<exercise checkit-seed="0000" 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 - 5 \, {w} \\
{w} \\
x + 2 \, y + 3 \, z - 5 \, {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 & -5 \\
0 & 0 & 0 & 1 \\
1 & 2 & 3 & -5
\end{array}\right]=\left[\begin{array}{cccc}
1 & 2 & 3 & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0
\end{array}\right]</me></p>
<ol>
<li>
<p>
<m>\operatorname{Im}\ T = \left\{ \left[\begin{array}{c}
a - 5 \, b \\
b \\
a - 5 \, b
\end{array}\right] \middle|\,a,b\in\mathbb{R}\right\}</m> and
<m>\operatorname{ker}\ T = \left\{ \left[\begin{array}{c}
-2 \, a - 3 \, b \\
a \\
b \\
0
\end{array}\right] \middle|\,a,b\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 \\
0 \\
1
\end{array}\right] , \left[\begin{array}{c}
-5 \\
1 \\
-5
\end{array}\right] \right\}</m>.
A basis of <m>\operatorname{ker}\ T</m> is <m>\left\{ \left[\begin{array}{c}
-2 \\
1 \\
0 \\
0
\end{array}\right] , \left[\begin{array}{c}
-3 \\
0 \\
1 \\
0
\end{array}\right] \right\}</m>.
</p>
</li>
<li>
<p>
The rank of <m>T</m> is <m>2</m>, the nullity of <m>T</m> is <m>2</m>,
and the dimension of the domain of <m>T</m> is <m>4</m>. The rank-nullity theorem asserts that
<m>2+2=4</m>, which we see to be true.
</p>
</li>
</ol>
</answer>
</exercise>
