A3-8.ptx

<exercise checkit-seed="0008" 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}
-5 \, x_{1} - 2 \, x_{2} - 3 \, x_{3} - 11 \, x_{4} \\
2 \, x_{1} + 5 \, x_{2} + 2 \, x_{3} + 7 \, x_{4} \\
-3 \, x_{1} - 2 \, x_{2} - 2 \, x_{3} - 7 \, x_{4}
\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}
-5 &amp; -2 &amp; -3 &amp; -11 \\
2 &amp; 5 &amp; 2 &amp; 7 \\
-3 &amp; -2 &amp; -2 &amp; -7
\end{array}\right]=\left[\begin{array}{cccc}
1 &amp; 0 &amp; 0 &amp; 3 \\
0 &amp; 1 &amp; 0 &amp; 1 \\
0 &amp; 0 &amp; 1 &amp; -2
\end{array}\right]</me></p>
      <ol>
          <li>
              <p>
              <m>\operatorname{Im}\ T =  \left\{ \left[\begin{array}{c}
-5 \, a - 2 \, b - 3 \, c \\
2 \, a + 5 \, b + 2 \, c \\
-3 \, a - 2 \, b - 2 \, c
\end{array}\right] \middle|\,a,b,c\in\mathbb{R}\right\}</m> and
              <m>\operatorname{ker}\ T = \left\{ \left[\begin{array}{c}
-3 \, a \\
-a \\
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}
-5 \\
2 \\
-3
\end{array}\right] , \left[\begin{array}{c}
-2 \\
5 \\
-2
\end{array}\right] , \left[\begin{array}{c}
-3 \\
2 \\
-2
\end{array}\right] \right\}</m>.
              A basis of <m>\operatorname{ker}\ T</m> is <m>\left\{ \left[\begin{array}{c}
-3 \\
-1 \\
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>