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ubuntu2004
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<exercise checkit-seed="0010" checkit-slug="A3" checkit-title="Image and kernel">
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<statement>
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Let <m>T:\mathbb{R}^ 3 \to \mathbb{R}^ 4 </m> be the linear transformation given by
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<me>T\left( \left[\begin{array}{c}
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x_{1} \\
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x_{2} \\
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x_{3}
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\end{array}\right] \right) = \left[\begin{array}{c}
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5 \, x_{1} + 4 \, x_{2} - 8 \, x_{3} \\
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-2 \, x_{1} - 3 \, x_{2} + 6 \, x_{3} \\
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2 \, x_{1} + x_{2} - 2 \, x_{3} \\
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-3 \, x_{1} - 4 \, x_{2} + 8 \, x_{3}
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\end{array}\right] .</me><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>
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<answer>
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<p>
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<me>\operatorname{RREF} \left[\begin{array}{ccc}
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5 &amp; 4 &amp; -8 \\
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-2 &amp; -3 &amp; 6 \\
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2 &amp; 1 &amp; -2 \\
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-3 &amp; -4 &amp; 8
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\end{array}\right] = \left[\begin{array}{ccc}
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1 &amp; 0 &amp; 0 \\
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0 &amp; 1 &amp; -2 \\
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0 &amp; 0 &amp; 0 \\
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0 &amp; 0 &amp; 0
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\end{array}\right] </me>
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</p>
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<ol>
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<li>
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<me>\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c}
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5 \\
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-2 \\
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2 \\
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-3
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\end{array}\right] , \left[\begin{array}{c}
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4 \\
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-3 \\
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1 \\
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-4
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\end{array}\right] \right\} </me>
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<me>\operatorname{ker}\ T = \left\{ \left[\begin{array}{c}
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0 \\
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2 \, a \\
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a
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\end{array}\right] \middle|\,a\in\mathbb{R}\right\} </me>
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</li>
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<li>
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A basis of <m>\operatorname{Im}\ T</m> is <m> \left\{ \left[\begin{array}{c}
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5 \\
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-2 \\
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2 \\
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-3
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\end{array}\right] , \left[\begin{array}{c}
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4 \\
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-3 \\
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1 \\
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-4
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\end{array}\right] \right\} </m>.
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A basis of <m>\operatorname{ker}\ T</m> is <m> \left\{ \left[\begin{array}{c}
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0 \\
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2 \\
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1
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\end{array}\right] \right\} </m></li>
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<li>
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The rank of <m>T</m> is <m> 2 </m>, the nullity of <m>T</m> is <m> 1 </m>,
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and the dimension of the domain of <m>T</m> is <m> 3 </m>. The rank-nullity theorem asserts that
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<m> 2 + 1 = 3 </m>, which we see to be true.
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</li>
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</ol>
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</answer>
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</exercise>
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