + 3x_3 + x_4 \amp= 0\\
-x_1 + x_3 - x_4 \amp= 0\\
2x_2 + 2x_3 - 2x_4 \amp= 0
\end{align*}
We solve this system using the augmented matrix:
\begin{align*}
\left[\begin{array}{rrrr|r}
1 \amp 1 \amp 0 \amp 0 \amp 0\\
0 \amp 1 \amp 3 \amp 1 \amp 0\\
-1 \amp 0 \amp 1 \amp -1 \amp 0\\
0 \amp 2 \amp 2 \amp -2 \amp 0
\end{array}\right]
\xrightarrow{\text{RREF}}
\left[\begin{array}{rrrr|r}
1 \amp 0 \amp 0 \amp 2 \amp 0\\
0 \amp 1 \amp 0 \amp -2 \amp 0\\
0 \amp 0 \amp 1 \amp 1 \amp 0\\
0 \amp 0 \amp 0 \amp 0 \amp 0
\end{array}\right]. \tag{1}
\end{align*}
From the RREF, we get:
\begin{align*}
x_1 + 2x_4 \amp= 0\\
x_2 - 2x_4 \amp= 0\\
x_3 + x_4 \amp= 0
\end{align*}
The only free variable is \(x_4\text{.}\) Let \(x_4 = t\text{,}\) then:
\begin{align*}
x_1 \amp= -2t\\
x_2 \amp= 2t\\
x_3 \amp= -t\\
x_4 \amp= t
\end{align*}
Therefore,
\begin{equation*}
\mathbf{x} = \begin{pmatrix}-2t \\ 2t \\ -t \\ t\end{pmatrix} = t\begin{pmatrix}-2\\2\\-1\\1\end{pmatrix}.
\end{equation*}
Answer for kernel:
\begin{equation*}
\operatorname{ker}(T) = \operatorname{span}\left\{\begin{pmatrix}-2\\2\\-1\\1\end{pmatrix}\right\}.
\end{equation*}
The dimension of the kernel is \(\dim(\operatorname{ker}(T)) = 1\text{.}\)
Step 2: Find the range. By definition, \(\operatorname{range}(T) = \{T(\mathbf{v}) \mid \mathbf{v} \in \mathbb{R}^4\}\text{.}\) For any \(\mathbf{v} = \begin{pmatrix}x_1\\x_2\\x_3\\x_4\end{pmatrix} \in \mathbb{R}^4\text{,}\) we have:
\begin{align*}
T(\mathbf{v}) \amp= \begin{pmatrix}
x_1+x_2 \\
x_2+3x_3+x_4 \\
-x_1+x_3-x_4\\
2x_2+2x_3-2x_4
\end{pmatrix}\\
\amp= x_1\begin{pmatrix}1\\0\\-1\\0\end{pmatrix} + x_2\begin{pmatrix}1\\1\\0\\2\end{pmatrix} + x_3\begin{pmatrix}0\\3\\1\\2\end{pmatrix} + x_4\begin{pmatrix}0\\1\\-1\\-2\end{pmatrix}.
\end{align*}
Therefore, \(\operatorname{range}(T)\) is spanned by the four vectors:
\begin{equation*}
\mathbf{v}_1 = \begin{pmatrix}1\\0\\-1\\0\end{pmatrix}, \quad \mathbf{v}_2 = \begin{pmatrix}1\\1\\0\\2\end{pmatrix}, \quad \mathbf{v}_3 = \begin{pmatrix}0\\3\\1\\2\end{pmatrix}, \quad \mathbf{v}_4 = \begin{pmatrix}0\\1\\-1\\-2\end{pmatrix}.
\end{equation*}
However, from the RREF in (1), we see that the first three columns are pivot columns, while the fourth column is not. This means \(\mathbf{v}_4\) is a linear combination of \(\mathbf{v}_1\text{,}\) \(\mathbf{v}_2\text{,}\) and \(\mathbf{v}_3\text{.}\) Indeed, reading from the fourth column of the RREF, we have:
\begin{align*}
\mathbf{v}_4 \amp= 2\mathbf{v}_1 - 2\mathbf{v}_2 + \mathbf{v}_3\\
\amp= 2\begin{pmatrix}1\\0\\-1\\0\end{pmatrix} - 2\begin{pmatrix}1\\1\\0\\2\end{pmatrix} + \begin{pmatrix}0\\3\\1\\2\end{pmatrix}\\
\amp= \begin{pmatrix}2\\0\\-2\\0\end{pmatrix} - \begin{pmatrix}2\\2\\0\\4\end{pmatrix} + \begin{pmatrix}0\\3\\1\\2\end{pmatrix}\\
\amp= \begin{pmatrix}0\\1\\-1\\-2\end{pmatrix}.
\end{align*}
Since \(\mathbf{v}_4\) is a linear combination of \(\mathbf{v}_1\text{,}\) \(\mathbf{v}_2\text{,}\) and \(\mathbf{v}_3\text{,}\) we have:
\begin{equation*}
\operatorname{span}\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3, \mathbf{v}_4\} = \operatorname{span}\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\}.
\end{equation*}
Answer for range:
\begin{equation*}
\operatorname{range}(T) = \operatorname{span}\left\{\begin{pmatrix}1\\0\\-1\\0\end{pmatrix}, \begin{pmatrix}1\\1\\0\\2\end{pmatrix}, \begin{pmatrix}0\\3\\1\\2\end{pmatrix}\right\}.
\end{equation*}
The dimension of the range is \(\dim(\operatorname{range}(T)) = 3\text{.}\)
Verification: Note that
\(\dim(\operatorname{ker}(T)) + \dim(\operatorname{range}(T)) = 1 + 3 = 4 = \dim(\mathbb{R}^4)\text{,}\) which confirms the rank-nullity theorem.
