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ubuntu2004
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<exercise checkit-seed="0005" checkit-slug="A3" checkit-title="Image and kernel">
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<statement>
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<p>
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Let <m>T:\mathbb{R}^4 \to \mathbb{R}^3</m>
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be the linear transformation given by
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<me>T\left( \left[\begin{array}{c}
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x \\
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y \\
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z \\
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{w}
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\end{array}\right] \right) = \left[\begin{array}{c}
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3 \, x - 6 \, y - 4 \, z + {w} \\
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-2 \, x + 4 \, y + 3 \, z - {w} \\
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3 \, x - 6 \, y - 3 \, z
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\end{array}\right].</me>
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</p>
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<ol>
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<li>Explain how to find the image of <m>T</m> and the kernel of <m>T</m>.</li>
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<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>
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<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>
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</ol>
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</statement>
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<answer>
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<p><me>\operatorname{RREF}\left[\begin{array}{cccc}
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3 &amp; -6 &amp; -4 &amp; 1 \\
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-2 &amp; 4 &amp; 3 &amp; -1 \\
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3 &amp; -6 &amp; -3 &amp; 0
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\end{array}\right]=\left[\begin{array}{cccc}
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1 &amp; -2 &amp; 0 &amp; -1 \\
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0 &amp; 0 &amp; 1 &amp; -1 \\
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0 &amp; 0 &amp; 0 &amp; 0
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\end{array}\right]</me></p>
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<ol>
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<li>
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<p>
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<m>\operatorname{Im}\ T = \left\{ \left[\begin{array}{c}
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3 \, a - 4 \, b \\
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-2 \, a + 3 \, b \\
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3 \, a - 3 \, b
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\end{array}\right] \middle|\,a,b\in\mathbb{R}\right\}</m> and
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<m>\operatorname{ker}\ T = \left\{ \left[\begin{array}{c}
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2 \, a + b \\
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a \\
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b \\
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b
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\end{array}\right] \middle|\,a,b\in\mathbb{R}\right\}</m>
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</p>
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</li>
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<li>
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<p>
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A basis of <m>\operatorname{Im}\ T</m> is <m>\left\{ \left[\begin{array}{c}
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3 \\
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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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-3
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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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2 \\
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1 \\
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0 \\
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0
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\end{array}\right] , \left[\begin{array}{c}
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1 \\
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0 \\
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1 \\
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1
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\end{array}\right] \right\}</m>.
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</p>
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</li>
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<li>
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<p>
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The rank of <m>T</m> is <m>2</m>, the nullity of <m>T</m> is <m>2</m>,
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and the dimension of the domain of <m>T</m> is <m>4</m>. The rank-nullity theorem asserts that
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<m>2+2=4</m>, which we see to be true.
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</p>
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</li>
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</ol>
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</answer>
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</exercise>
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