<exercise checkit-seed="0004" 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 \\
y \\
z \\
{w}
\end{array}\right] \right) = \left[\begin{array}{c}
-x - 2 \, y - 3 \, z - 4 \, {w} \\
x + y - 2 \, {w} \\
-2 \, x - 2 \, y + z + 6 \, {w}
\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 & -4 \\
1 & 1 & 0 & -2 \\
-2 & -2 & 1 & 6
\end{array}\right]=\left[\begin{array}{cccc}
1 & 0 & 0 & -2 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 2
\end{array}\right]</me></p>
<ol>
<li>
<p>
<m>\operatorname{Im}\ T = \left\{ \left[\begin{array}{c}
-a - 2 \, b - 3 \, c \\
a + b \\
-2 \, a - 2 \, b + c
\end{array}\right] \middle|\,a,b,c\in\mathbb{R}\right\}</m> and
<m>\operatorname{ker}\ T = \left\{ \left[\begin{array}{c}
2 \, a \\
0 \\
-2 \, a \\
a
\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 \\
-2
\end{array}\right] , \left[\begin{array}{c}
-2 \\
1 \\
-2
\end{array}\right] , \left[\begin{array}{c}
-3 \\
0 \\
1
\end{array}\right] \right\}</m>.
A basis of <m>\operatorname{ker}\ T</m> is <m>\left\{ \left[\begin{array}{c}
2 \\
0 \\
-2 \\
1
\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>
