ubuntu2004
<exercise checkit-seed="0004" 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}11-x - 2 \, y - 3 \, z - 4 \, {w} \\12x + y - 2 \, {w} \\13-2 \, x - 2 \, y + z + 6 \, {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}24-1 & -2 & -3 & -4 \\251 & 1 & 0 & -2 \\26-2 & -2 & 1 & 627\end{array}\right]=\left[\begin{array}{cccc}281 & 0 & 0 & -2 \\290 & 1 & 0 & 0 \\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}36-a - 2 \, b - 3 \, c \\37a + b \\38-2 \, a - 2 \, b + c39\end{array}\right] \middle|\,a,b,c\in\mathbb{R}\right\}</m> and40<m>\operatorname{ker}\ T = \left\{ \left[\begin{array}{c}412 \, a \\420 \\43-2 \, 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}51-1 \\521 \\53-254\end{array}\right] , \left[\begin{array}{c}55-2 \\561 \\57-258\end{array}\right] , \left[\begin{array}{c}59-3 \\600 \\61162\end{array}\right] \right\}</m>.63A basis of <m>\operatorname{ker}\ T</m> is <m>\left\{ \left[\begin{array}{c}642 \\650 \\66-2 \\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