matrix-trans.xml

<?xml version="1.0" encoding="UTF-8"?>

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<section xml:id="matrix-transformations">
  <title>Matrix Transformations</title>

  <objectives>
    <ol>
      <li>Learn to view a matrix geometrically as a function.</li>
      <li>Learn examples of matrix transformations: reflection, dilation, rotation, shear, projection.</li>
      <li>Understand the vocabulary surrounding transformations: domain, codomain, range.</li>
      <li>Understand the domain, codomain, and range of a matrix transformation.</li>
      <li><em>Pictures:</em> common matrix transformations.</li>
      <li><em>Vocabulary words:</em> <term>transformation</term> / <term>function</term>, <term>domain</term>, <term>codomain</term>, <term>range</term>, <term>identity transformation</term>, <term>matrix transformation</term>.</li>
    </ol>
  </objectives>

  <introduction>
    <p>
      In this section we learn to understand matrices geometrically as <em>functions</em>, or <em>transformations</em>.  We briefly discuss transformations in general, then specialize to matrix transformations, which are transformations that come from matrices.
    </p>
  </introduction>

  <subsection xml:id="matrix-trans-matrices-functions">
    <title>Matrices as Functions</title>

    <p>
      Informally, a <em>function</em> is a rule that accepts inputs and produces outputs.  For instance, <m>f(x) = x^2</m> is a function that accepts one number <m>x</m> as its input, and outputs the square of that number: <m>f(2) = 4</m>.  In this subsection, we interpret matrices as functions.
    </p>

  <p>
    <idx><h>Matrix</h><h>as a function</h></idx>
    <idx><h>Matrix</h><h>as a function</h><see>Matrix transformation</see></idx>
    Let <m>A</m> be a matrix with <m>m</m> rows and <m>n</m> columns.  Consider the matrix equation <m>b=Ax</m> (we write it this way instead of <m>Ax=b</m> to remind the reader of the notation <m>y=f(x)</m>).  If we vary <m>x</m>, then <m>b</m> will also vary; in this way, we think of <m>A</m> as a function with independent variable <m>x</m> and dependent variable <m>b</m>.
    <ul>
      <li>
        The independent variable (the input) is <m>x</m>, which is a vector in <m>\R^n</m>.
      </li>
      <li>
        The dependent variable (the output) is <m>b</m>, which is a vector in <m>\R^m</m>.
      </li>
    </ul>
    The set of all possible output vectors are the vectors <m>b</m> such that <m>Ax=b</m> has some solution; this is the same as the column space of <m>A</m> by this <xref ref="matrixeq-spans-consistency"/>.
    <latex-code>
\begin{tikzpicture}[thin border nodes]
  \path[use as bounding box] (-3,-5.5) -- (15,3);
  \draw[grid lines, light gray] (-3,-3) grid (3,3);

  \node (A) at (0,-3.5) {$\R^n$};
  \node (B) at (8,-3.5) {$\R^m$};
  \draw[->] (A.east) +(5mm,0) -- node[midway,above=1mm] {$b=Ax$}
    ($(B.west)-(5mm,0)$);

  \point["$x$" left] (P) at (1,1,5);

  \begin{scope}[myxyz, xshift=8cm]
    \path[clip, resetxy] (-3,-3) rectangle (3,3);

    \def\v{(-1,2,1)}
    \def\w{(0,1,-1)}

    \node[coordinate] (X) at \v {};
    \node[coordinate] (Y) at \w {};

    \draw[very thin] (-2,0,0) -- (0,0,0);
    \draw[very thin] (0,-2,0) -- (0,0,0);
    \draw[very thin] (0,0,-2) -- (0,0,0);

    \begin{scope}[x=(X), y=(Y), transformxy]
      \fill[seq4!10, nearly opaque] (-1,-1) rectangle (1,1);
      \draw[step=.5cm, seq4!30, very thin] (-1,-1) grid (1,1);
      \point["$Ax$" {fill=none,below}] (Q) at (-.5,.5);
      \node[coordinate,
          pin={[pin edge={very thin,-},pin distance=3mm,anchor=north]-70:$\Col(A)$}]
        at (0.1,1) {};
    \end{scope}

    \draw[->, very thin] (0,0,0) -- (2,0,0);
    \draw[->, very thin] (0,0,0) -- (0,2,0);
    \draw[->, very thin] (0,0,0) -- (0,0,2);

    \draw[resetxy] (-3,-3) rectangle (3,3);

  \end{scope}

  \draw[|->, shorten >=.35mm, shorten &lt;=.35mm]
    (P.east) .. controls +(0:3cm) and +(190:3cm) .. (Q.west);

\end{tikzpicture}
    </latex-code>
  </p>

  <example hide-type="true">
    <title>Interactive: A <m>2\times 3</m> matrix</title>
    <figure>
      <caption>A picture of a <m>2\times 3</m> matrix, regarded as a function.  The input vector is <m>x</m>, which is a vector in <m>\R^3</m>, and the output vector is <m>b=Ax</m>, which is a vector in <m>\R^2</m>.  The violet line on the right is the column space; as you vary <m>x</m>, the output <m>b</m> is constrained to lie on this line.</caption>
      <mathbox source="demos/Axequalsb.html?show=false" height="500px"/>
    </figure>
  </example>

  <example hide-type="true">
    <title>Interactive: A <m>3\times 2</m> matrix</title>
    <figure>
      <caption>A picture of a <m>3\times 2</m> matrix, regarded as a function.  The input vector is <m>x</m>, which is a vector in <m>\R^2</m>, and the output vector is <m>b=Ax</m>, which is a vector in <m>\R^3</m>.  The violet plane on the right is the column space; as you vary <m>x</m>, the output <m>b</m> is constrained to lie on this plane.</caption>
      <mathbox source="demos/Axequalsb.html?mat=1,0:0,1:1,0&amp;range2=5&amp;closed=true&amp;show=false" height="500px"/>
    </figure>
  </example>

  <example xml:id="matrix-trans-eg-projection">
    <title>Projection onto the <m>xy</m>-plane</title>
    <idx><h>Projection</h><see>Orthogonal projection</see></idx>
    <idx><h>Orthogonal projection</h><h>onto the <m>xy</m>-plane</h></idx>
    <statement>
      <p>Let <me>A = \mat{1 0 0; 0 1 0; 0 0 0}.</me>  Describe the function <m>b=Ax</m> geometrically.</p>
    </statement>
    <solution>
      <p>
        In the equation <m>Ax=b</m>, the input vector <m>x</m> and the output vector <m>b</m> are both in <m>\R^3</m>.
        First we multiply <m>A</m> by a vector to see what it does:
        <me>A\vec{x y z} = \mat{1 0 0; 0 1 0; 0 0 0}\vec{x y z} = \vec{x y 0}.</me>
        Multiplication by <m>A</m> simply sets the <m>z</m>-coordinate equal to zero: it <em>projects vertically onto the <m>xy</m>-plane</em>.
        <latex-code>
\begin{tikzpicture}[myxyz]
  \begin{scope}[arrows={|[width=4pt]->[width=4pt]}, shorten=4pt]
    \draw (-1,-1,-2.5) node[point, seq2] {}
      -- (-1,-1,0);
    \draw (2,-2,-1) node[point, seq5] {}
      -- (2,-2,0);
  \end{scope}

  \begin{scope}[transformxy]
    \fill[white, nearly opaque] (-3, -3) rectangle (3, 3);
    \draw[step=1cm, help lines] (-3, -3) grid (3, 3);
  \end{scope}

  \begin{scope}[arrows={|[width=4pt]->[width=4pt]}, shorten=4pt]
    \draw (2,1,3) node[point,seq1] {}
      -- (2,1,0) node[point,seq1] {};
    \point[seq2] at (-1,-1,0);
    \point[seq5] at (2,-2,0);
    \draw (-1,2,2) node[point,seq3] {}
      -- (-1,2,0) node[point,seq3] {};
    \draw (-2,0,1) node[point,seq4] {}
      -- (-2,0,0) node[point,seq4] {};
  \end{scope}

\end{tikzpicture}
        </latex-code>
      </p>
      <figure>
        <caption>Multiplication by the matrix <m>A</m> projects a vector onto the <m>xy</m>-plane.  Move the input vector <m>x</m> to see how the output vector <m>b</m> changes.</caption>
        <mathbox source="demos/Axequalsb.html?mat=1,0,0:0,1,0:0,0,0&amp;range2=5&amp;closed=true" height="500px"/>
      </figure>
    </solution>
  </example>

  <example xml:id="matrix-trans-eg-reflection">
    <title>Reflection</title>
    <idx><h>Reflection</h><h>over the <m>y</m>-axis</h></idx>
    <statement>
      <p>Let <me>A = \mat{-1 0; 0 1}.</me>  Describe the function <m>b=Ax</m> geometrically.</p>
    </statement>
    <solution>
      <p>
        In the equation <m>Ax=b</m>, the input vector <m>x</m> and the output vector <m>b</m> are both in <m>\R^2</m>.
        First we multiply <m>A</m> by a vector to see what it does:
        <me>A\vec{x y} = \mat{-1 0; 0 1}\vec{x y} = \vec{-x y}.</me>
        Multiplication by <m>A</m> negates the <m>x</m>-coordinate: it <em>reflects over the <m>y</m>-axis</em>.
        <latex-code>
\def\theo{\includegraphics[width=4cm]{theo2.jpg}}
\begin{tikzpicture}
  \node (theo1) at (0,0,0) {\theo};
  \draw[->,opacity=.3] (-1.8,0) -- (1.8,0);
  \draw[->,opacity=.3] (0,-1.8) -- (0,1.8);

  \begin{scope}[xshift=7cm]
  \node[xscale=-1] (theo2) at (0,0,0) {\theo};
  \draw[->,opacity=.3] (-1.8,0) -- (1.8,0);
  \draw[->,opacity=.3] (0,-1.8) -- (0,1.8);
  \end{scope}

  \draw[->] (theo1.20)
    to[out=20,in=160] node[midway,above] {$b=Ax$} (theo2.20);
\end{tikzpicture}
        </latex-code>
      </p>
      <figure>
        <caption>Multiplication by the matrix <m>A</m> reflects over the <m>y</m>-axis.  Move the input vector <m>x</m> to see how the output vector <m>b</m> changes.</caption>
        <mathbox source="demos/twobytwo.html?mat=-1,0,0,1&amp;closed=true" height="500px"/>
      </figure>
    </solution>
  </example>

  <example xml:id="matrix-trans-eg-dilation">
    <title>Dilation</title>
    <idx><h>Dilation</h></idx>
    <idx><h>Scale</h><see>Dilation</see></idx>
    <idx><h>Scale</h><see>Row operations, scaling</see></idx>
    <statement>
      <p>Let <me>A = \mat{1.5 0; 0 1.5}.</me>  Describe the function <m>b=Ax</m> geometrically.</p>
    </statement>
    <solution>
      <p>
        In the equation <m>Ax=b</m>, the input vector <m>x</m> and the output vector <m>b</m> are both in <m>\R^2</m>.
        First we multiply <m>A</m> by a vector to see what it does:
        <me>A\vec{x y} = \mat{1.5 0; 0 1.5}\vec{x y} = \vec{1.5x 1.5y}
        = 1.5\vec{x y}.</me>
        Multiplication by <m>A</m> is the same as scalar multiplication by <m>1.5</m>: it <em>scales</em> or <em>dilates</em> the plane by a factor of <m>1.5</m>.
        <latex-code>
\def\theo{\includegraphics[width=3.5cm]{theo3.jpg}}
\begin{tikzpicture}
  \node (theo1) at (0,0,0) {\theo};
  \draw[->,opacity=.6] (-1.3,0) -- (1.3,0);
  \draw[->,opacity=.6] (0,-1.3) -- (0,1.3);

  \begin{scope}[xshift=7cm]
  \node[scale=1.5] (theo2) at (0,0,0) {\theo};
  \draw[->,opacity=.6] (-1.3,0) -- (1.3,0);
  \draw[->,opacity=.6] (0,-1.3) -- (0,1.3);
  \end{scope}

  \draw[->] ($(theo1.east)+(0,.3)$)
    to[bend left, "$b = Ax$"] ($(theo2.west)+(0,.3)$);
\end{tikzpicture}
        </latex-code>
      </p>
      <figure>
        <caption>Multiplication by the matrix <m>A</m> dilates the plane by a factor of <m>1.5</m>.  Move the input vector <m>x</m> to see how the output vector <m>b</m> changes.</caption>
        <mathbox source="demos/twobytwo.html?mat=1.5,0,0,1.5&amp;closed=true&amp;pic=theo3.jpg" height="500px"/>
      </figure>
    </solution>
  </example>

  <example xml:id="matrix-trans-eg-identity">
    <title>Identity</title>
    <idx><h>Identity matrix</h><h>as a function</h></idx>
    <statement>
      <p>Let <me>A = \mat{1 0; 0 1}.</me>  Describe the function <m>b=Ax</m> geometrically.</p>
    </statement>
    <solution>
      <p>
        In the equation <m>Ax=b</m>, the input vector <m>x</m> and the output vector <m>b</m> are both in <m>\R^2</m>.
        First we multiply <m>A</m> by a vector to see what it does:
        <me>A\vec{x y} = \mat{1 0; 0 1}\vec{x y} = \vec{x y}.</me>
        Multiplication by <m>A</m> does not change the input vector at all: it is the <em>identity transformation which does nothing</em>.
        <latex-code>
\def\theo{\includegraphics[width=4cm]{theo11.jpg}}
\begin{tikzpicture}
  \node (theo1) at (0,0,0) {\theo};
  \draw[->,opacity=.6] (-1.3,0) -- (1.3,0);
  \draw[->,opacity=.6] (0,-1.3) -- (0,1.3);

  \begin{scope}[xshift=7cm]
  \node (theo2) at (0,0,0) {\theo};
  \draw[->,opacity=.6] (-1.3,0) -- (1.3,0);
  \draw[->,opacity=.6] (0,-1.3) -- (0,1.3);
  \end{scope}

  \draw[->] ($(theo1.east)+(0,.3)$)
    to[bend left, "$b = Ax$"] ($(theo2.west)+(0,.3)$);
\end{tikzpicture}
        </latex-code>
      </p>
      <figure>
        <caption>Multiplication by the matrix <m>A</m> does not move the vector <m>x</m>: that is, <m>b=Ax=x</m>.  Move the input vector <m>x</m> to see how the output vector <m>b</m> changes.</caption>
        <mathbox source="demos/twobytwo.html?mat=1,0,0,1&amp;closed=true&amp;pic=theo11.jpg" height="500px"/>
      </figure>
    </solution>
  </example>

  <example xml:id="matrix-trans-eg-rotation">
    <title>Rotation</title>
    <idx><h>Rotation</h><h>counterclockwise by <m>90^\circ</m></h></idx>
    <statement>
      <p>Let <me>A = \mat{0 -1; 1 0}.</me>  Describe the function <m>b=Ax</m> geometrically.</p>
    </statement>
    <solution>
      <p>
        In the equation <m>Ax=b</m>, the input vector <m>x</m> and the output vector <m>b</m> are both in <m>\R^2</m>.
        First we multiply <m>A</m> by a vector to see what it does:
        <me>A\vec{x y} = \mat{0 -1; 1 0}\vec{x y} = \vec{-y x}.</me>
        We substitute a few test points in order to understand the geometry of the transformation:
        <latex-code>
\def\arcarrow#1;{
  \pgfmathanglebetweenpoints{\pgfpoint{0cm}{0cm}}{\pgfpointanchor{#1}{center}}
  \let\sangle=\pgfmathresult
  \draw[|->, shorten=2mm]
    let \p1 = (#1.center), \n1={veclen(\x1,\y1)} in
      (#1) arc[radius=\n1, delta angle=90, start angle=\sangle];
}

  \begin{minipage}{5cm}
    \[\begin{split}
      A\color{seq1}\vec{1 2} &amp;= \color{seq1}\vec{-2 1} \\
      A\color{seq2}\vec{-1 1} &amp;= \color{seq2}\vec{-1 -1} \\
      A\color{seq3}\vec{0 -2} &amp;= \color{seq3}\vec{2 0} \\
    \end{split}\]
  \end{minipage}
\qquad
\begin{tikzpicture}[baseline=0]
  \draw[help lines] (-3,-3) grid (3,3);
  \draw[->] (-3,0) -- (3,0);
  \draw[->] (0,-3) -- (0,3);

  \point[seq1] (X1) at (1,2);
  \point[seq1] (TX1) at (-2,1);
  \arcarrow X1;

  \point[seq2] (X2) at (-1,1);
  \point[seq2] (TX2) at (-1,-1);
  \arcarrow X2;

  \point[seq3] (TX3) at (2,0);
  \point[seq3] (X3) at (0,-2);
  \arcarrow X3;
\end{tikzpicture}
        </latex-code>
        Multiplication by <m>A</m> is <em>counterclockwise rotation by <m>90^\circ</m></em>.
        <latex-code>
\def\theo{\includegraphics[width=4cm]{theo8.jpg}}
\begin{tikzpicture}
  \node (theo1) at (0,0,0) {\theo};
  \draw[->,opacity=.6] (-1.3,0) -- (1.3,0);
  \draw[->,opacity=.6] (0,-1.3) -- (0,1.3);

  \begin{scope}[xshift=7cm]
  \node[rotate=90] (theo2) at (0,0,0) {\theo};
  \draw[->,opacity=.6] (-1.3,0) -- (1.3,0);
  \draw[->,opacity=.6] (0,-1.3) -- (0,1.3);
  \end{scope}

  \draw[->] ($(theo1.east)+(0,.3)$)
    to[bend left, "$b = Ax$"] ($(theo2.north)+(0,.3)$);
\end{tikzpicture}
        </latex-code>
      </p>
      <figure>
        <caption>Multiplication by the matrix <m>A</m> rotates the vector <m>x</m> counterclockwise by <m>90^\circ</m>.  Move the input vector <m>x</m> to see how the output vector <m>b</m> changes.</caption>
        <mathbox source="demos/twobytwo.html?mat=0,-1,1,0&amp;closed=true&amp;pic=theo8.jpg" height="500px"/>
      </figure>
    </solution>
  </example>

  <example xml:id="matrix-trans-shear">
    <title>Shear</title>
    <idx><h>Shear</h><h>in the <m>x</m>-direction</h></idx>
    <statement>
      <p>Let <me>A = \mat{1 1; 0 1}.</me>  Describe the function <m>b=Ax</m> geometrically.</p>
    </statement>
    <solution>
      <p>
        In the equation <m>Ax=b</m>, the input vector <m>x</m> and the output vector <m>b</m> are both in <m>\R^2</m>.
        First we multiply <m>A</m> by a vector to see what it does:
        <me>A\vec{x y} = \mat{1 1; 0 1}\vec{x y} = \vec{x+y y}.</me>
        Multiplication by <m>A</m> adds the <m>y</m>-coordinate to the <m>x</m>-coordinate; this is called a <em>shear in the <m>x</m>-direction</em>.
        <latex-code>
\def\sheep{\includegraphics[width=4cm]{sheep.jpg}}
\begin{tikzpicture}
  \node (sheep1) at (0,0,0) {\sheep};
  \draw[->,opacity=.6] (-1.3,0) -- (1.3,0);
  \draw[->,opacity=.6] (0,-1.3) -- (0,1.3);

  \begin{scope}[xshift=7cm]
  \node[cm={1,0,1,1,(0,0)}] (sheep2) at (0,0,0) {\sheep};
  \draw[->,opacity=.6] (-1.3,0) -- (1.3,0);
  \draw[->,opacity=.6] (0,-1.3) -- (0,1.3);
  \end{scope}

  \draw[->] ($(sheep1.east)+(0,.3)$)
    to[bend left, "$b = Ax$"] ($(sheep2.west)+(0,.3)$);
\end{tikzpicture}
        </latex-code>
      </p>
      <figure>
        <caption>Multiplication by the matrix <m>A</m> adds the <m>y</m>-coordinate to the <m>x</m>-coordinate.  Move the input vector <m>x</m> to see how the output vector <m>b</m> changes.</caption>
        <mathbox source="demos/twobytwo.html?mat=1,1,0,1&amp;closed=true&amp;pic=sheep.jpg" height="500px"/>
      </figure>
    </solution>
  </example>
</subsection>

<subsection xml:id="matrix-trans-transformations">
  <title>Transformations</title>

  <p>
    At this point it is convenient to fix our ideas and terminology regarding functions, which we will call <em>transformations</em> in this book.  This allows us to systematize our discussion of matrices as functions.
  </p>

  <definition>
    <idx><h>Transformation</h><h>definition of</h></idx>
    <idx><h>Transformation</h><h>domain of</h></idx>
    <idx><h>Transformation</h><h>codomain of</h></idx>
    <idx><h>Transformation</h><h>range of</h></idx>
    <idx><h>Function</h><see>Transformation</see></idx>
    <idx><h>Domain</h><see>Transformation</see></idx>
    <idx><h>Codomain</h><see>Transformation</see></idx>
    <idx><h>Range</h><see>Transformation</see></idx>
    <notation><usage>T\colon\R^n\to\R^m</usage><description>transformation with domain <m>\R^n</m> and codomain <m>\R^m</m></description></notation>
    <statement>
      <p>
        A <term>transformation</term> from <m>\R^n</m> to <m>\R^m</m> is a rule <m>T</m> that assigns to each vector <m>x</m> in <m>\R^n</m> a vector <m>T(x)</m> in <m>\R^m</m>.
        <ul>
          <li>
            <m>\R^n</m> is called the <term>domain</term> of <m>T</m>.
          </li>
          <li>
            <m>\R^m</m> is called the <term>codomain</term> of <m>T</m>.
          </li>
          <li>
            For <m>x</m> in <m>\R^n</m>, the vector <m>T(x)</m> in <m>\R^m</m> is the <term>image</term> of <m>x</m> under <m>T</m>.
          </li>
          <li>
            The set of all images <m>\{T(x)\mid x\text{ in }\R^n\}</m> is the <term>range</term> of <m>T</m>.
          </li>
        </ul>
        The notation <m>T\colon\R^n\To\R^m</m>
        means <q><m>T</m> is a transformation from <m>\R^n</m> to <m>\R^m</m>.</q>
      </p>
    </statement>
  </definition>

  <p>
    <idx><h>Transformation</h><h>as a machine</h></idx>
    It may help to think of <m>T</m> as a <q>machine</q> that takes <m>x</m> as an input, and gives you <m>T(x)</m> as the output.
    <latex-code>
            <![CDATA[
\tikzset{
  pics/gear/.style args={#1/#2}{
    code = { % #1 = number of gears, #2 = tooth length
      \filldraw[fill=black!30] (1cm-#2/2,0)
        let \n{angle} = {360/#1} in
          \foreach \gear [evaluate=\gear as \startangle using \gear*\n{angle}]
              in {1,...,#1}
            {
              arc[radius=1cm-#2/2, start angle=\startangle-\n{angle},
                  delta angle=\n{angle}/4]
              -- (\startangle-3*\n{angle}/4+\n{angle}/10:1cm+#2/2)
              arc[radius=1cm+#2/2,
                  start angle=\startangle-3*\n{angle}/4+\n{angle}/10,
                  end angle  =\startangle-  \n{angle}/4-\n{angle}/10]
              -- (\startangle-  \n{angle}/4:1cm-#2/2)
              arc[radius=1cm-#2/2, start angle=\startangle-\n{angle}/4,
                  delta angle=\n{angle}/4]
            };
          \draw (0,0) circle[radius=.4cm];
        }
    },
  % machine is an input/output machine for illustrating functions
  machine/.pic = {
    \filldraw[rounded corners=.3mm, fill=steel!30] (1.5, -1)
      -- (1.5, -.5) -- (1.5-.2, -.5) -- (1.5-.2, .4) -- (1.5, .4)
      -- (1.5, 1) -- (-1.5, 1) -- (-1.5, -1) -- cycle;
    \filldraw[fill=steel!30] (-1.5, -.2)
      -- (-1.5-.5, -.5) -- (-1.5-.5, .5) -- (-1.5, .2);
    \coordinate (-input) at (-1.5-.5, 0);
    %\fill (-1.5-.4, -.05) rectangle (-1.5+.5, .05);
    %\fill (-1.5+.3, -.15) -- (-1.5+.7, 0) -- (-1.5+.3, .15);
    \filldraw[yshift=-.1cm, fill=black!30] (1.5-.2, -.3)
      -- (1.5+.4, -.3)
      arc[radius=.15, start angle=-90, end angle=90]
      -- (1.5-.2,0);
    \draw[yshift=-.1cm] (1.5+.4, -.15) circle[radius=.1];
    %\fill (1.5-.7, -.05) rectangle (1.5+.2, .05);
    %\fill (1.5+.0, -.15) -- (1.5+.4, 0) -- (1.5+.0, .15);
    \coordinate (-output) at (1.5+.5, 0);
    \pic[transform shape, scale=.4] at (-.4, -.4) {gear={15/.2cm}};
    \pic[transform shape, scale=.4] at ( .4, -.4) {gear={15/.2cm}};
    % Need to expand \tikzpictextoptions *first* so as not to confuse \pgfkeys
    \expandafter\node\expandafter[\tikzpictextoptions] at (0, .5) {\tikzpictext};
    },
}
\begin{tikzpicture}[thin border nodes, scale=.9]
  \draw[grid lines, light gray] (-3,-3) grid (3,3);

  \node (A) at (0,-3.5) {$\R^n$};
  \node (B) at (10,-3.5) {$\R^m$};
  \node at (0,-4) {domain};
  \node at (10,-4) {codomain};
  \draw[->] (A.east) +(5mm,0) -- node[midway,above=1mm] {$T$}
    ($(B.west)-(5mm,0)$);

  \point["$x$" left] (P) at (1,1,5);

  \begin{scope}[myxyz, xshift=10cm]
    \path[clip, resetxy] (-3,-3) rectangle (3,3);

    \def\v{(-1,2,1)}
    \def\w{(0,1,-1)}

    \node[coordinate] (X) at \v {};
    \node[coordinate] (Y) at \w {};

    \draw[very thin] (-2,0,0) -- (0,0,0);
    \draw[very thin] (0,-2,0) -- (0,0,0);
    \draw[very thin] (0,0,-2) -- (0,0,0);

    \begin{scope}[x=(X), y=(Y), transformxy]
      \fill[seq4!10, nearly opaque] (-1,-1) rectangle (1,1);
      \draw[step=.5cm, seq4!30, very thin] (-1,-1) grid (1,1);
      \point["$T(x)$" {fill=none,below}] (Q) at (-.5,.5);
      \node[coordinate,
          pin={[pin edge={very thin,-},pin distance=3mm,anchor=north]-70:range}]
        at (0.1,1) {};
    \end{scope}

    \draw[->, very thin] (0,0,0) -- (2,0,0);
    \draw[->, very thin] (0,0,0) -- (0,2,0);
    \draw[->, very thin] (0,0,0) -- (0,0,2);

    \draw[resetxy] (-3,-3) rectangle (3,3);

  \end{scope}

  \pic[scale=.75, "$T$"] (machine) at (5,0) {machine};
  \node[coordinate] (IN) at (5cm-1.7cm,0) {};
  \node[coordinate] (OUT) at (5cm+1.7cm,0) {};
  \draw[->, shorten >=.35mm, shorten <=.35mm] (P.east)
    .. controls +(0:1cm) and +(180:1cm) .. (IN);
  \draw[->, shorten >=.35mm, shorten <=.35mm] (OUT)
    .. controls +(0:1cm) and +(190:2cm) .. (Q.west);

\end{tikzpicture}
            ]]>
    </latex-code>
    The points of the domain <m>\R^n</m> are the <em>inputs</em> of <m>T</m>: this simply means that it makes sense to evaluate <m>T</m> on vectors with <m>n</m> entries, i.e., lists of <m>n</m> numbers.  Likewise, the points of the codomain <m>\R^m</m> are the <em>outputs</em> of <m>T</m>: this means that the result of evaluating <m>T</m> is always a vector with <m>m</m> entries.
</p>

<p>
  The <em>range</em> of <m>T</m> is the set of all vectors in the codomain that actually arise as outputs of the function <m>T</m>, for some input.  In other words, the range is all vectors <m>b</m> in the codomain such that <m>T(x)=b</m> has a solution <m>x</m> in the domain.
</p>

  <example>
    <title>A Function of one variable</title>
    <idx><h>Transformation</h><h>of one variable</h></idx>
    <p>
      Most of the functions you may have seen previously have domain and codomain equal to <m>\R = \R^1</m>.  For example,
      <me>\sin\colon\R\To\R \qquad \sin(x) =
  \left(\,\parbox{5cm}{
    the length of the opposite edge over the hypotenuse of a right
    triangle with angle $x$ in radians}\,\right).
      </me>
      Notice that we have defined <m>\sin</m> by a rule: a function is defined by specifying what the output of the function is for any possible input.
    </p>

    <p>
      You may be used to thinking of such functions in terms of their graphs:
      <latex-code>
\usetikzlibrary{datavisualization.formats.functions}
\begin{tikzpicture}
  \datavisualization[school book axes,
      all axes={ticks=none},
      visualize as smooth line/.list={sin},
      sin={style=seq1}]

    data[format=function, set=sin] {
      var x : interval[-2:2] ;
      func y = sin(\value x r);
    };

  \point[minimum size=1.5mm, "$x$" below] at (1,0);
  \point[minimum size=1.5mm, "${(x,\sin x)}$" {label distance=1.2mm, above}] at (1,{sin(1 r)});
\end{tikzpicture}
      </latex-code>
      In this case, the horizontal axis is the domain, and the vertical axis is the codomain.  This is useful when the domain and codomain are <m>\R</m>, but it is hard to do when, for instance, the domain is <m>\R^2</m> and the codomain is <m>\R^3</m>.  The graph of such a function is a subset of <m>\R^5</m>, which is difficult to visualize.  For this reason, we will rarely graph a transformation.
    </p>

    <p>
      Note that the <em>range</em> of <m>\sin</m> is the interval <m>[-1,1]</m>: this is the set of all possible outputs of the <m>\sin</m> function.
    </p>
  </example>

  <example>
    <title>Functions of several variables</title>
    <idx><h>Transformation</h><h>of several variables</h></idx>
    <p>
      Here is an example of a function from <m>\R^2</m> to <m>\R^3</m>:
      <me>f\vec{x y} = \vec{x+y \cos(y) y-x^2}.</me>
      The inputs of <m>f</m> each have two entries, and the outputs have three entries.  In this case, we have defined <m>f</m> by a formula, so we evaluate <m>f</m> by substituting values for the variables:
      <me>f\vec{2 3} = \vec{2+3 \cos(3) 3-2^2} = \vec{5 \cos(3) -1}.</me>
    </p>
    <p>
      Here is an example of a function from <m>\R^3</m> to <m>\R^3</m>:
      <me>f(v) = \left(\,\parbox{5.5cm}{the counterclockwise rotation of $v$ by an angle of $42^\circ$ about the $z$-axis}\,\right).</me>
      In other words, <m>f</m> takes a vector with three entries, then rotates it; hence the ouput of <m>f</m> also has three entries.  In this case, we have defined <m>f</m> by a geometric rule.
    </p>
  </example>

  <definition xml:id="matrix-trans-identity">
    <idx><h>Identity transformation</h><h>definition of</h></idx>
    <idx><h>Transformation</h><h>identity</h><see>Identity transformation</see></idx>
    <notation><usage>\Id_{\R^n}</usage><description>Identity transformation</description></notation>
    <statement>
      <p>
        The <term>identity transformation</term> <m>\Id_{\R^n}\colon\R^n\to\R^n</m> is the transformation defined by the rule
        <me>\Id_{\R^n}(x) = x \qquad\text{for all $x$ in $\R^n$}.</me>
      </p>
    </statement>
  </definition>

  <p>
    In other words, the identity transformation does not move its input vector: the output is the same as the input.  Its domain and codomain are both <m>\R^n</m>, and its range is <m>\R^n</m> as well, since every vector in <m>\R^n</m> is the output of itself.
  </p>

  <example xml:id="matrix-trans-eg-robot">
    <title>A real-word transformation: robotics</title>
    <p>
      The definition of <em>transformation</em> and its associated vocabulary may seem quite abstract, but transformations are extremely common in real life.  Here is an example from the fields of robotics and computer graphics.
    </p>
    <p>
      Suppose you are building a robot arm with three joints that can move its hand around a plane, as in the following picture.
      <latex-code>
\usetikzlibrary{ipe}
\usetikzlibrary{angles}
\tikzstyle{ipe stylesheet} = [
  ipe import,
  even odd rule,
  line join=round,
  line cap=butt,
  ipe pen normal/.style={line width=0.4},
  ipe pen heavier/.style={line width=0.8},
  ipe pen fat/.style={line width=1.2},
  ipe pen ultrafat/.style={line width=2},
  ipe pen normal,
  ipe mark normal/.style={ipe mark scale=3},
  ipe mark large/.style={ipe mark scale=5},
  ipe mark small/.style={ipe mark scale=2},
  ipe mark tiny/.style={ipe mark scale=1.1},
  ipe mark normal,
  /pgf/arrow keys/.cd,
  ipe arrow normal/.style={scale=7},
  ipe arrow large/.style={scale=10},
  ipe arrow small/.style={scale=5},
  ipe arrow tiny/.style={scale=3},
  ipe arrow normal,
  /tikz/.cd,
  ipe arrows, % update arrows
  &lt;->/.tip = ipe normal,
  ipe dash normal/.style={dash pattern=},
  ipe dash dashed/.style={dash pattern=on 4bp off 4bp},
  ipe dash dotted/.style={dash pattern=on 1bp off 3bp},
  ipe dash dash dotted/.style={dash pattern=on 4bp off 2bp on 1bp off 2bp},
  ipe dash dash dot dotted/.style={dash pattern=on 4bp off 2bp on 1bp off 2bp on 1bp off 2bp},
  ipe dash normal,
  ipe node/.append style={font=\normalsize},
  ipe stretch normal/.style={ipe node stretch=1},
  ipe stretch normal,
  ipe opacity 10/.style={opacity=0.1},
  ipe opacity 30/.style={opacity=0.3},
  ipe opacity 50/.style={opacity=0.5},
  ipe opacity 75/.style={opacity=0.75},
  ipe opacity opaque/.style={opacity=1},
  ipe opacity opaque,
]
\begin{tikzpicture}[ipe stylesheet, scale=.7]
  \draw[fill=white!80!black]
    (100, 528) -- (196, 656) -- (220, 656) -- (124, 528) -- cycle;
  \draw[fill=white!80!black]
    (208, 664) -- (320, 712) -- (320, 696) -- (208, 648) -- cycle;
  \begin{scope}[shift={(321.91, 715.847)}, rotate=-9.1623]
  \draw[fill=white!90!black]
    (0, 0)
     -- (48, 12)
     -- (96, 0)
     -- (96, -4)
     -- (20, -4)
     -- (20, -20)
     -- (96, -20)
     -- (96, -24)
     -- (48, -36)
     -- (0, -24) -- cycle;
  \draw[fill=red] (96, -12) circle[radius=1mm]
     node[right=2mm, red] {$\vec{x y} = f(\theta,\phi,\psi)$};
  \end{scope}
  \filldraw[fill=white]
    (208, 656) circle[radius=16];
  \filldraw[fill=white]
    (320, 704) circle[radius=16];
  \filldraw[fill=white]
    (112, 528) circle[radius=24];
  \filldraw[white!20!black]
    (112, 528) circle[radius=4]
    (208, 656) circle[radius=4]
    (320, 704) circle[radius=4];
  \coordinate (c1) at (112, 528);
  \coordinate (c2) at (208, 656);
  \coordinate (c3) at (320, 704);
  \coordinate (x1) at ($(c1) + (2cm,0)$);
  \coordinate (x3) at ($(c3) + (2cm,0)$);

  \draw (c1) -- (x1);
  \draw (c1) -- ($(c1)!2cm!(c2)$);
  \pic[draw, "$\theta$", angle radius=1cm, angle eccentricity=.8] {angle=x1--c1--c2};

  \draw (c2) -- ($(c2)!2cm!(c1)$);
  \draw (c2) -- ($(c2)!2cm!(c3)$);
  \pic[draw, "$\phi$", angle radius=1cm, angle eccentricity=.8] {angle=c1--c2--c3};

  \coordinate (x3p) at ($(c3)!2cm!-9.1623:(x3)$);
  \draw (c3) -- ($(c3)!2cm!(c2)$);
  \draw (c3) -- (x3p);
  \pic[draw, "$\psi$", angle radius=1cm, angle eccentricity=.8] {angle=c2--c3--x3p};

\end{tikzpicture}
      </latex-code>
      Define a transformation <m>f\colon\R^3\to\R^2</m> as follows: <m>f(\theta,\phi,\psi)</m> is the <m>(x,y)</m> position of the hand when the joints are rotated by angles <m>\theta, \phi, \psi</m>, respectively.  Evaluating <m>f</m> tells you where the hand will be on the plane when the joints are set at the given angles.
    </p>

    <p>
      It is relatively straightforward to find a formula for <m>f(\theta,\phi,\psi)</m> using some basic trigonometry.  If you want the robot to fetch your coffee cup, however, you have to <em>find</em> the angles <m>\theta,\phi,\psi</m> that will put the hand at the position of your beverage.  It is not at all obvious how to do this, and it is not even clear if the answer is unique!  You can ask yourself: <q>which positions on the table can my robot arm reach?</q> or <q>what is the arm<rsq/>s range of motion?</q>  This is the same as asking: <q>what is the range of <m>f</m>?</q>
    </p>
    <p>
      Unfortunately, this kind of function does not come from a matrix, so one cannot use linear algebra to answer these kinds of questions.  In fact, these functions are rather complicated; their study is the subject of <url href="https://en.wikipedia.org/wiki/Inverse_kinematics">inverse kinematics</url>.
    </p>
  </example>

</subsection>

<subsection>
  <title>Matrix Transformations</title>

  <p>
    Now we specialize the general notions and vocabulary from the previous <xref ref="matrix-trans-transformations"/> to the functions defined by matrices that we considered in the first <xref ref="matrix-trans-matrices-functions"/>.
  </p>

  <definition xml:id="matrix-trans-defn-of">
    <idx><h>Matrix transformation</h><h>definition of</h></idx>
    <idx><h>Transformation</h><h>associated to a matrix</h><see>Matrix transformation</see></idx>
    <statement>
      <p>
        Let <m>A</m> be an <m>m\times n</m> matrix.  The <term>matrix transformation</term> associated to <m>A</m> is the transformation
        <me>
          T\colon \R^n\To\R^m \sptxt{defined by} T(x) = Ax.
        </me>
        This is the transformation that takes a vector <m>x</m> in <m>\R^n</m> to the vector <m>Ax</m> in <m>\R^m</m>.
      </p>
    </statement>
  </definition>

  <p>
    If <m>A</m> has <m>n</m> columns, then it only makes sense to multiply <m>A</m> by vectors with <m>n</m> entries.  This is why the domain of <m>T(x)=Ax</m> is <m>\R^n</m>.  If <m>A</m> has <m>n</m> rows, then <m>Ax</m> has <m>m</m> entries for any vector <m>x</m> in <m>\R^n</m>; this is why the codomain of <m>T(x)=Ax</m> is <m>\R^m</m>.
  </p>

  <p>
    The definition of a matrix transformation <m>T</m> tells us how to evaluate <m>T</m> on any given vector: we multiply the input vector by a matrix.  For instance, let
    <me>
      A = \mat{1 2 3; 4 5 6}
    </me>
    and let <m>T(x)=Ax</m> be the associated matrix transformation.  Then
    <me>T\vec{-1 -2 -3} = A\vec{-1 -2 -3} = \mat{1 2 3; 4 5 6}\vec{-1 -2 -3}
    = \vec{-14 -32}.
    </me>
  </p>


  <p>
    Suppose that <m>A</m> has columns <m>v_1,v_2,\ldots,v_n</m>.  If we multiply <m>A</m> by a general vector <m>x</m>, we get
    <me>
      Ax = \mat{| | ,, |; v_1 v_2 \cdots, v_n; | | ,, |}\vec{x_1 x_2 \vdots, x_n}
      = x_1v_1 + x_2v_2 + \cdots + x_nv_n.
    </me>
    This is just a general linear combination of <m>v_1,v_2,\ldots,v_n</m>.  Therefore, the outputs of <m>T(x) = Ax</m> are exactly the linear combinations of the columns of <m>A</m>: the <em>range</em> of <m>T</m> is the column space of <m>A</m>.  See this <xref ref="matrixeq-spans-consistency"/>.
  </p>

  <bluebox xml:id="matrix-trans-dictionary">
    <idx><h>Matrix transformation</h><h>domain of</h></idx>
    <idx><h>Matrix transformation</h><h>codomain of</h></idx>
    <idx><h>Matrix transformation</h><h>range of</h></idx>
    <idx><h>Column space</h><h>range of a transformation</h></idx>
    <p>
      Let <m>A</m> be an <m>m\times n</m> matrix, and let <m>T(x)=Ax</m> be the associated matrix transformation.
      <ul>
        <li>
          The <em>domain</em> of <m>T</m> is <m>\R^n</m>, where <m>n</m> is the number of <em>columns</em> of <m>A</m>.
        </li>
        <li>
          The <em>codomain</em> of <m>T</m> is <m>\R^m</m>, where <m>m</m> is the number of <em>rows</em> of <m>A</m>.
        </li>
        <li>
          The <em>range</em> of <m>T</m> is the <em>column space</em> of <m>A</m>.
        </li>
      </ul>
    </p>
  </bluebox>

  <example hide-type="true">
    <title>Interactive: A <m>2\times 3</m> matrix: reprise</title>
    <p>
      Let
      <me>
        A = \mat{1 -1 2; -2 2 4},
      </me>
      and define <m>T(x) = Ax</m>.  The domain of <m>T</m> is <m>\R^3</m>, and the codomain is <m>\R^2</m>.  The range of <m>T</m> is the column space; since all three columns are collinear, the range is a line in <m>\R^2</m>.
    </p>
    <figure>
      <caption>A picture of the matrix transformation <m>T</m>.  The input vector is <m>x</m>, which is a vector in <m>\R^3</m>, and the output vector is <m>b=T(x)=Ax</m>, which is a vector in <m>\R^2</m>.  The violet line on the right is the range of <m>T</m>; as you vary <m>x</m>, the output <m>b</m> is constrained to lie on this line.</caption>
      <mathbox source="demos/Axequalsb.html?show=false" height="500px"/>
    </figure>
  </example>

  <example hide-type="true">
    <title>Interactive: A <m>3\times 2</m> matrix: reprise</title>
    <p>
      Let
      <me>
        A = \mat{1 0; 0 1; 1 0},
      </me>
      and define <m>T(x) = Ax</m>.  The domain of <m>T</m> is <m>\R^2</m>, and the codomain is <m>\R^3</m>.  The range of <m>T</m> is the column space; since <m>A</m> has two columns which are not collinear, the range is a plane in <m>\R^3</m>.
    </p>
    <figure>
      <caption>A picture of the matrix transformation <m>T</m>.  The input vector is <m>x</m>, which is a vector in <m>\R^2</m>, and the output vector is <m>b=T(x)=Ax</m>, which is a vector in <m>\R^3</m>.  The violet plane on the right is the range of <m>T</m>; as you vary <m>x</m>, the output <m>b</m> is constrained to lie on this plane.</caption>
      <mathbox source="demos/Axequalsb.html?mat=1,0:0,1:1,0&amp;range2=5&amp;closed=true&amp;show=false" height="500px"/>
    </figure>
  </example>

  <example>
    <title>Projection onto the <m>xy</m>-plane: reprise</title>
    <idx><h>Orthogonal projection</h><h>onto the <m>xy</m>-plane</h></idx>
    <statement>
      <p>
        Let <me>A = \mat{1 0 0; 0 1 0; 0 0 0},</me> and let <m>T(x) = Ax</m>.  What are the domain, the codomain, and the range of <m>T</m>?
      </p>
    </statement>
    <solution>
      <p>
        Geometrically, the transformation <m>T</m> projects a vector directly <q>down</q> onto the <m>xy</m>-plane in <m>\R^3</m>.
        <latex-code>
\begin{tikzpicture}[myxyz]
  \begin{scope}[arrows={|[width=4pt]->[width=4pt]}, shorten=4pt]
    \draw (-1,-1,-2.5) node[point, seq2] {}
      -- (-1,-1,0);
    \draw (2,-2,-1) node[point, seq5] {}
      -- (2,-2,0);
  \end{scope}

  \begin{scope}[transformxy]
    \fill[white, nearly opaque] (-3, -3) rectangle (3, 3);
    \draw[step=1cm, help lines] (-3, -3) grid (3, 3);
  \end{scope}

  \begin{scope}[arrows={|[width=4pt]->[width=4pt]}, shorten=4pt]
    \draw (2,1,3) node[point,seq1] {}
      -- (2,1,0) node[point,seq1] {};
    \point[seq2] at (-1,-1,0);
    \point[seq5] at (2,-2,0);
    \draw (-1,2,2) node[point,seq3] {}
      -- (-1,2,0) node[point,seq3] {};
    \draw (-2,0,1) node[point,seq4] {}
      -- (-2,0,0) node[point,seq4] {};
  \end{scope}

\end{tikzpicture}
        </latex-code>
        The inputs and outputs have three entries, so the domain and codomain are both <m>\R^3</m>.  The possible outputs all lie on the <m>xy</m>-plane, and every point on the <m>xy</m>-plane is an output of <m>T</m> (with itself as the input), so the range of <m>T</m> is the <m>xy</m>-plane.
      </p>
      <p>
        Be careful not to confuse the codomain with the range here.  The range is a plane, but it is a plane <em>in <m>\R^3</m></em>, so the codomain is still <m>\R^3</m>.  The outputs of <m>T</m> all have three entries; the last entry is simply always zero.
      </p>
    </solution>
  </example>

  <p>
    In the case of an <m>n\times n</m> square matrix, the domain and codomain of <m>T(x) = Ax</m> are both <m>\R^n</m>.  In this situation, one can regard <m>T</m> as <em>operating on</em> <m>\R^n</m>: it moves the vectors around in the same space.
  </p>

  <example>
    <title>Matrix transformations of <m>\R^2</m></title>
    <idx><h>Matrix transformation</h><h>of <m>\R^2</m></h></idx>
    <idx><h>Reflection</h><h>over the <m>y</m>-axis</h></idx>
    <idx><h>Dilation</h></idx>
    <idx><h>Identity matrix</h><h>as a matrix transformation</h></idx>
    <idx><h>Rotation</h><h>counterclockwise by <m>90^\circ</m></h></idx>
    <idx><h>Shear</h><h>in the <m>x</m>-direction</h></idx>
    <p>
      In the first <xref ref="matrix-trans-matrices-functions"/> we discussed the transformations defined by several <m>2\times 2</m> matrices, namely:
      <md>
        <mrow>
          \text{Reflection:} &amp;\qquad A=\mat{-1 0; 0 1}
        </mrow>
        <mrow>
          \text{Dilation:} &amp;\qquad A=\mat{1.5 0; 0 1.5}
        </mrow>
        <mrow>
          \text{Identity:} &amp;\qquad A=\mat{1 0; 0 1}
        </mrow>
        <mrow>
          \text{Rotation:} &amp;\qquad A=\mat{0 -1; 1 0}
        </mrow>
        <mrow>
          \text{Shear:} &amp;\qquad A=\mat{1 1; 0 1}.
        </mrow>
      </md>
      In each case, the associated matrix transformation <m>T(x)=Ax</m> has domain and codomain equal to <m>\R^2</m>.  The range is also <m>\R^2</m>, as can be seen geometrically (what is the input for a given output?), or using the fact that the columns of <m>A</m> are not collinear (so they form a basis for <m>\R^2</m>).
    </p>
  </example>

  <example xml:id="matrix-trans-questions">
    <title>Questions about a [matrix] transformation</title>
    <statement>
      <p>
        Let
        <me>
          A = \mat{1 1; 0 1; 1 1},
        </me>
        and let <m>T(x)=Ax</m>, so <m>T\colon\R^2\to\R^3</m> is a matrix transformation.
        <ol>
          <li>
            Evaluate <m>T(u)</m> for <m>u=\vec{3 4}</m>.
          </li>
          <li>
            Let
            <me>
              b = \vec{7 5 7}.
            </me>
            Find a vector <m>v</m> in <m>\R^2</m> such that <m>T(v)=b</m>.  Is there more than one?
          </li>
          <li>
            Does there exist a vector <m>w</m> in <m>\R^3</m> such that there is more than one <m>v</m> in <m>\R^2</m> with <m>T(v)=w</m>?
          </li>
          <li>
            Find a vector <m>w</m> in <m>\R^3</m> which is not in the range of <m>T</m>.
          </li>
        </ol>
      </p>

      <p>
        <em>Note:</em> all of the above questions are intrinsic to the transformation <m>T</m>: they make sense to ask whether or not <m>T</m> is a matrix transformation.  See the next <xref ref="matrix-trans-questions-nonlinear"/>.  As <m>T</m> is in fact a matrix transformation, all of these questions will translate into questions about the corresponding matrix <m>A</m>.
      </p>
    </statement>
    <solution>
      <p>
        <ol>
          <li>
            We evaluate <m>T(u)</m> by substituting the definition of <m>T</m> in terms of matrix multiplication:
            <me>
              T\vec{3 4} = \mat{1 1; 0 1; 1 1}\vec{3 4} = \vec{7 4 7}.
            </me>
          </li>

          <li>
            We want to find a vector <m>v</m> such that <m>b = T(v) = Av</m>.  In other words, we want to solve the matrix equation <m>Av = b</m>.  We form an augmented matrix and row reduce:
            <me>
              \mat{1 1; 0 1; 1 1} v = \vec{7 5 7}
              \;\xrightarrow{\parbox{\widthof{\small augmented}}
              {\small\centering augmented matrix}}\;
              \amat{1 1 7; 0 1 5; 1 1 7}
              \;\xrightarrow{\parbox{\widthof{\small reduce}}
              {\small\centering row reduce}}\;
              \amat{1 0 2; 0 1 5; 0 0 0}.
            </me>
            This gives <m>x=2</m> and <m>y=5</m>, so that there is a unique vector
            <me>v = \vec{2 5}</me>
            such that <m>T(v) = b</m>.
          </li>

          <li>
            <em>Translation:</em> is there any vector <m>w</m> in <m>\R^3</m> such that the solution set of <m>Av=w</m> has more than one vector in it?  The solution set of <m>Ax=w</m>, if non-empty, is a translate of the solution set of <m>Av=b</m> above, which has one vector in it.  See this <xref ref="solnsets-translate-span"/>.  It follows that the solution set of <m>Av=w</m> can have at most one vector.
          </li>

          <li>
            <em>Translation:</em> find a vector <m>w</m> such that the matrix equation <m>Av=w</m> is not consistent.  Notice that if we take
            <me>w = \vec{1 2 3}</me>,
            then the matrix equation <m>Av=w</m> translates into the system of equations
            <me>
              \syseq{x + y = 1; \. \+ y = 2; x + y = 3\rlap{,}}
            </me>
            which is clearly inconsistent.
          </li>
        </ol>
      </p>
    </solution>
  </example>

  <example xml:id="matrix-trans-questions-nonlinear">
    <title>Questions about a [non-matrix] transformation</title>
    <statement>
      <p>
        Define a transformation <m>T\colon\R^2\to\R^3</m> by the formula
        <me>
          T\vec{x y} = \vec{\ln(x) \cos(y) \ln(x)}.
        </me>
        <ol>
          <li>
            Evaluate <m>T(u)</m> for <m>u=\vec{1 \pi}</m>.
          </li>
          <li>
            Let
            <me>
              b = \vec{7 1 7}.
            </me>
            Find a vector <m>v</m> in <m>\R^2</m> such that <m>T(v)=b</m>.  Is there more than one?
          </li>
          <li>
            Does there exist a vector <m>w</m> in <m>\R^3</m> such that there is more than one <m>v</m> in <m>\R^2</m> with <m>T(v)=w</m>?
          </li>
          <li>
            Find a vector <m>w</m> in <m>\R^3</m> which is not in the range of <m>T</m>.
          </li>
        </ol>
      </p>
      <p>
        <em>Note:</em> we asked (almost) the exact same questions about a matrix transformation in the previous <xref ref="matrix-trans-questions"/>.  The point of this example is to illustrate the fact that the questions make sense for a transformation that has no hope of coming from a matrix.  In this case, these questions do not translate into questions about a matrix; they have to be answered in some other way.
      </p>
    </statement>
    <solution>
      <p>
        <ol>
          <li>
            We evaluate <m>T(u)</m> using the defining formula:
            <me>
              T\vec{1 \pi} = \vec{\ln(1) \cos(\pi) \ln(1)} = \vec{0 -1 0}.
            </me>
          </li>

          <li>
            We have
            <me>
              T\vec{e^7 2\pi n e^7} = \vec{\ln(e^7) \cos(2\pi n) \ln(e^7)} = \vec{7 1 7}
            </me>
            for any whole number <m>n</m>.  Hence there are infinitely many such vectors.
          </li>

          <li>
            The vector <m>b</m> from the previous part is an example of such a vector.
          </li>

          <li>
            Since <m>\cos(y)</m> is always between <m>-1</m> and <m>1</m>, the vector
            <me>
              w = \vec{0 2 0}
            </me>
            is not in the range of <m>T</m>.
          </li>
        </ol>
      </p>
    </solution>
  </example>

</subsection>

</section>