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<!DOCTYPE html>
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\newcommand{\IR}{\mathbb{R}}
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<header id="masthead" class="smallbuttons"><div class="banner"><div class="container">
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<h1 class="heading"><a href="linear-algebra-for-team-based-inquiry-learning.html"><span class="title">Linear Algebra for Team-Based Inquiry Learning</span></a></h1>
<p class="byline">Steven Clontz, Drew Lewis</p>
</div>
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<li class="link frontmatter"><a href="frontmatter.html" data-scroll="frontmatter"><span class="title">Front Matter</span></a></li>
<li class="link">
<a href="E.html" data-scroll="E"><span class="codenumber">1</span> <span class="title">Systems of Linear Equations (E)</span></a><ul>
<li><a href="E1.html" data-scroll="E1">Linear Systems, Vector Equations, and Augmented Matrices (E1)</a></li>
<li><a href="E2.html" data-scroll="E2">Row Reduction of Matrices (E2)</a></li>
<li><a href="E3.html" data-scroll="E3">Solving Linear Systems (E3)</a></li>
</ul>
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<a href="V.html" data-scroll="V"><span class="codenumber">2</span> <span class="title">Vector Spaces (V)</span></a><ul>
<li><a href="V1.html" data-scroll="V1">Vector Spaces (V1)</a></li>
<li><a href="V2.html" data-scroll="V2">Linear Combinations (V2)</a></li>
<li><a href="V3.html" data-scroll="V3">Spanning Sets (V3)</a></li>
<li><a href="V4.html" data-scroll="V4">Subspaces (V4)</a></li>
<li><a href="V5.html" data-scroll="V5">Linear Independence (V5)</a></li>
<li><a href="V6.html" data-scroll="V6">Identifying a Basis (V6)</a></li>
<li><a href="V7.html" data-scroll="V7">Subspace Basis and Dimension (V7)</a></li>
<li><a href="V8.html" data-scroll="V8">Polynomial and Matrix Spaces (V8)</a></li>
<li><a href="V9.html" data-scroll="V9">Homogeneous Linear Systems (V9)</a></li>
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<a href="A.html" data-scroll="A"><span class="codenumber">3</span> <span class="title">Algebraic Properties of Linear Maps (A)</span></a><ul>
<li><a href="A1.html" data-scroll="A1">Linear Transformations (A1)</a></li>
<li><a href="A2.html" data-scroll="A2">Standard Matrices (A2)</a></li>
<li><a href="A3.html" data-scroll="A3" class="active">Image and Kernel (A3)</a></li>
<li><a href="A4.html" data-scroll="A4">Injective and Surjective Linear Maps (A4)</a></li>
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<a href="M.html" data-scroll="M"><span class="codenumber">4</span> <span class="title">Matrices (M)</span></a><ul>
<li><a href="M1.html" data-scroll="M1">Matrices and Multiplication (M1)</a></li>
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<li><a href="M3.html" data-scroll="M3">The Inverse of a Matrix (M3)</a></li>
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<a href="G.html" data-scroll="G"><span class="codenumber">5</span> <span class="title">Geometric Properties of Linear Maps (G)</span></a><ul>
<li><a href="G1.html" data-scroll="G1">Row Operations and Determinants (G1)</a></li>
<li><a href="G2.html" data-scroll="G2">Computing Determinants (G2)</a></li>
<li><a href="G3.html" data-scroll="G3">Eigenvalues and Characteristic Polynomials (G3)</a></li>
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<main class="main"><div id="content" class="pretext-content"><section xmlns:svg="http://www.w3.org/2000/svg" class="section" id="A3"><h2 class="heading hide-type">
<span class="type">Section</span> <span class="codenumber">3.3</span> <span class="title">Image and Kernel (A3)</span>
</h2>
<article class="definition definition-like" id="definition-20"><h6 class="heading">
<span class="type">Definition</span><span class="space"> </span><span class="codenumber">3.3.1</span><span class="period">.</span>
</h6>
<p id="p-535">Let \(T: V \rightarrow W\) be a linear transformation. The <dfn class="terminology">kernel</dfn> of \(T\) is an important subspace of \(V\) defined by</p>
<div class="displaymath">
\begin{equation*}
\ker T = \left\{ \vec{v} \in V\ \big|\ T(\vec{v})=\vec{z}\right\}
\end{equation*}
</div>
<pre class="code-block tex2jax_ignore">
\begin{tikzpicture}[x=0.15in,y=0.15in]
\begin{scope}[shift={(0,0)}]
\draw (0,0) node[anchor=north west] {\(\ker T\)} -- (3,0);
\draw (0,0) -- (0,3);
\draw (0,0) -- (-2,-1);
\draw[thick,latex-latex,blue] (-3,-3) -- (3,3);
\end{scope}
\draw[dashed,-latex] (3.5,3) to [bend left=30] (7,3);
\begin{scope}[shift={(9,0.5)}]
\draw (-2,0) -- (2,0);
\draw (0,-2) -- (0,2);
\fill[blue] (0,0) circle (0.2)
node[anchor=south east] {\(\vec{0}\)};
\end{scope}
\end{tikzpicture}
</pre></article><article class="activity project-like" id="activity-79"><h6 class="heading">
<span class="type">Activity</span><span class="space"> </span><span class="codenumber">3.3.1</span><span class="period">.</span>
</h6>
<p id="p-536">Let \(T: \IR^2 \rightarrow \IR^3\) be given by</p>
<div class="displaymath">
\begin{equation*}
T\left(\left[\begin{array}{c}x \\ y \end{array}\right] \right)
=
\left[\begin{array}{c} x \\ y \\ 0 \end{array}\right]
\hspace{3em}
\text{with standard matrix }
\left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \\ 0 & 0 \end{array}\right]
\end{equation*}
</div>
<p data-braille="continuation">Which of these subspaces of \(\IR^2\) describes \(\ker T\text{,}\) the set of all vectors that transform into \(\vec 0\text{?}\)</p>
<ol class="lower-alpha">
<li id="li-373"><p id="p-537">\(\displaystyle \setBuilder{\left[\begin{array}{c}a \\ a\end{array}\right]}{a\in\IR}\)</p></li>
<li id="li-374"><p id="p-538">\(\displaystyle \setList{\left[\begin{array}{c}0\\0\end{array}\right]}\)</p></li>
<li id="li-375"><p id="p-539">\(\displaystyle \IR^2=\setBuilder{\left[\begin{array}{c}x \\ y\end{array}\right]}{x,y\in\IR}\)</p></li>
</ol></article><article class="activity project-like" id="activity-80"><h6 class="heading">
<span class="type">Activity</span><span class="space"> </span><span class="codenumber">3.3.2</span><span class="period">.</span>
</h6>
<p id="p-540">Let \(T: \IR^3 \rightarrow \IR^2\) be given by</p>
<div class="displaymath">
\begin{equation*}
T\left(\left[\begin{array}{c}x \\ y\\z \end{array}\right] \right)
=
\left[\begin{array}{c} x \\ y \end{array}\right]
\hspace{3em}
\text{with standard matrix }
\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \end{array}\right]
\end{equation*}
</div>
<p data-braille="continuation">Which of these subspaces of \(\IR^3\) describes \(\ker T\text{,}\) the set of all vectors that transform into \(\vec 0\text{?}\)</p>
<ol class="lower-alpha">
<li id="li-376"><p id="p-541">\(\displaystyle \setBuilder{\left[\begin{array}{c}0 \\ 0\\ a\end{array}\right]}{a\in\IR}\)</p></li>
<li id="li-377"><p id="p-542">\(\displaystyle \setBuilder{\left[\begin{array}{c}a \\ a\\ 0\end{array}\right]}{a\in\IR}\)</p></li>
<li id="li-378"><p id="p-543">\(\displaystyle \setList{\left[\begin{array}{c}0\\0\\0\end{array}\right]}\)</p></li>
<li id="li-379"><p id="p-544">\(\displaystyle \IR^3=\setBuilder{\left[\begin{array}{c}x \\ y\\z\end{array}\right]}{x,y,z\in\IR}\)</p></li>
</ol></article><article class="activity project-like" id="activity-81"><h6 class="heading">
<span class="type">Activity</span><span class="space"> </span><span class="codenumber">3.3.3</span><span class="period">.</span>
</h6>
<div class="introduction" id="introduction-28">
<p id="p-545">Let \(T: \IR^3 \rightarrow \IR^2\) be the linear transformation given by the standard matrix</p>
<div class="displaymath">
\begin{equation*}
T\left( \left[\begin{array}{c} x \\ y \\ z \end{array}\right]\right) = \left[\begin{array}{c} 3x+4y-z \\ x+2y+z \end{array}\right]
\end{equation*}
</div>
</div>
<article class="task exercise-like" id="task-65"><h6 class="heading"><span class="codenumber">(a)</span></h6>
<p id="p-546">Set \(T\left(\left[\begin{array}{c}x\\y\\z\end{array}\right]\right)
=
\left[\begin{array}{c}0\\0\end{array}\right]\) to find a linear system of equations whose solution set is the kernel.</p></article><article class="task exercise-like" id="task-66"><h6 class="heading"><span class="codenumber">(b)</span></h6>
<p id="p-547">Use \(\RREF(A)\) to solve this homogeneous system of equations and find a basis for the kernel of \(T\text{.}\)</p></article></article><article class="activity project-like" id="activity-82"><h6 class="heading">
<span class="type">Activity</span><span class="space"> </span><span class="codenumber">3.3.4</span><span class="period">.</span>
</h6>
<p id="p-548">Let \(T: \IR^4 \rightarrow \IR^3\) be the linear transformation given by</p>
<div class="displaymath">
\begin{equation*}
T\left(\left[\begin{array}{c} x \\ y \\ z \\ w \end{array}\right] \right) =
\left[\begin{array}{c} 2x+4y+2z-4w \\ -2x-4y+z+w \\ 3x+6y-z-4w\end{array}\right].
\end{equation*}
</div>
<p id="p-549">Find a basis for the kernel of \(T\text{.}\)</p></article><article class="definition definition-like" id="definition-21"><h6 class="heading">
<span class="type">Definition</span><span class="space"> </span><span class="codenumber">3.3.2</span><span class="period">.</span>
</h6>
<p id="p-550">Let \(T: V \rightarrow W\) be a linear transformation. The <dfn class="terminology">image</dfn> of \(T\) is an important subspace of \(W\) defined by</p>
<div class="displaymath">
\begin{equation*}
\Im T = \left\{ \vec{w} \in W\ \big|\ \text{there is some }\vec v\in V \text{ with } T(\vec{v})=\vec{w}\right\}
\end{equation*}
</div>
<p id="p-551">In the examples below, the left example's image is all of \(\IR^2\text{,}\) but the right example's image is a planar subspace of \(\IR^3\text{.}\)</p>
<pre class="code-block tex2jax_ignore">
\begin{tikzpicture}[x=0.15in,y=0.15in]
\begin{scope}[shift={(0,0)}]
\draw (0,0) -- (3,0);
\draw (0,0) -- (0,3);
\draw (0,0) -- (-2,-1);
\draw[thick,-latex,blue] (0,0) -- (2,1);
\draw[thick,-latex,blue] (0,0) -- (1,2);
\draw[thick,-latex,blue] (0,0) -- (0,2);
\draw[thick,-latex,blue] (0,0) -- (1,-1);
\draw[thick,-latex,blue] (0,0) -- (-2,3);
\draw[thick,-latex,blue] (0,0) -- (-3,-2);
\end{scope}
\draw[dashed,-latex] (3,3) to [bend left=30] (7,3);
\begin{scope}[shift={(9,1)}]
\draw (-2,0) -- (2,0);
\draw (0,-2) -- (0,2);
\draw[thick,-latex,blue] (0,0) -- (0.5,2);
\draw[thick,-latex,blue] (0,0) -- (2,1);
\draw[thick,-latex,blue] (0,0) -- (-1.5,1);
\draw[thick,-latex,blue] (0,0) -- (0,-1.5);
\draw[thick,-latex,blue] (0,0) -- (2,-2);
\fill[color=blue, opacity=0.5] (-2,-2) rectangle (2,2);
\end{scope}
\end{tikzpicture}
\hspace{3em}
\begin{tikzpicture}[x=0.15in,y=0.15in]
\begin{scope}[shift={(0,1)}]
\draw (-2,0) -- (2,0);
\draw (0,-2) -- (0,2);
\draw[thick,-latex,blue] (0,0) -- (0.5,2);
\draw[thick,-latex,blue] (0,0) -- (2,1);
\draw[thick,-latex,blue] (0,0) -- (-1.5,1);
\draw[thick,-latex,blue] (0,0) -- (0,-1.5);
\draw[thick,-latex,blue] (0,0) -- (2,-2);
\end{scope}
\draw[dashed,-latex] (3,3) to [bend left=30] (7,3);
\begin{scope}[shift={(9,0)}]
\draw (0,0) -- (3,0);
\draw (0,0) -- (0,3);
\draw (0,0) -- (-2,-1);
\draw[thick,-latex,blue] (0,0) -- (0.5,1.5);
\draw[thick,-latex,blue] (0,0) -- (2,1);
\draw[thick,-latex,blue] (0,0) -- (-2.5,1);
\draw[thick,-latex,blue] (0,0) -- (-0.5,-1.5);
\draw[thick,-latex,blue] (0,0) -- (2.5,-0.5);
\fill[color=blue, opacity=0.5] (-2,-2) -- (3,-1) -- (2,2) -- (-3,1) -- (-2,-2);
\end{scope}
\end{tikzpicture}
</pre></article><article class="activity project-like" id="activity-83"><h6 class="heading">
<span class="type">Activity</span><span class="space"> </span><span class="codenumber">3.3.5</span><span class="period">.</span>
</h6>
<p id="p-552">Let \(T: \IR^2 \rightarrow \IR^3\) be given by</p>
<div class="displaymath">
\begin{equation*}
T\left(\left[\begin{array}{c}x \\ y \end{array}\right] \right)
=
\left[\begin{array}{c} x \\ y \\ 0 \end{array}\right]
\hspace{3em}
\text{with standard matrix }
\left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \\ 0 & 0 \end{array}\right]
\end{equation*}
</div>
<p data-braille="continuation">Which of these subspaces of \(\IR^3\) describes \(\Im T\text{,}\) the set of all vectors that are the result of using \(T\) to transform \(\IR^2\) vectors?</p>
<ol class="lower-alpha">
<li id="li-380"><p id="p-553">\(\displaystyle \setBuilder{\left[\begin{array}{c}0 \\ 0\\ a\end{array}\right]}{a\in\IR}\)</p></li>
<li id="li-381"><p id="p-554">\(\displaystyle \setBuilder{\left[\begin{array}{c}a \\ b\\ 0\end{array}\right]}{a,b\in\IR}\)</p></li>
<li id="li-382"><p id="p-555">\(\displaystyle \setList{\left[\begin{array}{c}0\\0\\0\end{array}\right]}\)</p></li>
<li id="li-383"><p id="p-556">\(\displaystyle \IR^3=\setBuilder{\left[\begin{array}{c}x \\ y\\z\end{array}\right]}{x,y,z\in\IR}\)</p></li>
</ol></article><article class="activity project-like" id="activity-84"><h6 class="heading">
<span class="type">Activity</span><span class="space"> </span><span class="codenumber">3.3.6</span><span class="period">.</span>
</h6>
<p id="p-557">Let \(T: \IR^3 \rightarrow \IR^2\) be given by</p>
<div class="displaymath">
\begin{equation*}
T\left(\left[\begin{array}{c}x \\ y\\z \end{array}\right] \right)
=
\left[\begin{array}{c} x \\ y \end{array}\right]
\hspace{3em}
\text{with standard matrix }
\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \end{array}\right]
\end{equation*}
</div>
<p data-braille="continuation">Which of these subspaces of \(\IR^2\) describes \(\Im T\text{,}\) the set of all vectors that are the result of using \(T\) to transform \(\IR^3\) vectors?</p>
<ol class="lower-alpha">
<li id="li-384"><p id="p-558">\(\displaystyle \setBuilder{\left[\begin{array}{c}a \\ a\end{array}\right]}{a\in\IR}\)</p></li>
<li id="li-385"><p id="p-559">\(\displaystyle \setList{\left[\begin{array}{c}0\\0\end{array}\right]}\)</p></li>
<li id="li-386"><p id="p-560">\(\displaystyle \IR^2=\setBuilder{\left[\begin{array}{c}x \\ y\end{array}\right]}{x,y\in\IR}\)</p></li>
</ol></article><article class="activity project-like" id="activity-85"><h6 class="heading">
<span class="type">Activity</span><span class="space"> </span><span class="codenumber">3.3.7</span><span class="period">.</span>
</h6>
<p id="p-561">Let \(T: \IR^4 \rightarrow \IR^3\) be the linear transformation given by the standard matrix</p>
<div class="displaymath">
\begin{equation*}
A
=
\left[\begin{array}{cccc} 3 & 4 & 7 & 1\\ -1 & 1 & 0 & 2 \\ 2 & 1 & 3 & -1 \end{array}\right]
=
\left[\begin{array}{cccc}T(\vec e_1)&T(\vec e_2)&T(\vec e_3)&T(\vec e_4)\end{array}\right]
.
\end{equation*}
</div>
<p id="p-562">Since \(T(\vec v)=T(x_1\vec e_1+x_2\vec e_2+x_3\vec e_3+x_4\vec e_4)\text{,}\) the set of vectors</p>
<div class="displaymath">
\begin{equation*}
\setList{
\left[\begin{array}{c}3\\-1\\2\end{array}\right],
\left[\begin{array}{c}4\\1\\1\end{array}\right],
\left[\begin{array}{c}7\\0\\3\end{array}\right],
\left[\begin{array}{c}1\\2\\-1\end{array}\right]
}
\end{equation*}
</div>
<ol class="lower-alpha">
<li id="li-387"><p id="p-563">spans \(\Im T\)</p></li>
<li id="li-388"><p id="p-564">is a linearly independent subset of \(\Im T\)</p></li>
<li id="li-389"><p id="p-565">is a basis for \(\Im T\)</p></li>
</ol></article><article class="observation remark-like" id="observation-17"><h6 class="heading">
<span class="type">Observation</span><span class="space"> </span><span class="codenumber">3.3.3</span><span class="period">.</span>
</h6>
<p id="p-566">Let \(T: \IR^4 \rightarrow \IR^3\) be the linear transformation given by the standard matrix</p>
<div class="displaymath">
\begin{equation*}
A
=
\left[\begin{array}{cccc} 3 & 4 & 7 & 1\\ -1 & 1 & 0 & 2 \\ 2 & 1 & 3 & -1 \end{array}\right]
.
\end{equation*}
</div>
<p id="p-567">Since the set \(\setList{
\left[\begin{array}{c}3\\-1\\2\end{array}\right],
\left[\begin{array}{c}4\\1\\1\end{array}\right],
\left[\begin{array}{c}7\\0\\3\end{array}\right],
\left[\begin{array}{c}1\\2\\-1\end{array}\right]
}\) spans \(\Im T\text{,}\) we can obtain a basis for \(\Im T\) by finding \(\RREF A
=
\left[\begin{array}{cccc} 1 & 0 & 1 & -1\\ 0 & 1 & 1 & 1 \\ 0 & 0 & 0 & 0 \end{array}\right]\) and only using the vectors corresponding to pivot columns:</p>
<div class="displaymath">
\begin{equation*}
\setList{
\left[\begin{array}{c}3\\-1\\2\end{array}\right],
\left[\begin{array}{c}4\\1\\1\end{array}\right]
}
\end{equation*}
</div></article><article class="fact theorem-like" id="fact-17"><h6 class="heading">
<span class="type">Fact</span><span class="space"> </span><span class="codenumber">3.3.4</span><span class="period">.</span>
</h6>
<p id="p-568">Let \(T:\IR^n\to\IR^m\) be a linear transformation with standard matrix \(A\text{.}\)</p>
<ul class="disc">
<li id="li-390"><p id="p-569">The kernel of \(T\) is the solution set of the homogeneous system given by the augmented matrix \(\left[\begin{array}{c|c}A&\vec 0\end{array}\right]\text{.}\) Use the coefficients of its free variables to get a basis for the kernel.</p></li>
<li id="li-391"><p id="p-570">The image of \(T\) is the span of the columns of \(A\text{.}\) Remove the vectors creating non-pivot columns in \(\RREF A\) to get a basis for the image.</p></li>
</ul></article><article class="activity project-like" id="activity-86"><h6 class="heading">
<span class="type">Activity</span><span class="space"> </span><span class="codenumber">3.3.8</span><span class="period">.</span>
</h6>
<p id="p-571">Let \(T: \IR^3 \rightarrow \IR^4\) be the linear transformation given by the standard matrix</p>
<div class="displaymath">
\begin{equation*}
A
=
\left[\begin{array}{ccc} 1 & -3 & 2\\ 2 & -6 & 0 \\ 0 & 0 & 1 \\ -1 & 3 & 1 \end{array}\right]
.
\end{equation*}
</div>
<p id="p-572">Find a basis for the kernel and a basis for the image of \(T\text{.}\)</p></article><article class="activity project-like" id="activity-87"><h6 class="heading">
<span class="type">Activity</span><span class="space"> </span><span class="codenumber">3.3.9</span><span class="period">.</span>
</h6>
<p id="p-573">Let \(T: \IR^n \rightarrow \IR^m\) be a linear transformation with standard matrix \(A\text{.}\) Which of the following is equal to the dimension of the kernel of \(T\text{?}\)</p>
<ol class="lower-alpha">
<li id="li-392"><p id="p-574">The number of pivot columns</p></li>
<li id="li-393"><p id="p-575">The number of non-pivot columns</p></li>
<li id="li-394"><p id="p-576">The number of pivot rows</p></li>
<li id="li-395"><p id="p-577">The number of non-pivot rows</p></li>
</ol></article><article class="activity project-like" id="activity-88"><h6 class="heading">
<span class="type">Activity</span><span class="space"> </span><span class="codenumber">3.3.10</span><span class="period">.</span>
</h6>
<p id="p-578">Let \(T: \IR^n \rightarrow \IR^m\) be a linear transformation with standard matrix \(A\text{.}\) Which of the following is equal to the dimension of the image of \(T\text{?}\)</p>
<ol class="lower-alpha">
<li id="li-396"><p id="p-579">The number of pivot columns</p></li>
<li id="li-397"><p id="p-580">The number of non-pivot columns</p></li>
<li id="li-398"><p id="p-581">The number of pivot rows</p></li>
<li id="li-399"><p id="p-582">The number of non-pivot rows</p></li>
</ol></article><article class="observation remark-like" id="observation-18"><h6 class="heading">
<span class="type">Observation</span><span class="space"> </span><span class="codenumber">3.3.5</span><span class="period">.</span>
</h6>
<p id="p-583">Combining these with the observation that the number of columns is the dimension of the domain of \(T\text{,}\) we have the <dfn class="terminology">rank-nullity theorem</dfn>:</p>
<blockquote class="blockquote" id="blockquote-1"><p id="p-584">The dimension of the domain of \(T\) equals \(\dim(\ker T)+\dim(\Im T)\text{.}\)</p></blockquote>
<p id="p-585">The dimension of the image is called the <dfn class="terminology">rank</dfn> of \(T\) (or \(A\)) and the dimension of the kernel is called the <dfn class="terminology">nullity</dfn>.</p></article><article class="activity project-like" id="activity-89"><h6 class="heading">
<span class="type">Activity</span><span class="space"> </span><span class="codenumber">3.3.11</span><span class="period">.</span>
</h6>
<p id="p-586">Let \(T: \IR^3 \rightarrow \IR^4\) be the linear transformation given by the standard matrix</p>
<div class="displaymath">
\begin{equation*}
A
=
\left[\begin{array}{ccc} 1 & -3 & 2\\ 2 & -6 & 0 \\ 0 & 0 & 1 \\ -1 & 3 & 1 \end{array}\right]
.
\end{equation*}
</div>
<p data-braille="continuation">Verify that the rank-nullity theorem holds for \(T\text{.}\)</p></article><section class="exercises" id="exercises-15"><h3 class="heading hide-type">
<span class="type">Exercises</span> <span class="codenumber">3.3.1</span> <span class="title">Exercises</span>
</h3>
<article class="exercise exercise-like" id="exercise-71"><h6 class="heading"><span class="codenumber">1<span class="period">.</span></span></h6>Let \(T:\mathbb{R}^ 3 \to \mathbb{R}^ 4 \) be the linear transformation given by<div class="displaymath">
\begin{equation*}
T\left( \left[\begin{array}{c}
x \\
y \\
z
\end{array}\right] \right) = \left[\begin{array}{c}
x + 3 \, y - z \\
-3 \, x - 8 \, y + 2 \, z \\
-4 \, x - 8 \, y \\
-x - 7 \, y + 5 \, z
\end{array}\right] .
\end{equation*}
</div>
<ol class="lower-alpha">
<li id="li-400">Explain how to find the image of \(T\) and the kernel of \(T\text{.}\)</li>
<li id="li-401">Explain how to find a basis of the image of \(T\) and a basis of the kernel of \(T\text{.}\)</li>
<li id="li-402">Explain how to find the rank and nullity of \(T\text{,}\) and why the rank-nullity theorem holds for \(T\text{.}\)</li>
</ol>
<div class="solutions">
<a data-knowl="" class="id-ref answer-knowl original" data-refid="hk-answer-71" id="answer-71"><span class="type">Answer</span></a><div class="hidden-content tex2jax_ignore" id="hk-answer-71"><div class="answer solution-like">
<div class="displaymath" id="p-587">
\begin{equation*}
\operatorname{RREF} \left[\begin{array}{ccc}
1 & 3 & -1 \\
-3 & -8 & 2 \\
-4 & -8 & 0 \\
-1 & -7 & 5
\end{array}\right] = \left[\begin{array}{ccc}
1 & 0 & 2 \\
0 & 1 & -1 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right]
\end{equation*}
</div>
<ol class="lower-alpha">
<li id="li-403">
<div class="displaymath">
\begin{equation*}
\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c}
1 \\
-3 \\
-4 \\
-1
\end{array}\right] , \left[\begin{array}{c}
3 \\
-8 \\
-8 \\
-7
\end{array}\right] \right\}
\end{equation*}
</div>
<div class="displaymath">
\begin{equation*}
\operatorname{ker}\ T = \left\{ \left[\begin{array}{c}
-2 \, a \\
a \\
a
\end{array}\right] \middle|\,a\in\mathbb{R}\right\}
\end{equation*}
</div>
</li>
<li id="li-404">A basis of \(\operatorname{Im}\ T\) is \(\left\{ \left[\begin{array}{c}
1 \\
-3 \\
-4 \\
-1
\end{array}\right] , \left[\begin{array}{c}
3 \\
-8 \\
-8 \\
-7
\end{array}\right] \right\} \text{.}\) A basis of \(\operatorname{ker}\ T\) is \(\left\{ \left[\begin{array}{c}
-2 \\
1 \\
1
\end{array}\right] \right\} \)</li>
<li id="li-405">The rank of \(T\) is \(2 \text{,}\) the nullity of \(T\) is \(1 \text{,}\) and the dimension of the domain of \(T\) is \(3 \text{.}\) The rank-nullity theorem asserts that \(2 + 1 = 3 \text{,}\) which we see to be true.</li>
</ol>
</div></div>
</div></article><article class="exercise exercise-like" id="exercise-72"><h6 class="heading"><span class="codenumber">2<span class="period">.</span></span></h6>Let \(T:\mathbb{R}^ 3 \to \mathbb{R}^ 4 \) be the linear transformation given by<div class="displaymath">
\begin{equation*}
T\left( \left[\begin{array}{c}
x \\
y \\
z
\end{array}\right] \right) = \left[\begin{array}{c}
3 \, x + 2 \, y - 3 \, z \\
-3 \, x - 2 \, y + 3 \, z \\
2 \, x + y - z \\
x - y + 4 \, z
\end{array}\right] .
\end{equation*}
</div>
<ol class="lower-alpha">
<li id="li-406">Explain how to find the image of \(T\) and the kernel of \(T\text{.}\)</li>
<li id="li-407">Explain how to find a basis of the image of \(T\) and a basis of the kernel of \(T\text{.}\)</li>
<li id="li-408">Explain how to find the rank and nullity of \(T\text{,}\) and why the rank-nullity theorem holds for \(T\text{.}\)</li>
</ol>
<div class="solutions">
<a data-knowl="" class="id-ref answer-knowl original" data-refid="hk-answer-72" id="answer-72"><span class="type">Answer</span></a><div class="hidden-content tex2jax_ignore" id="hk-answer-72"><div class="answer solution-like">
<div class="displaymath" id="p-588">
\begin{equation*}
\operatorname{RREF} \left[\begin{array}{ccc}
3 & 2 & -3 \\
-3 & -2 & 3 \\
2 & 1 & -1 \\
1 & -1 & 4
\end{array}\right] = \left[\begin{array}{ccc}
1 & 0 & 1 \\
0 & 1 & -3 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right]
\end{equation*}
</div>
<ol class="lower-alpha">
<li id="li-409">
<div class="displaymath">
\begin{equation*}
\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c}
3 \\
-3 \\
2 \\
1
\end{array}\right] , \left[\begin{array}{c}
2 \\
-2 \\
1 \\
-1
\end{array}\right] \right\}
\end{equation*}
</div>
<div class="displaymath">
\begin{equation*}
\operatorname{ker}\ T = \left\{ \left[\begin{array}{c}
-a \\
3 \, a \\
a
\end{array}\right] \middle|\,a\in\mathbb{R}\right\}
\end{equation*}
</div>
</li>
<li id="li-410">A basis of \(\operatorname{Im}\ T\) is \(\left\{ \left[\begin{array}{c}
3 \\
-3 \\
2 \\
1
\end{array}\right] , \left[\begin{array}{c}
2 \\
-2 \\
1 \\
-1
\end{array}\right] \right\} \text{.}\) A basis of \(\operatorname{ker}\ T\) is \(\left\{ \left[\begin{array}{c}
-1 \\
3 \\
1
\end{array}\right] \right\} \)</li>
<li id="li-411">The rank of \(T\) is \(2 \text{,}\) the nullity of \(T\) is \(1 \text{,}\) and the dimension of the domain of \(T\) is \(3 \text{.}\) The rank-nullity theorem asserts that \(2 + 1 = 3 \text{,}\) which we see to be true.</li>
</ol>
</div></div>
</div></article><article class="exercise exercise-like" id="exercise-73"><h6 class="heading"><span class="codenumber">3<span class="period">.</span></span></h6>Let \(T:\mathbb{R}^ 3 \to \mathbb{R}^ 4 \) be the linear transformation given by<div class="displaymath">
\begin{equation*}
T\left( \left[\begin{array}{c}
x_{1} \\
x_{2} \\
x_{3}
\end{array}\right] \right) = \left[\begin{array}{c}
x_{1} - 2 \, x_{2} \\
x_{2} \\
x_{1} + 2 \, x_{2} \\
2 \, x_{1} - 7 \, x_{2}
\end{array}\right] .
\end{equation*}
</div>
<ol class="lower-alpha">
<li id="li-412">Explain how to find the image of \(T\) and the kernel of \(T\text{.}\)</li>
<li id="li-413">Explain how to find a basis of the image of \(T\) and a basis of the kernel of \(T\text{.}\)</li>
<li id="li-414">Explain how to find the rank and nullity of \(T\text{,}\) and why the rank-nullity theorem holds for \(T\text{.}\)</li>
</ol>
<div class="solutions">
<a data-knowl="" class="id-ref answer-knowl original" data-refid="hk-answer-73" id="answer-73"><span class="type">Answer</span></a><div class="hidden-content tex2jax_ignore" id="hk-answer-73"><div class="answer solution-like">
<div class="displaymath" id="p-589">
\begin{equation*}
\operatorname{RREF} \left[\begin{array}{ccc}
1 & -2 & 0 \\
0 & 1 & 0 \\
1 & 2 & 0 \\
2 & -7 & 0
\end{array}\right] = \left[\begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right]
\end{equation*}
</div>
<ol class="lower-alpha">
<li id="li-415">
<div class="displaymath">
\begin{equation*}
\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c}
1 \\
0 \\
1 \\
2
\end{array}\right] , \left[\begin{array}{c}
-2 \\
1 \\
2 \\
-7
\end{array}\right] \right\}
\end{equation*}
</div>
<div class="displaymath">
\begin{equation*}
\operatorname{ker}\ T = \left\{ \left[\begin{array}{c}
0 \\
0 \\
a
\end{array}\right] \middle|\,a\in\mathbb{R}\right\}
\end{equation*}
</div>
</li>
<li id="li-416">A basis of \(\operatorname{Im}\ T\) is \(\left\{ \left[\begin{array}{c}
1 \\
0 \\
1 \\
2
\end{array}\right] , \left[\begin{array}{c}
-2 \\
1 \\
2 \\
-7
\end{array}\right] \right\} \text{.}\) A basis of \(\operatorname{ker}\ T\) is \(\left\{ \left[\begin{array}{c}
0 \\
0 \\
1
\end{array}\right] \right\} \)</li>
<li id="li-417">The rank of \(T\) is \(2 \text{,}\) the nullity of \(T\) is \(1 \text{,}\) and the dimension of the domain of \(T\) is \(3 \text{.}\) The rank-nullity theorem asserts that \(2 + 1 = 3 \text{,}\) which we see to be true.</li>
</ol>
</div></div>
</div></article><article class="exercise exercise-like" id="exercise-74"><h6 class="heading"><span class="codenumber">4<span class="period">.</span></span></h6>Let \(T:\mathbb{R}^ 3 \to \mathbb{R}^ 4 \) be the linear transformation given by<div class="displaymath">
\begin{equation*}
T\left( \left[\begin{array}{c}
x \\
y \\
z
\end{array}\right] \right) = \left[\begin{array}{c}
x - 2 \, y + 5 \, z \\
-x + 3 \, y - 7 \, z \\
-3 \, x + 2 \, y - 7 \, z \\
4 \, y - 8 \, z
\end{array}\right] .
\end{equation*}
</div>
<ol class="lower-alpha">
<li id="li-418">Explain how to find the image of \(T\) and the kernel of \(T\text{.}\)</li>
<li id="li-419">Explain how to find a basis of the image of \(T\) and a basis of the kernel of \(T\text{.}\)</li>
<li id="li-420">Explain how to find the rank and nullity of \(T\text{,}\) and why the rank-nullity theorem holds for \(T\text{.}\)</li>
</ol>
<div class="solutions">
<a data-knowl="" class="id-ref answer-knowl original" data-refid="hk-answer-74" id="answer-74"><span class="type">Answer</span></a><div class="hidden-content tex2jax_ignore" id="hk-answer-74"><div class="answer solution-like">
<div class="displaymath" id="p-590">
\begin{equation*}
\operatorname{RREF} \left[\begin{array}{ccc}
1 & -2 & 5 \\
-1 & 3 & -7 \\
-3 & 2 & -7 \\
0 & 4 & -8
\end{array}\right] = \left[\begin{array}{ccc}
1 & 0 & 1 \\
0 & 1 & -2 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right]
\end{equation*}
</div>
<ol class="lower-alpha">
<li id="li-421">
<div class="displaymath">
\begin{equation*}
\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c}
1 \\
-1 \\
-3 \\
0
\end{array}\right] , \left[\begin{array}{c}
-2 \\
3 \\
2 \\
4
\end{array}\right] \right\}
\end{equation*}
</div>
<div class="displaymath">
\begin{equation*}
\operatorname{ker}\ T = \left\{ \left[\begin{array}{c}
-a \\
2 \, a \\
a
\end{array}\right] \middle|\,a\in\mathbb{R}\right\}
\end{equation*}
</div>
</li>
<li id="li-422">A basis of \(\operatorname{Im}\ T\) is \(\left\{ \left[\begin{array}{c}
1 \\
-1 \\
-3 \\
0
\end{array}\right] , \left[\begin{array}{c}
-2 \\
3 \\
2 \\
4
\end{array}\right] \right\} \text{.}\) A basis of \(\operatorname{ker}\ T\) is \(\left\{ \left[\begin{array}{c}
-1 \\
2 \\
1
\end{array}\right] \right\} \)</li>
<li id="li-423">The rank of \(T\) is \(2 \text{,}\) the nullity of \(T\) is \(1 \text{,}\) and the dimension of the domain of \(T\) is \(3 \text{.}\) The rank-nullity theorem asserts that \(2 + 1 = 3 \text{,}\) which we see to be true.</li>
</ol>
</div></div>
</div></article><article class="exercise exercise-like" id="exercise-75"><h6 class="heading"><span class="codenumber">5<span class="period">.</span></span></h6>Let \(T:\mathbb{R}^ 3 \to \mathbb{R}^ 4 \) be the linear transformation given by<div class="displaymath">
\begin{equation*}
T\left( \left[\begin{array}{c}
x \\
y \\
z
\end{array}\right] \right) = \left[\begin{array}{c}
x - z \\
-2 \, x + y + z \\
-5 \, x + 3 \, y + 2 \, z \\
-2 \, x - 2 \, y + 4 \, z
\end{array}\right] .
\end{equation*}
</div>
<ol class="lower-alpha">
<li id="li-424">Explain how to find the image of \(T\) and the kernel of \(T\text{.}\)</li>
<li id="li-425">Explain how to find a basis of the image of \(T\) and a basis of the kernel of \(T\text{.}\)</li>
<li id="li-426">Explain how to find the rank and nullity of \(T\text{,}\) and why the rank-nullity theorem holds for \(T\text{.}\)</li>
</ol>
<div class="solutions">
<a data-knowl="" class="id-ref answer-knowl original" data-refid="hk-answer-75" id="answer-75"><span class="type">Answer</span></a><div class="hidden-content tex2jax_ignore" id="hk-answer-75"><div class="answer solution-like">
<div class="displaymath" id="p-591">
\begin{equation*}
\operatorname{RREF} \left[\begin{array}{ccc}
1 & 0 & -1 \\
-2 & 1 & 1 \\
-5 & 3 & 2 \\
-2 & -2 & 4
\end{array}\right] = \left[\begin{array}{ccc}
1 & 0 & -1 \\
0 & 1 & -1 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right]
\end{equation*}
</div>
<ol class="lower-alpha">
<li id="li-427">
<div class="displaymath">
\begin{equation*}
\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c}
1 \\
-2 \\
-5 \\
-2
\end{array}\right] , \left[\begin{array}{c}
0 \\
1 \\
3 \\
-2
\end{array}\right] \right\}
\end{equation*}
</div>
<div class="displaymath">
\begin{equation*}
\operatorname{ker}\ T = \left\{ \left[\begin{array}{c}
a \\
a \\
a
\end{array}\right] \middle|\,a\in\mathbb{R}\right\}
\end{equation*}
</div>
</li>
<li id="li-428">A basis of \(\operatorname{Im}\ T\) is \(\left\{ \left[\begin{array}{c}
1 \\
-2 \\
-5 \\
-2
\end{array}\right] , \left[\begin{array}{c}
0 \\
1 \\
3 \\
-2
\end{array}\right] \right\} \text{.}\) A basis of \(\operatorname{ker}\ T\) is \(\left\{ \left[\begin{array}{c}
1 \\
1 \\
1
\end{array}\right] \right\} \)</li>
<li id="li-429">The rank of \(T\) is \(2 \text{,}\) the nullity of \(T\) is \(1 \text{,}\) and the dimension of the domain of \(T\) is \(3 \text{.}\) The rank-nullity theorem asserts that \(2 + 1 = 3 \text{,}\) which we see to be true.</li>
</ol>
</div></div>
</div></article><p id="p-592"><em class="emphasis">Additional exercises available at <a class="external" href="https://checkit.clontz.org/banks/tbil-la/A3/" target="_blank">checkit.clontz.org</a>.</em></p></section></section></div></main>
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