ubuntu2004
<exercise checkit-seed="0006" checkit-slug="A3" checkit-title="Image and kernel">1<statement>2<p>3Let <m>T:\mathbb{R}^4 \to \mathbb{R}^3</m>4be the linear transformation given by5<me>T\left( \left[\begin{array}{c}6x \\7y \\8z \\9{w}10\end{array}\right] \right) = \left[\begin{array}{c}11y + 5 \, z - 11 \, {w} \\12-x + z - 4 \, {w} \\13-y - 4 \, z + 9 \, {w}14\end{array}\right].</me>15</p>16<ol>17<li>Explain how to find the image of <m>T</m> and the kernel of <m>T</m>.</li>18<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>19<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>20</ol>21</statement>22<answer>23<p><me>\operatorname{RREF}\left[\begin{array}{cccc}240 & 1 & 5 & -11 \\25-1 & 0 & 1 & -4 \\260 & -1 & -4 & 927\end{array}\right]=\left[\begin{array}{cccc}281 & 0 & 0 & 2 \\290 & 1 & 0 & -1 \\300 & 0 & 1 & -231\end{array}\right]</me></p>32<ol>33<li>34<p>35<m>\operatorname{Im}\ T = \left\{ \left[\begin{array}{c}36b + 5 \, c \\37-a + c \\38-b - 4 \, c39\end{array}\right] \middle|\,a,b,c\in\mathbb{R}\right\}</m> and40<m>\operatorname{ker}\ T = \left\{ \left[\begin{array}{c}41-2 \, a \\42a \\432 \, a \\44a45\end{array}\right] \middle|\,a\in\mathbb{R}\right\}</m>46</p>47</li>48<li>49<p>50A basis of <m>\operatorname{Im}\ T</m> is <m>\left\{ \left[\begin{array}{c}510 \\52-1 \\53054\end{array}\right] , \left[\begin{array}{c}551 \\560 \\57-158\end{array}\right] , \left[\begin{array}{c}595 \\601 \\61-462\end{array}\right] \right\}</m>.63A basis of <m>\operatorname{ker}\ T</m> is <m>\left\{ \left[\begin{array}{c}64-2 \\651 \\662 \\67168\end{array}\right] \right\}</m>.69</p>70</li>71<li>72<p>73The rank of <m>T</m> is <m>3</m>, the nullity of <m>T</m> is <m>1</m>,74and the dimension of the domain of <m>T</m> is <m>4</m>. The rank-nullity theorem asserts that75<m>3+1=4</m>, which we see to be true.76</p>77</li>78</ol>79</answer>80</exercise>818283