<exercise checkit-seed="0005" 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 \\
y \\
z
\end{array}\right] \right) = \left[\begin{array}{c}
-x - 4 \, y + 5 \, z \\
-x - 4 \, y + 3 \, z \\
-2 \, x - 8 \, y + 7 \, z \\
x + 4 \, y - 5 \, z
\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}
-1 & -4 & 5 \\
-1 & -4 & 3 \\
-2 & -8 & 7 \\
1 & 4 & -5
\end{array}\right] = \left[\begin{array}{ccc}
1 & 4 & 0 \\
0 & 0 & 1 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right] </me>
</p>
<ol>
<li>
<me>\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c}
-1 \\
-1 \\
-2 \\
1
\end{array}\right] , \left[\begin{array}{c}
5 \\
3 \\
7 \\
-5
\end{array}\right] \right\} </me>
<me>\operatorname{ker}\ T = \left\{ \left[\begin{array}{c}
-4 \, a \\
a \\
0
\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}
-1 \\
-1 \\
-2 \\
1
\end{array}\right] , \left[\begin{array}{c}
5 \\
3 \\
7 \\
-5
\end{array}\right] \right\} </m>.
A basis of <m>\operatorname{ker}\ T</m> is <m> \left\{ \left[\begin{array}{c}
-4 \\
1 \\
0
\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>