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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="A1"><h2 class="heading hide-type">
<span class="type">Section</span> <span class="codenumber">3.1</span> <span class="title">Linear Transformations (A1)</span>
</h2>
<article class="definition definition-like" id="definition-17"><h6 class="heading">
<span class="type">Definition</span><span class="space"> </span><span class="codenumber">3.1.1</span><span class="period">.</span>
</h6>
<p id="p-451">A <dfn class="terminology">linear transformation</dfn> (also known as a <dfn class="terminology">linear map</dfn>) is a map between vector spaces that preserves the vector space operations. More precisely, if \(V\) and \(W\) are vector spaces, a map \(T:V\rightarrow W\) is called a linear transformation if</p>
<ol class="decimal">
<li id="li-325"><p id="p-452">\(T(\vec{v}+\vec{w}) = T(\vec{v})+T(\vec{w})\) for any \(\vec{v},\vec{w} \in V\text{.}\)</p></li>
<li id="li-326"><p id="p-453">\(T(c\vec{v}) = cT(\vec{v})\) for any \(c \in \IR,\vec{v} \in V\text{.}\)</p></li>
</ol>
<p data-braille="continuation">In other words, a map is linear when vector space operations can be applied before or after the transformation without affecting the result.</p></article><article class="definition definition-like" id="definition-18"><h6 class="heading">
<span class="type">Definition</span><span class="space"> </span><span class="codenumber">3.1.2</span><span class="period">.</span>
</h6>
<p id="p-454">Given a linear transformation \(T:V\to W\text{,}\) \(V\) is called the <dfn class="terminology">domain</dfn> of \(T\) and \(W\) is called the <dfn class="terminology">co-domain</dfn> of \(T\text{.}\)</p>
<pre class="code-block tex2jax_ignore">
\begin{tikzpicture}[x=0.2in,y=0.2in]
\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)
node[anchor=south west] {\(\vec v\)};
\node[anchor=west] at (-1,-1) {domain \(\IR^3\)};
\end{scope}
\draw[dashed,-latex] (3,3) to [bend left=30] (7,3);
\node[anchor=south] at (5,4) {Linear transformation \(T:\IR^3\to\IR^2\)};
\begin{scope}[shift={(9,0.5)}]
\draw (-2,0) -- (2,0);
\draw (0,-2) -- (0,2);
\draw[thick,-latex,red] (0,0) -- (-1.5,1)
node[anchor=south east] {\(T(\vec v)\)};
\node[anchor=west] at (0,-1) {co-domain \(\IR^2\)};
\end{scope}
\end{tikzpicture}
</pre></article><article class="example example-like" id="example-2"><a data-knowl="" class="id-ref example-knowl original" data-refid="hk-example-2"><h6 class="heading">
<span class="type">Example</span><span class="space"> </span><span class="codenumber">3.1.3</span><span class="period">.</span>
</h6></a></article><div class="hidden-content tex2jax_ignore" id="hk-example-2"><article class="example example-like"><p id="p-455">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-z \\ 3y \end{array}\right]
\end{equation*}
</div>
<p id="p-456">To show that \(T\) is linear, we must verify...</p>
<div class="displaymath">
\begin{equation*}
T\left(
\left[\begin{array}{c} x \\ y \\ z \end{array}\right] +
\left[\begin{array}{c} u \\ v \\ w \end{array}\right]
\right)
=
T\left(
\left[\begin{array}{c} x+u \\ y+v \\ z+w \end{array}\right]
\right) =
\left[\begin{array}{c} (x+u)-(z+w) \\ 3(y+v) \end{array}\right]
\end{equation*}
</div>
<div class="displaymath">
\begin{equation*}
T\left(
\left[\begin{array}{c} x \\ y \\ z \end{array}\right]
\right) + T\left(
\left[\begin{array}{c} u \\ v \\ w \end{array}\right]
\right)
=
\left[\begin{array}{c} x-z \\ 3y \end{array}\right] +
\left[\begin{array}{c} u-w \\ 3v \end{array}\right]=
\left[\begin{array}{c} (x+u)-(z+w) \\ 3(y+v) \end{array}\right]
\end{equation*}
</div>
<p data-braille="continuation">And also...</p>
<div class="displaymath">
\begin{equation*}
T\left(c\left[\begin{array}{c} x \\ y \\ z \end{array}\right] \right)
=
T\left(\left[\begin{array}{c} cx \\ cy \\ cz \end{array}\right] \right)
=
\left[\begin{array}{c} cx-cz \\ 3cy \end{array}\right]
\text{ and }
cT\left(\left[\begin{array}{c} x \\ y \\ z \end{array}\right] \right)
=
c\left[\begin{array}{c} x-z \\ 3y \end{array}\right]
=
\left[\begin{array}{c} cx-cz \\ 3cy \end{array}\right]
\end{equation*}
</div>
<p id="p-457">Therefore \(T\) is a linear transformation.</p></article></div>
<article class="example example-like" id="example-3"><a data-knowl="" class="id-ref example-knowl original" data-refid="hk-example-3"><h6 class="heading">
<span class="type">Example</span><span class="space"> </span><span class="codenumber">3.1.4</span><span class="period">.</span>
</h6></a></article><div class="hidden-content tex2jax_ignore" id="hk-example-3"><article class="example example-like"><p id="p-458">Let \(T : \IR^2 \rightarrow \IR^4\) 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 \\ x^2 \\ y+3 \\ y-2^x \end{array}\right]
\end{equation*}
</div>
<p id="p-459">To show that \(T\) is not linear, we only need to find one counterexample.</p>
<div class="displaymath">
\begin{equation*}
T\left(
\left[\begin{array}{c} 0 \\ 1 \end{array}\right] +
\left[\begin{array}{c} 2 \\ 3 \end{array}\right]
\right)
=
T\left(
\left[\begin{array}{c} 2 \\ 4 \end{array}\right]
\right) =
\left[\begin{array}{c} 6 \\ 4 \\ 7 \\ 0 \end{array}\right]
\end{equation*}
</div>
<div class="displaymath">
\begin{equation*}
T\left(
\left[\begin{array}{c} 0 \\ 1 \end{array}\right]
\right) + T\left(
\left[\begin{array}{c} 2 \\ 3\end{array}\right]
\right)
=
\left[\begin{array}{c} 1 \\ 0 \\ 4 \\ 0 \end{array}\right] +
\left[\begin{array}{c} 5 \\ 4 \\ 6 \\ -1 \end{array}\right]
=
\left[\begin{array}{c} 6 \\ 4 \\ 10 \\ -1 \end{array}\right]
\end{equation*}
</div>
<p id="p-460">Since the resulting vectors are different, \(T\) is not a linear transformation.</p></article></div>
<article class="fact theorem-like" id="fact-13"><h6 class="heading">
<span class="type">Fact</span><span class="space"> </span><span class="codenumber">3.1.5</span><span class="period">.</span>
</h6>
<p id="p-461">A map between Euclidean spaces \(T:\IR^n\to\IR^m\) is linear exactly when every component of the output is a linear combination of the variables of \(\IR^n\text{.}\)</p>
<p id="p-462">For example, the following map is definitely linear because \(x-z\) and \(3y\) are linear combinations of \(x,y,z\text{:}\)</p>
<div class="displaymath">
\begin{equation*}
T\left(\left[\begin{array}{c} x \\ y \\ z \end{array}\right] \right)
=
\left[\begin{array}{c} x-z \\ 3y \end{array}\right]
=
\left[\begin{array}{c} 1x+0y-1z \\ 0x+3y+0z \end{array}\right]
\end{equation*}
</div>
<p id="p-463">But this map is not linear because \(x^2\text{,}\) \(y+3\text{,}\) and \(y-2^x\) are not linear combinations (even though \(x+y\) is):</p>
<div class="displaymath">
\begin{equation*}
T\left(\left[\begin{array}{c} x \\ y \end{array}\right] \right)
=
\left[\begin{array}{c} x+y \\ x^2 \\ y+3 \\ y-2^x \end{array}\right]
\end{equation*}
</div></article><article class="activity project-like" id="activity-67"><h6 class="heading">
<span class="type">Activity</span><span class="space"> </span><span class="codenumber">3.1.1</span><span class="period">.</span>
</h6>
<p id="p-464">Recall the following rules from calculus, where \(D:\P\to\P\) is the derivative map defined by \(D(f(x))=f'(x)\) for each polynomial \(f\text{.}\)</p>
<div class="displaymath">
\begin{equation*}
D(f+g)=f'(x)+g'(x)
\end{equation*}
</div>
<div class="displaymath">
\begin{equation*}
D(cf(x))=cf'(x)
\end{equation*}
</div>
<p id="p-465">What can we conclude from these rules?</p>
<ol class="lower-alpha">
<li id="li-327"><p id="p-466">\(\P\) is not a vector space</p></li>
<li id="li-328"><p id="p-467">\(D\) is a linear map</p></li>
<li id="li-329"><p id="p-468">\(D\) is not a linear map</p></li>
</ol></article><article class="activity project-like" id="activity-68"><h6 class="heading">
<span class="type">Activity</span><span class="space"> </span><span class="codenumber">3.1.2</span><span class="period">.</span>
</h6>
<p id="p-469">Let the polynomial maps \(S: \P^4 \rightarrow \P^3\) and \(T: \P^4 \rightarrow \P^3\) be defined by</p>
<div class="displaymath">
\begin{equation*}
S(f(x)) = 2f'(x)-f''(x) \hspace{3em} T(f(x)) = f'(x)+x^3\text{.}
\end{equation*}
</div>
<p id="p-470">Compute \(S(x^4+x)\text{,}\) \(S(x^4)+S(x)\text{,}\) \(T(x^4+x)\text{,}\) and \(T(x^4)+T(x)\text{.}\) Which of these maps is definitely not linear?</p></article><article class="fact theorem-like" id="fact-14"><h6 class="heading">
<span class="type">Fact</span><span class="space"> </span><span class="codenumber">3.1.6</span><span class="period">.</span>
</h6>
<p id="p-471">If \(L:V\to W\) is linear, then \(L(\vec z)=L(0\vec v)=0L(\vec v)=\vec z\) where \(\vec z\) is the additive identity of the vector spaces \(V,W\text{.}\)</p>
<p id="p-472">Put another way, an easy way to prove that a map like \(T(f(x)) = f'(x)+x^3\) can't be linear is because</p>
<div class="displaymath">
\begin{equation*}
T(0)=\frac{d}{dx}[0]+x^3=0+x^3=x^3\not=0.
\end{equation*}
</div></article><article class="observation remark-like" id="observation-16"><h6 class="heading">
<span class="type">Observation</span><span class="space"> </span><span class="codenumber">3.1.7</span><span class="period">.</span>
</h6>
<p id="p-473">Showing \(L:V\to W\) is not a linear transformation can be done by finding an example for any one of the following.</p>
<ul class="disc">
<li id="li-330"><p id="p-474">Show \(L(\vec z)\not=\vec z\) (where \(\vec z\) is the additive identity of \(L\) and \(W\)).</p></li>
<li id="li-331"><p id="p-475">Find \(\vec v,\vec w\in V\) such that \(L(\vec v+\vec w)\not=L(\vec v)+L(\vec w)\text{.}\)</p></li>
<li id="li-332"><p id="p-476">Find \(\vec v\in V\) and \(c\in \IR\) such that \(L(c\vec v)\not=cL(\vec v)\text{.}\)</p></li>
</ul>
<p id="p-477">Otherwise, \(L\) can be shown to be linear by proving the following in general.</p>
<ul class="disc">
<li id="li-333"><p id="p-478">For all \(\vec v,\vec w\in V\text{,}\) \(L(\vec v+\vec w)=L(\vec v)+L(\vec w)\text{.}\)</p></li>
<li id="li-334"><p id="p-479">For all \(\vec v\in V\) and \(c\in \IR\text{,}\) \(L(c\vec v)=cL(\vec v)\text{.}\)</p></li>
</ul>
<p id="p-480">Note the similarities between this process and showing that a subset of a vector space is/isn't a subspace.</p></article><article class="activity project-like" id="activity-69"><h6 class="heading">
<span class="type">Activity</span><span class="space"> </span><span class="codenumber">3.1.3</span><span class="period">.</span>
</h6>
<div class="introduction" id="introduction-24">
<p id="p-481">Continue to consider \(S: \P^4 \rightarrow \P^3\) defined by</p>
<div class="displaymath">
\begin{equation*}
S(f(x)) = 2f'(x)-f''(x)
\end{equation*}
</div>
</div>
<article class="task exercise-like" id="task-56"><h6 class="heading"><span class="codenumber">(a)</span></h6>
<p id="p-482">Verify that</p>
<div class="displaymath">
\begin{equation*}
S(f(x)+g(x))=2f'(x)+2g'(x)-f''(x)-g''(x)
\end{equation*}
</div>
<p data-braille="continuation">is equal to \(S(f(x))+S(g(x))\) for all polynomials \(f,g\text{.}\)</p></article><article class="task exercise-like" id="task-57"><h6 class="heading"><span class="codenumber">(b)</span></h6>
<p id="p-483">Verify that \(S(cf(x))\) is equal to \(cS(f(x))\) for all real numbers \(c\) and polynomials \(f\text{.}\)</p></article><article class="task exercise-like" id="task-58"><h6 class="heading"><span class="codenumber">(c)</span></h6>
<p id="p-484">Is \(S\) linear?</p></article></article><article class="activity project-like" id="activity-70"><h6 class="heading">
<span class="type">Activity</span><span class="space"> </span><span class="codenumber">3.1.4</span><span class="period">.</span>
</h6>
<div class="introduction" id="introduction-25">
<p id="p-485">Let the polynomial maps \(S: \P \rightarrow \P\) and \(T: \P \rightarrow \P\) be defined by</p>
<div class="displaymath">
\begin{equation*}
S(f(x)) = (f(x))^2 \hspace{3em} T(f(x)) = 3xf(x^2)
\end{equation*}
</div>
</div>
<article class="task exercise-like" id="task-59"><h6 class="heading"><span class="codenumber">(a)</span></h6>
<p id="p-486">Note that \(S(0)=0\) and \(T(0)=0\text{.}\) So instead, show that \(S(x+1)\not= S(x)+S(1)\) to verify that \(S\) is not linear.</p></article><article class="task exercise-like" id="task-60"><h6 class="heading"><span class="codenumber">(b)</span></h6>
<p id="p-487">Prove that \(T\) is linear by verifying that \(T(f(x)+g(x))=T(f(x))+T(g(x))\) and \(T(cf(x))=cT(f(x))\text{.}\)</p></article></article><section class="exercises" id="exercises-13"><h3 class="heading hide-type">
<span class="type">Exercises</span> <span class="codenumber">3.1.1</span> <span class="title">Exercises</span>
</h3>
<article class="exercise exercise-like" id="exercise-61"><h6 class="heading"><span class="codenumber">1<span class="period">.</span></span></h6>
<p id="p-488">Consider the following maps of polynomials \(S:\mathcal{P}\rightarrow\mathcal{P}\) and \(T:\mathcal{P}\rightarrow\mathcal{P}\) defined by</p>
<div class="displaymath">
\begin{align*}
S(f(x))= -4 \, x + 4 \, f\left(5\right) & \text{and} & T(f)= 3 \, f\left(x\right) - 4 \, f'\left(-5\right) .
\end{align*}
</div>
<p data-braille="continuation">Explain why one these maps is a linear transformation and why the other map is not.</p>
<div class="solutions">
<a data-knowl="" class="id-ref answer-knowl original" data-refid="hk-answer-61" id="answer-61"><span class="type">Answer</span></a><div class="hidden-content tex2jax_ignore" id="hk-answer-61"><div class="answer solution-like"><p id="p-489">\(S\) is not linear and \(T\) is linear.</p></div></div>
</div></article><article class="exercise exercise-like" id="exercise-62"><h6 class="heading"><span class="codenumber">2<span class="period">.</span></span></h6>
<p id="p-490">Consider the following maps of polynomials \(S:\mathcal{P}\rightarrow\mathcal{P}\) and \(T:\mathcal{P}\rightarrow\mathcal{P}\) defined by</p>
<div class="displaymath">
\begin{align*}
S(g(x))= -3 \, g\left(1\right) + 5 \, g\left(x^{2}\right) & \text{and} & T(g)= -3 \, g\left(x\right)^{3} - 2 \, g'\left(x\right) .
\end{align*}
</div>
<p data-braille="continuation">Explain why one these maps is a linear transformation and why the other map is not.</p>
<div class="solutions">
<a data-knowl="" class="id-ref answer-knowl original" data-refid="hk-answer-62" id="answer-62"><span class="type">Answer</span></a><div class="hidden-content tex2jax_ignore" id="hk-answer-62"><div class="answer solution-like"><p id="p-491">\(S\) is linear and \(T\) is not linear.</p></div></div>
</div></article><article class="exercise exercise-like" id="exercise-63"><h6 class="heading"><span class="codenumber">3<span class="period">.</span></span></h6>
<p id="p-492">Consider the following maps of polynomials \(S:\mathcal{P}\rightarrow\mathcal{P}\) and \(T:\mathcal{P}\rightarrow\mathcal{P}\) defined by</p>
<div class="displaymath">
\begin{align*}
S(h(x))= -x^{3} h\left(x\right) - 3 \, h\left(x\right) & \text{and} & T(h)= x^{3} - h\left(-4\right) .
\end{align*}
</div>
<p data-braille="continuation">Explain why one these maps is a linear transformation and why the other map is not.</p>
<div class="solutions">
<a data-knowl="" class="id-ref answer-knowl original" data-refid="hk-answer-63" id="answer-63"><span class="type">Answer</span></a><div class="hidden-content tex2jax_ignore" id="hk-answer-63"><div class="answer solution-like"><p id="p-493">\(S\) is linear and \(T\) is not linear.</p></div></div>
</div></article><article class="exercise exercise-like" id="exercise-64"><h6 class="heading"><span class="codenumber">4<span class="period">.</span></span></h6>
<p id="p-494">Consider the following maps of polynomials \(S:\mathcal{P}\rightarrow\mathcal{P}\) and \(T:\mathcal{P}\rightarrow\mathcal{P}\) defined by</p>
<div class="displaymath">
\begin{align*}
S(g(x))= x^{3} g\left(x\right) + 2 \, g\left(x\right)^{3} & \text{and} & T(g)= -3 \, g\left(x^{3}\right) + 2 \, g'\left(-2\right) .
\end{align*}
</div>
<p data-braille="continuation">Explain why one these maps is a linear transformation and why the other map is not.</p>
<div class="solutions">
<a data-knowl="" class="id-ref answer-knowl original" data-refid="hk-answer-64" id="answer-64"><span class="type">Answer</span></a><div class="hidden-content tex2jax_ignore" id="hk-answer-64"><div class="answer solution-like"><p id="p-495">\(S\) is not linear and \(T\) is linear.</p></div></div>
</div></article><article class="exercise exercise-like" id="exercise-65"><h6 class="heading"><span class="codenumber">5<span class="period">.</span></span></h6>
<p id="p-496">Consider the following maps of polynomials \(S:\mathcal{P}\rightarrow\mathcal{P}\) and \(T:\mathcal{P}\rightarrow\mathcal{P}\) defined by</p>
<div class="displaymath">
\begin{align*}
S(h(x))= 2 \, h\left(-5\right) + 4 \, h'\left(-1\right) & \text{and} & T(h)= 3 \, x - 3 \, h'\left(x\right) .
\end{align*}
</div>
<p data-braille="continuation">Explain why one these maps is a linear transformation and why the other map is not.</p>
<div class="solutions">
<a data-knowl="" class="id-ref answer-knowl original" data-refid="hk-answer-65" id="answer-65"><span class="type">Answer</span></a><div class="hidden-content tex2jax_ignore" id="hk-answer-65"><div class="answer solution-like"><p id="p-497">\(S\) is linear and \(T\) is not linear.</p></div></div>
</div></article><p id="p-498"><em class="emphasis">Additional exercises available at <a class="external" href="https://checkit.clontz.org/banks/tbil-la/A1/" target="_blank">checkit.clontz.org</a>.</em></p></section></section></div></main>
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