ubuntu2004
<exercise checkit-seed="0009" 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_{1} \\7x_{2} \\8x_{3} \\9x_{4}10\end{array}\right] \right) = \left[\begin{array}{c}11x_{1} + 5 \, x_{3} - 13 \, x_{4} \\12x_{2} + 5 \, x_{3} - 18 \, x_{4} \\13x_{3} - 3 \, x_{4}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}241 & 0 & 5 & -13 \\250 & 1 & 5 & -18 \\260 & 0 & 1 & -327\end{array}\right]=\left[\begin{array}{cccc}281 & 0 & 0 & 2 \\290 & 1 & 0 & -3 \\300 & 0 & 1 & -331\end{array}\right]</me></p>32<ol>33<li>34<p>35<m>\operatorname{Im}\ T = \left\{ \left[\begin{array}{c}36a + 5 \, c \\37b + 5 \, c \\38c39\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 \\423 \, a \\433 \, 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}511 \\520 \\53054\end{array}\right] , \left[\begin{array}{c}550 \\561 \\57058\end{array}\right] , \left[\begin{array}{c}595 \\605 \\61162\end{array}\right] \right\}</m>.63A basis of <m>\operatorname{ker}\ T</m> is <m>\left\{ \left[\begin{array}{c}64-2 \\653 \\663 \\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