\begin{exercise}{A3}{Image and kernel}{0004}
\begin{exerciseStatement} Let \(T:\mathbb{R}^ 3 \to \mathbb{R}^ 4 \) be the linear transformation given by \[T\left( \left[\begin{array}{c}
x \\
y \\
z
\end{array}\right] \right) = \left[\begin{array}{c}
x - z \\
-2 \, x + y + z \\
-5 \, x + 3 \, y + 2 \, z \\
-2 \, x - 2 \, y + 4 \, z
\end{array}\right] .\]
\begin{enumerate}[(a)]
\item Explain how to find the image of \(T\) and the kernel of \(T\).
\item Explain how to find a basis of the image of \(T\) and a basis of the kernel of \(T\).
\item Explain how to find the rank and nullity of \(T\), and why the rank-nullity theorem holds for \(T\).
\end{enumerate}
\end{exerciseStatement}
\begin{exerciseAnswer}
\[\operatorname{RREF} \left[\begin{array}{ccc}
1 & 0 & -1 \\
-2 & 1 & 1 \\
-5 & 3 & 2 \\
-2 & -2 & 4
\end{array}\right] = \left[\begin{array}{ccc}
1 & 0 & -1 \\
0 & 1 & -1 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right] \]
\begin{enumerate}[(a)]
\item \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c}
1 \\
-2 \\
-5 \\
-2
\end{array}\right] , \left[\begin{array}{c}
0 \\
1 \\
3 \\
-2
\end{array}\right] \right\} \]\[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c}
a \\
a \\
a
\end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]
\item A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c}
1 \\
-2 \\
-5 \\
-2
\end{array}\right] , \left[\begin{array}{c}
0 \\
1 \\
3 \\
-2
\end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c}
1 \\
1 \\
1
\end{array}\right] \right\} \)
\item The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.
\end{enumerate}
\end{exerciseAnswer}
\end{exercise}
