ubuntu2004
<exercise checkit-seed="0005" 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}113 \, x - 6 \, y - 4 \, z + {w} \\12-2 \, x + 4 \, y + 3 \, z - {w} \\133 \, x - 6 \, y - 3 \, z14\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}243 & -6 & -4 & 1 \\25-2 & 4 & 3 & -1 \\263 & -6 & -3 & 027\end{array}\right]=\left[\begin{array}{cccc}281 & -2 & 0 & -1 \\290 & 0 & 1 & -1 \\300 & 0 & 0 & 031\end{array}\right]</me></p>32<ol>33<li>34<p>35<m>\operatorname{Im}\ T = \left\{ \left[\begin{array}{c}363 \, a - 4 \, b \\37-2 \, a + 3 \, b \\383 \, a - 3 \, b39\end{array}\right] \middle|\,a,b\in\mathbb{R}\right\}</m> and40<m>\operatorname{ker}\ T = \left\{ \left[\begin{array}{c}412 \, a + b \\42a \\43b \\44b45\end{array}\right] \middle|\,a,b\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}513 \\52-2 \\53354\end{array}\right] , \left[\begin{array}{c}55-4 \\563 \\57-358\end{array}\right] \right\}</m>.59A basis of <m>\operatorname{ker}\ T</m> is <m>\left\{ \left[\begin{array}{c}602 \\611 \\620 \\63064\end{array}\right] , \left[\begin{array}{c}651 \\660 \\671 \\68169\end{array}\right] \right\}</m>.70</p>71</li>72<li>73<p>74The rank of <m>T</m> is <m>2</m>, the nullity of <m>T</m> is <m>2</m>,75and the dimension of the domain of <m>T</m> is <m>4</m>. The rank-nullity theorem asserts that76<m>2+2=4</m>, which we see to be true.77</p>78</li>79</ol>80</answer>81</exercise>828384