<exercise checkit-seed="0002" checkit-slug="A3" checkit-title="Image and kernel">
<statement>
<p>
Let <m>T:\mathbb{R}^4 \to \mathbb{R}^3</m>
be the linear transformation given by
<me>T\left( \left[\begin{array}{c}
x_{1} \\
x_{2} \\
x_{3} \\
x_{4}
\end{array}\right] \right) = \left[\begin{array}{c}
-x_{1} - 2 \, x_{2} - 3 \, x_{3} + x_{4} \\
-x_{1} - 3 \, x_{2} - 4 \, x_{3} \\
-x_{2} - x_{3}
\end{array}\right].</me>
</p>
<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}{cccc}
-1 & -2 & -3 & 1 \\
-1 & -3 & -4 & 0 \\
0 & -1 & -1 & 0
\end{array}\right]=\left[\begin{array}{cccc}
1 & 0 & 1 & 0 \\
0 & 1 & 1 & 0 \\
0 & 0 & 0 & 1
\end{array}\right]</me></p>
<ol>
<li>
<p>
<m>\operatorname{Im}\ T = \left\{ \left[\begin{array}{c}
-a - 2 \, b + c \\
-a - 3 \, b \\
-b
\end{array}\right] \middle|\,a,b,c\in\mathbb{R}\right\}</m> and
<m>\operatorname{ker}\ T = \left\{ \left[\begin{array}{c}
-a \\
-a \\
a \\
0
\end{array}\right] \middle|\,a\in\mathbb{R}\right\}</m>
</p>
</li>
<li>
<p>
A basis of <m>\operatorname{Im}\ T</m> is <m>\left\{ \left[\begin{array}{c}
-1 \\
-1 \\
0
\end{array}\right] , \left[\begin{array}{c}
-2 \\
-3 \\
-1
\end{array}\right] , \left[\begin{array}{c}
1 \\
0 \\
0
\end{array}\right] \right\}</m>.
A basis of <m>\operatorname{ker}\ T</m> is <m>\left\{ \left[\begin{array}{c}
-1 \\
-1 \\
1 \\
0
\end{array}\right] \right\}</m>.
</p>
</li>
<li>
<p>
The rank of <m>T</m> is <m>3</m>, the nullity of <m>T</m> is <m>1</m>,
and the dimension of the domain of <m>T</m> is <m>4</m>. The rank-nullity theorem asserts that
<m>3+1=4</m>, which we see to be true.
</p>
</li>
</ol>
</answer>
</exercise>
