<exercise checkit-seed="0006" checkit-slug="A3" checkit-title="Image and kernel">
<statement>
Let <m>T:\mathbb{R}^ 3 \to \mathbb{R}^ 4 </m> be the linear transformation given by
<me>T\left( \left[\begin{array}{c}
x_{1} \\
x_{2} \\
x_{3}
\end{array}\right] \right) = \left[\begin{array}{c}
-3 \, x_{1} + 5 \, x_{2} + 4 \, x_{3} \\
4 \, x_{1} - 7 \, x_{2} - 6 \, x_{3} \\
-x_{1} + x_{2} \\
4 \, x_{1} - 4 \, x_{2}
\end{array}\right] .</me><ol><li>Explain how to find the image of <m>T</m> and the kernel of <m>T</m>.</li><li>Explain how to find a basis of the image of <m>T</m> and a basis of the kernel of <m>T</m>.</li><li>Explain how to find the rank and nullity of <m>T</m>, and why the rank-nullity theorem holds for <m>T</m>.</li></ol></statement>
<answer>
<p>
<me>\operatorname{RREF} \left[\begin{array}{ccc}
-3 & 5 & 4 \\
4 & -7 & -6 \\
-1 & 1 & 0 \\
4 & -4 & 0
\end{array}\right] = \left[\begin{array}{ccc}
1 & 0 & 2 \\
0 & 1 & 2 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right] </me>
</p>
<ol>
<li>
<me>\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c}
-3 \\
4 \\
-1 \\
4
\end{array}\right] , \left[\begin{array}{c}
5 \\
-7 \\
1 \\
-4
\end{array}\right] \right\} </me>
<me>\operatorname{ker}\ T = \left\{ \left[\begin{array}{c}
-2 \, a \\
-2 \, a \\
a
\end{array}\right] \middle|\,a\in\mathbb{R}\right\} </me>
</li>
<li>
A basis of <m>\operatorname{Im}\ T</m> is <m> \left\{ \left[\begin{array}{c}
-3 \\
4 \\
-1 \\
4
\end{array}\right] , \left[\begin{array}{c}
5 \\
-7 \\
1 \\
-4
\end{array}\right] \right\} </m>.
A basis of <m>\operatorname{ker}\ T</m> is <m> \left\{ \left[\begin{array}{c}
-2 \\
-2 \\
1
\end{array}\right] \right\} </m></li>
<li>
The rank of <m>T</m> is <m> 2 </m>, the nullity of <m>T</m> is <m> 1 </m>,
and the dimension of the domain of <m>T</m> is <m> 3 </m>. The rank-nullity theorem asserts that
<m> 2 + 1 = 3 </m>, which we see to be true.
</li>
</ol>
</answer>
</exercise>