one-one-onto.xml

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

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<section xml:id="one-to-one-onto">
  <title>One-to-one and Onto Transformations</title>

  <objectives>
    <ol>
      <li>Understand the definitions of one-to-one and onto transformations.</li>
      <li><em>Recipes:</em> verify whether a matrix transformation is one-to-one and/or onto.</li>
      <li><em>Pictures:</em> examples of matrix transformations that are/are not one-to-one and/or onto.</li>
      <li><em>Vocabulary words:</em> <term>one-to-one</term>, <term>onto</term>.</li>
    </ol>
  </objectives>

  <introduction>
    <p>
      In this section, we discuss two of the most basic questions one can ask about a transformation: whether it is <em>one-to-one</em> and/or <em>onto.</em>  For a matrix transformation, we translate these questions into the language of matrices.
    </p>
  </introduction>

<subsection>
  <title>One-to-one Transformations</title>

  <definition>
    <title>One-to-one transformations</title>
    <idx><h>One-to-one</h><h>definition of</h></idx>
    <idx><h>Transformation</h><h>one-to-one</h><see>One-to-one</see></idx>
    <statement>
      <p>
        A transformation <m>T\colon\R^n\to\R^m</m> is <term>one-to-one</term> if, for every vector <m>b</m> in <m>\R^m</m>, the equation <m>T(x)=b</m> has <em>at most one</em> solution <m>x</m> in <m>\R^n</m>.
      </p>
    </statement>
  </definition>

  <remark>
    <p>
      Another word for <em>one-to-one</em> is <em>injective</em>.
    </p>
  </remark>

  <p>
    <idx><h>One-to-one</h><h>equivalent formulations</h></idx>
    Here are some equivalent ways of saying that <m>T</m> is one-to-one:
    <ul>
      <li>
        For every vector <m>b</m> in <m>\R^m</m>, the equation <m>T(x) = b</m> has <em>zero or one</em> solution <m>x</m> in <m>\R^n</m>.
      </li>
      <li>
        Different inputs of <m>T</m> have different outputs.
      </li>
      <li>
        If <m>T(u) = T(v)</m> then <m>u=v</m>.
      </li>
    </ul>
    <latex-code>
\begin{tikzpicture}[thin border nodes, scale=.75]
  \draw[grid lines] (-3,-3) grid (3,3);

  \node (A) at (0,-3.6) {$\R^n$};
  \node (B) at (12,-3.6) {$\R^m$};
  \draw[->] (A) -- (B) node[midway,above=1pt] {$T$};

  \point["$x$" left] (x) at (1,1.5);
  \point["$y$" left] (y) at (-1,0);
  \point["$z$" left] (z) at (0,-2);

  \begin{scope}[myxyz, xshift=12cm]
    \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,
        every label/.style={font=\small,fill=none,inner sep=.5pt}]
      \fill[seq4!10, nearly opaque] (-1,-1) rectangle (1,1);
      \draw[step=.5cm, seq4!30, very thin] (-1,-1) grid (1,1);
      \point["$T(x)$" above] (Tx) at (.5,-.5);
      \point["$T(y)$" above] (Ty) at (.5,.5);
      \point["$T(z)$" above] (Tz) at (-.5,.75);
      \node[coordinate,
          pin={[pin edge={very thin,-},
            pin distance=3mm,anchor=north,text=seq4]-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}

  \begin{scope}[arrows={|->}, shorten=.35mm, every to/.style={out=0, in=180}]
    \draw (x) to (Tx);
    \draw (y) to (Ty);
    \draw (z) to (Tz);
  \end{scope}

  \node[seq-blue] at (6,2.5) {one-to-one};

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

  <p>
    <idx><h>One-to-one</h><h>negation of</h></idx>
    Here are some equivalent ways of saying that <m>T</m> is <em>not</em> one-to-one:
    <ul>
      <li>
        There exists some vector <m>b</m> in <m>\R^m</m> such that the equation <m>T(x)=b</m> has <em>more than one</em> solution <m>x</m> in <m>\R^n</m>.
      </li>
      <li>
        There are two different inputs of <m>T</m> with the same output.
      </li>
      <li>
        There exist vectors <m>u,v</m> such that <m>u\neq v</m> but <m>T(u)=T(v)</m>.
      </li>
    </ul>
    <latex-code>
\begin{tikzpicture}[thin border nodes, scale=.75]
  \draw[grid lines] (-3,-3) grid (3,3);

  \node (A) at (0,-3.6) {$\R^n$};
  \node (B) at (12,-3.6) {$\R^m$};
  \draw[->] (A) -- (B) node[midway,above=1pt] {$T$};

  \point["$x$" left] (x) at (1,1.5);
  \point["$y$" left] (y) at (-1,0);
  \point["$z$" left] (z) at (0,-2);

  \begin{scope}[myxyz, xshift=12cm,
      every label/.style={font=\small, fill=none, inner sep=.5pt}]
    \path[clip, resetxy] (-3,-3) rectangle (3,3);

    \def\v{(-1,1,2)}
    \node[coordinate] (V) at \v {};

    \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);

    \draw[seq4] ($-4*(V)$) -- ($4*(V)$);
    \point[label={[xshift=.25cm,fill=white]above:{$T(x)=T(y)$}}]
      (Tx) at ($.5*(V)$);
    \point["$T(z)$" right] (Tz) at ($-1*(V)$);
    \node[coordinate,
        pin={[pin edge={very thin,-},
          pin distance=8mm,anchor=north,text=seq4]-30:range}]
        at ($-.2*(V)$) {};

    \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}

  \begin{scope}[arrows={|->}, shorten=.35mm, every to/.style={out=0, in=180}]
    \draw (x) to (Tx);
    \draw[|-] (y) to (Tx);
    \draw (z) to (Tz);
  \end{scope}

  \node[seq-blue] at (6,2.5) {not one-to-one};

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

  <example>
    <title>Functions of one variable</title>
    <idx><h>One-to-one</h><h>functions of one variable</h></idx>
    <p>
      The function <m>\sin\colon\R\to\R</m> is not one-to-one.  Indeed, <m>\sin(0) = \sin(\pi) = 0</m>, so the inputs <m>0</m> and <m>\pi</m> have the same output <m>0</m>.  In fact, the equation <m>\sin(x) = 0</m> has infinitely many solutions <m>\ldots,-2\pi,-\pi,0,\pi,2\pi,\ldots</m>.
    </p>

    <p>
      The function <m>\exp\colon\R\to\R</m> defined by <m>\exp(x) = e^x</m> is one-to-one.  Indeed, if <m>T(x) = T(y)</m>, then <m>e^x = e^y</m>, so <m>\ln(e^x) = \ln(e^y)</m>, and hence <m>x = y</m>.  The equation <m>T(x) = C</m> has one solution <m>x = \ln(C)</m> if <m>C > 0</m>, and it has zero solutions if <m>C\leq 0</m>.
    </p>

    <p>
      The function <m>f\colon\R\to\R</m> defined by <m>f(x)=x^3</m> is one-to-one.  Indeed, if <m>f(x) = f(y)</m> then <m>x^3 = y^3</m>; taking cube roots gives <m>x = y</m>.  In other words, the only solution of <m>f(x) = C</m> is <m>x=\sqrt[3]C</m>.
    </p>

    <p>
      The function <m>f\colon\R\to\R</m> defined by <m>f(x)=x^3-x</m> is not one-to-one.   Indeed, <m>f(0) = f(1) = f(-1) = 0</m>, so the inputs <m>0,1,-1</m> all have the same output <m>0</m>.  The solutions of the equation <m>x^3 - x = 0</m> are exactly the roots of <m>f(x) = x(x-1)(x+1)</m>, and this equation has three roots.
    </p>

    <p>
      The function <m>f\colon\R\to\R</m> defined by <m>f(x)=x^2</m> is not one-to-one.  Indeed, <m>f(1) = 1 = f(-1)</m>, so the inputs <m>1</m> and <m>-1</m> have the same outputs.  The function <m>g\colon\R\to\R</m> defined by <m>g(x)=|x|</m> is not one-to-one for the same reason.
    </p>
  </example>

  <example xml:id="one-to-one-eg-robot">
    <title>A real-word transformation: robotics</title>
    <p>
      Suppose you are building a robot arm with three joints that can move its hand around a plane, as in this <xref ref="matrix-trans-eg-robot"/>.
      <latex-code>
\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.  Asking whether <m>f</m> is one-to-one is the same as asking whether there is more than one way to move the arm in order to reach your coffee cup.  (There is.)
    </p>
  </example>

  <theorem xml:id="matrix-trans-one-to-one">
    <title>One-to-one matrix transformations</title>
    <idx><h>One-to-one</h><h>criteria for matrix transformations</h></idx>
    <idx><h>Matrix transformation</h><h>one-to-one criteria</h></idx>
    <statement>
      <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.  The following statements are equivalent:
        <ol>
          <li><m>T</m> is one-to-one.</li>
          <li>For every <m>b</m> in <m>\R^m</m>, the equation <m>T(x)=b</m> has at most one solution.</li>
          <li>For every <m>b</m> in <m>\R^m</m>, the equation <m>Ax=b</m> has a unique solution or is inconsistent.</li>
          <li><m>Ax=0</m> has <em>only</em> the trivial solution.</li>
          <li>The columns of <m>A</m> are linearly independent.</li>
          <li><m>A</m> has a pivot in every column.</li>
          <li>The range of <m>T</m> has dimension <m>n</m>.</li>
        </ol>
      </p>
    </statement>
    <proof>
      <p>
        Statements 1, 2, and 3 are translations of each other.  The equivalence of 3 and 4 follows from this <xref ref="solnsets-translate-span"/>: if <m>Ax=0</m> has only one solution, then <m>Ax=b</m> has only one solution as well, or it is inconsistent.  The equivalence of 4, 5, and 6 is a consequence of this <xref ref="linindep-matrix-cols"/>, and the equivalence of 6 and 7 follows from the fact that the rank of a matrix is equal to the number of columns with pivots.
      </p>
    </proof>
  </theorem>

  <p>
    Recall that <em>equivalent</em> means that, for a given matrix, either all of the statements are true simultaneously, or they are all false.
  </p>

  <example xml:id="matrix-trans-11-notonto1">
    <title>A matrix transformation that is one-to-one</title>
    <statement>
      <p>
        Let <m>A</m> be the matrix
        <me>A = \mat{1 0; 0 1; 1 0},</me>
        and define <m>T\colon\R^2\to\R^3</m> by <m>T(x) = Ax</m>.  Is <m>T</m> one-to-one?
      </p>
    </statement>
    <solution>
      <p>
        The reduced row echelon form of <m>A</m> is
        <me>\mat{1 0; 0 1; 0 0}.</me>
        Hence <m>A</m> has a pivot in every column, so <m>T</m> is one-to-one.
      </p>
      <figure>
        <caption>A picture of the matrix transformation <m>T</m>.  As you drag the input vector on the left side, you see that different input vectors yield different output vectors on the right side.</caption>
        <mathbox source="demos/Axequalsb.html?mat=1,0:0,1:1,0&amp;range2=5&amp;closed=true&amp;show=true" height="500px"/>
      </figure>
    </solution>
  </example>

  <example>
    <title>A matrix transformation that is not one-to-one</title>
    <statement>
      <p>
        Let
        <me>A = \mat{1 0 0; 0 1 0; 0 0 0},</me>
        and define <m>T\colon\R^3\to\R^3</m> by <m>T(x) = Ax</m>.  Is <m>T</m> one-to-one?  If not, find two different vectors <m>u,v</m> such that <m>T(u)=T(v)</m>.
      </p>
    </statement>
    <solution>
      <p>
        The matrix <m>A</m> is already in reduced row echelon form.  It does not have a pivot in every column, so <m>T</m> is not one-to-one.  Therefore, we know from the <xref ref="matrix-trans-one-to-one"/> that <m>Ax=0</m> has nontrivial solutions.  If <m>v</m> is a nontrivial (i.e., nonzero) solution of <m>Av = 0</m>, then <m>T(v) = Av = 0 = A0 = T(0)</m>, so <m>0</m> and <m>v</m> are different vectors with the same output.  For instance,
        <me>T\vec{0 0 1} = \mat{1 0 0; 0 1 0; 0 0 0}\vec{0 0 1} = 0 = T\vec{0 0 0}.</me>
      </p>
      <p>
        Geometrically, <m>T</m> is projection onto the <m>xy</m>-plane.  Any two vectors that lie on the same vertical line will have the same projection.  For <m>b</m> on the <m>xy</m>-plane, the solution set of <m>T(x) = b</m> is the entire vertical line containing <m>b</m>.  In particular, <m>T(x) = b</m> has <em>infinitely many</em> solutions.
      </p>
      <figure>
        <caption>A picture of the matrix transformation <m>T</m>.  The transformation <m>T</m> projects a vector onto the <m>xy</m>-plane.  The violet line is the solution set of <m>T(x)=0</m>.  If you drag <m>x</m> along the violet line, the output <m>T(x) = Ax</m> does not change.  This demonstrates that <m>T(x)=0</m> has more than one solution, so <m>T</m> is not one-to-one.</caption>
        <mathbox source="demos/Axequalsb.html?mat=1,0,0:0,1,0:0,0,0&amp;range2=5&amp;closed=true&amp;lock=true&amp;x=0,0,0" height="500px"/>
      </figure>
    </solution>
  </example>

  <example xml:id="one-to-one-not11-onto1">
    <title>A matrix transformation that is not one-to-one</title>
    <statement>
      <p>
        Let <m>A</m> be the matrix
        <me>A = \mat{1 1 0; 0 1 1},</me>
        and define <m>T\colon\R^3\to\R^2</m> by <m>T(x) = Ax</m>.  Is <m>T</m> one-to-one?  If not, find two different vectors <m>u,v</m> such that <m>T(u)=T(v)</m>.
      </p>
    </statement>
    <solution>
      <p>
        The reduced row echelon form of <m>A</m> is
        <me>\mat{1 0 -1; 0 1 1}.</me>
        There is not a pivot in every column, so <m>T</m> is not one-to-one.  Therefore, we know from the <xref ref="matrix-trans-one-to-one"/> that <m>Ax=0</m> has nontrivial solutions.  If <m>v</m> is a nontrivial (i.e., nonzero) solution of <m>Av = 0</m>, then <m>T(v) = Av = 0 = A0 = T(0)</m>, so <m>0</m> and <m>v</m> are different vectors with the same output.  In order to find a nontrivial solution, we find the parametric form of the solutions of <m>Ax=0</m> using the reduced matrix above:
        <me>\syseq{x \+ \. - z = 0; \. \+ y + z = 0}
        \quad\implies\quad
        \syseq{x = z; y = -z}
        </me>
        The free variable is <m>z</m>.  Taking <m>z=1</m> gives the nontrivial solution
        <me>T\vec{1 -1 1} = \mat{1 1 0; 0 1 1}\vec{1 -1 1} = 0 = T\vec{0 0 0}.</me>
      </p>
      <figure>
        <caption>A picture of the matrix transformation <m>T</m>.  The violet line is the null space of <m>A</m>, i.e., solution set of <m>T(x)=0</m>.  If you drag <m>x</m> along the violet line, the output <m>T(x) = Ax</m> does not change.  This demonstrates that <m>T(x)=0</m> has more than one solution, so <m>T</m> is not one-to-one.</caption>
        <mathbox source="demos/Axequalsb.html?mat=1,1,0:0,1,1&amp;range2=5&amp;closed=true&amp;show=true&amp;lock=true&amp;x=0,0,0" height="500px"/>
      </figure>
    </solution>
  </example>

  <example xml:id="one-to-one-eg-neither1">
    <title>A matrix transformation that is not one-to-one</title>
    <statement>
      <p>
        Let
        <me>A = \mat{1 -1 2; -2 2 -4},</me>
        and define <m>T\colon\R^3\to\R^2</m> by <m>T(x) = Ax</m>.
        Is <m>T</m> one-to-one?  If not, find two different vectors <m>u,v</m> such that <m>T(u)=T(v)</m>.
      </p>
    </statement>
    <solution>
      <p>
        The reduced row echelon form of <m>A</m> is
        <me>\mat{1 -1 2; 0 0 0}</me>.
        There is not a pivot in every column, so <m>T</m> is not one-to-one.  Therefore, we know from the <xref ref="matrix-trans-one-to-one"/> that <m>Ax=0</m> has nontrivial solutions.  If <m>v</m> is a nontrivial (i.e., nonzero) solution of <m>Av = 0</m>, then <m>T(v) = Av = 0 = A0 = T(0)</m>, so <m>0</m> and <m>v</m> are different vectors with the same output.  In order to find a nontrivial solution, we find the parametric form of the solutions of <m>Ax=0</m> using the reduced matrix above:
        <me>x - y + 2z = 0 \quad\implies\quad x = y - 2z.</me>
        The free variables are <m>y</m> and <m>z</m>.  Taking <m>y=1</m> and <m>z=0</m> gives the nontrivial solution
        <me>T\vec{1 1 0} = \mat{1 -1 2; -2 2 -4}\vec{1 1 0} = 0 = T\vec{0 0 0}.</me>
      </p>
      <figure>
        <caption>A picture of the matrix transformation <m>T</m>.  The violet plane is the solution set of <m>T(x)=0</m>.  If you drag <m>x</m> along the violet plane, the output <m>T(x) = Ax</m> does not change.  This demonstrates that <m>T(x)=0</m> has more than one solution, so <m>T</m> is not one-to-one.</caption>
        <mathbox source="demos/Axequalsb.html?show=true&amp;closed=true&amp;lock=true&amp;x=0,0,0" height="500px"/>
      </figure>
    </solution>
  </example>

  <p>
    <idx><h>One-to-one</h><h>finding two vectors with the same image</h></idx>
    The previous three examples can be summarized as follows.  Suppose that <m>T(x)=Ax</m> is a matrix transformation that is <em>not</em> one-to-one.  By the <xref ref="matrix-trans-one-to-one"/>, there is a nontrivial solution of <m>Ax=0</m>.  This means that the null space of <m>A</m> is not the zero space.  All of the vectors in the null space are solutions to <m>T(x)=0</m>.  If you compute a nonzero vector <m>v</m> in the null space (by row reducing and finding the parametric form of the solution set of <m>Ax=0</m>, for instance), then <m>v</m> and <m>0</m> both have the same output: <m>T(v) = Av = 0 = T(0).</m>
  </p>

  <note hide-type="true" xml:id="one-to-one-wide-matrices">
    <title>Wide matrices do not have one-to-one transformations</title>
    <idx><h>One-to-one</h><h>wide matrices</h></idx>
    <idx><h>Matrix transformation</h><h>wide matrices</h></idx>
    <idx><h>Wide matrix</h><see>Matrix transformation</see></idx>
    <idx><h>Wide matrix</h><see>Linear independence</see></idx>
    <p>
      If <m>T\colon\R^n\to\R^m</m> is a one-to-one matrix transformation, what can we say about the relative sizes of <m>n</m> and <m>m</m>?
    </p>
    <p>
      The matrix associated to <m>T</m> has <m>n</m> columns and <m>m</m> rows.  Each row and each column can only contain one pivot, so in order for <m>A</m> to have a pivot in every column, it must have <em>at least as many rows</em> as columns: <m>n\leq m</m>.
    </p>
    <p>
      This says that, for instance, <m>\R^3</m> is <q>too big</q> to admit a one-to-one linear transformation into <m>\R^2</m>.
    </p>
    <p>
      Note that there exist tall matrices that are not one-to-one: for example,
      <me>\mat{1 0 0; 0 1 0; 0 0 0; 0 0 0}</me>
      does not have a pivot in every column.
    </p>
  </note>

</subsection>

<subsection>
  <title>Onto Transformations</title>

  <definition>
    <title>Onto transformations</title>
    <idx><h>Onto</h><h>definition of</h></idx>
    <idx><h>Transformation</h><h>onto</h><see>Onto</see></idx>
    <statement>
      <p>A transformation <m>T\colon\R^n\to\R^m</m> is <term>onto</term> if, for every vector <m>b</m> in <m>\R^m</m>, the equation <m>T(x)=b</m> has <em>at least one</em> solution <m>x</m> in <m>\R^n</m>.</p>
    </statement>
  </definition>

  <remark>
    <p>
      Another word for <em>onto</em> is <em>surjective</em>.
    </p>
  </remark>

  <p>
    <idx><h>Onto</h><h>equivalent formulations</h></idx>
    Here are some equivalent ways of saying that <m>T</m> is onto:
    <ul>
      <li>The range of <m>T</m> is equal to the codomain of <m>T</m>.</li>
      <li>Every vector in the codomain is the output of some input vector.</li>
    </ul>
    <latex-code>
\begin{tikzpicture}[thin border nodes, scale=.75]
  \draw (-3,-3) rectangle (3,3);

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

  \node (source) at (0,-3.6) {$\R^n$};
  \point["$x$" above left] (x) at (1,2);

  \begin{scope}[xshift=12cm, every label/.append style={fill=seq4!10}]
    \fill[seq4!10] (-3,-3) rectangle (3,3);
    \draw[grid lines] (-3,-3) grid (3,3);
    \point["$T(x)$" above right] (Tx) at (1,-1);
    \node[fill=seq4!10, text=seq4]
      at (0,2) {$\text{range}(T)$};

    \node (target) at (0,-3.6) {$\R^m =$ codomain};
  \end{scope}

  \draw[->] (source) -- (target) node[midway,above=1pt] {$T$};
  \draw[|->, shorten=.35mm] (x) to[out=0, in=180] (Tx);

  \node[seq-blue] at (6,2.5) {onto};

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

  <p>
    <idx><h>Onto</h><h>negation of</h></idx>
    Here are some equivalent ways of saying that <m>T</m> is <em>not</em> onto:
    <ul>
      <li>The range of <m>T</m> is smaller than the codomain of <m>T</m>.</li>
      <li>There exists a vector <m>b</m> in <m>\R^m</m> such that the equation <m>T(x)=b</m> does not have a solution.</li>
      <li>There is a vector in the codomain that is not the output of any input vector.</li>
    </ul>
    <latex-code>
\begin{tikzpicture}[thin border nodes, scale=.75]
  \draw (-3,-3) rectangle (3,3);

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

  \node (source) at (0,-3.6) {$\R^n$};
  \point["$x$" above left] (x) at (1,-2);

  \begin{scope}[xshift=12cm]
    \draw[grid lines] (-3,-3) grid (3,3);
    \draw[seq4] (-3,-1.5) -- (3,1.5);
   \point["$T(x)$" below right] (Tx) at (1,.5);
    \node[fill=white, text=seq4]
      at (-1,-1.8) {$\text{range}(T)$};

    \node (target) at (0,-3.6) {$\R^m =$ codomain};
  \end{scope}

  \draw[->] (source) -- (target) node[midway,above=1pt] {$T$};
  \draw[|->, shorten=.35mm] (x) to[out=0, in=150] (Tx);

  \node[seq-blue] at (6,2.5) {not onto};
\end{tikzpicture}
    </latex-code>
  </p>

  <example>
    <title>Functions of one variable</title>
    <idx><h>Onto</h><h>functions of one variable</h></idx>
    <p>
      The function <m>\sin\colon\R\to\R</m> is not onto.  Indeed, taking <m>b=2</m>, the equation <m>\sin(x)=2</m> has no solution.  The range of <m>\sin</m> is the closed interval <m>[-1,1]</m>, which is smaller than the codomain <m>\R</m>.
    </p>

    <p>
      The function <m>\exp\colon\R\to\R</m> defined by <m>\exp(x) = e^x</m> is not onto.  Indeed, taking <m>b=-1</m>, the equation <m>\exp(x) = e^x = -1</m> has no solution.  The range of <m>\exp</m> is the set <m>(0,\infty)</m> of all <em>positive</em> real numbers.
    </p>

    <p>
      The function <m>f\colon\R\to\R</m> defined by <m>f(x)=x^3</m> is onto.  Indeed, the equation <m>f(x) = x^3 = b</m> always has the solution <m>x = \sqrt[3]b</m>.
    </p>

    <p>
      The function <m>f\colon\R\to\R</m> defined by <m>f(x)=x^3-x</m> is onto.  Indeed, the solutions of the equation <m>f(x) = x^3 - x = b</m> are the roots of the polynomial <m>x^3 - x - b</m>; as this is a cubic polynomial, it has at least one real root.
    </p>
  </example>

  <example>
    <title>A real-word transformation: robotics</title>
    <p>
      The robot arm transformation of this <xref ref="one-to-one-eg-robot"/> is not onto.  The robot cannot reach objects that are very far away.
    </p>
  </example>

  <theorem xml:id="matrix-trans-onto">
    <title>Onto matrix transformations</title>
    <idx><h>Onto</h><h>criteria for matrix transformations</h></idx>
    <idx><h>Matrix transformation</h><h>onto criteria</h></idx>
    <statement>
      <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.  The following statements are equivalent:
        <ol>
          <li><m>T</m> is onto.</li>
          <li><m>T(x)=b</m> has at least one solution for every <m>b</m> in <m>\R^m</m>.</li>
          <li><m>Ax=b</m> is consistent for every <m>b</m> in <m>\R^m</m>.</li>
          <li>The columns of <m>A</m> span <m>\R^m</m>.</li>
          <li><m>A</m> has a pivot in every row.</li>
          <li>The range of <m>T</m> has dimension <m>m</m>.</li>
        </ol>
      </p>
    </statement>
    <proof>
      <p>
        Statements 1, 2, and 3 are translations of each other.  The equivalence of 3, 4, 5, and 6 follows from this <xref ref="matrixeq-thm-full-span"/>.
      </p>
    </proof>
  </theorem>

  <example xml:id="one-to-one-not11-onto2">
    <title>A matrix transformation that is onto</title>
    <statement>
      <p>
        Let <m>A</m> be the matrix
        <me>A = \mat{1 1 0; 0 1 1},</me>
        and define <m>T\colon\R^3\to\R^2</m> by <m>T(x) = Ax</m>.  Is <m>T</m> onto?
      </p>
    </statement>
    <solution>
      <p>
        The reduced row echelon form of <m>A</m> is
        <me>\mat{1 0 -1; 0 1 1}.</me>
        Hence <m>A</m> has a pivot in every row, so <m>T</m> is onto.
      </p>
      <figure>
        <caption>A picture of the matrix transformation <m>T</m>.  Every vector on the right side is the output of <m>T</m> for a suitable input.  If you drag <m>b</m>, the demo will find an input vector <m>x</m> with output <m>b</m>.</caption>
        <mathbox source="demos/Axequalsb.html?mat=1,1,0:0,1,1&amp;range2=5&amp;closed=true&amp;show=false&amp;x=1,2,.5" height="500px"/>
      </figure>
    </solution>
  </example>

  <example xml:id="matrix-trans-11-notonto2">
    <title>A matrix transformation that is not onto</title>
    <statement>
      <p>
        Let <m>A</m> be the matrix
        <me>A = \mat{1 0; 0 1; 1 0},</me>
        and define <m>T\colon\R^2\to\R^3</m> by <m>T(x) = Ax</m>.  Is <m>T</m> onto?  If not, find a vector <m>b</m> in <m>\R^3</m> such that <m>T(x)=b</m> has no solution.
      </p>
    </statement>
    <solution>
      <p>
        The reduced row echelon form of <m>A</m> is
        <me>\mat{1 0; 0 1; 0 0}.</me>
        Hence <m>A</m> does not have a pivot in every row, so <m>T</m> is not onto.  In fact, since
        <me>T\vec{x y} = \mat{1 0; 0 1; 1 0}\vec{x y} = \vec{x y x},</me>
        we see that for every output vector of <m>T</m>, the third entry is equal to the first.  Therefore, <me>b=(1,2,3)</me> is not in the range of <m>T</m>.
      </p>
      <figure>
        <caption>A picture of the matrix transformation <m>T</m>.  The range of <m>T</m> is the violet plane on the right; this is smaller than the codomain <m>\R^3</m>.  If you drag <m>b</m> off of the violet plane, then the equation <m>Ax=b</m> becomes inconsistent; this means <m>T(x)=b</m> has no solution.</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>
    </solution>
  </example>

  <example xml:id="one-to-one-eg-neither2">
    <title>A matrix transformation that is not onto</title>
    <statement>
      <p>
        Let
        <me>A = \mat{1 -1 2; -2 2 -4},</me>
        and define <m>T\colon\R^3\to\R^2</m> by <m>T(x) = Ax</m>.
        Is <m>T</m> onto?  If not, find a vector <m>b</m> in <m>\R^2</m> such that <m>T(x)=b</m> has no solution.
      </p>
    </statement>
    <solution>
      <p>
        The reduced row echelon form of <m>A</m> is
        <me>\mat{1 -1 2; 0 0 0}</me>.
        There is not a pivot in every row, so <m>T</m> is not onto.  The range of <m>T</m> is the column space of <m>A</m>, which is equal to
        <me>\Span\left\{\vec{1 -2},\;\vec{-1 2},\;\vec{2 -4}\right\}
        = \Span\left\{\vec{1 -2}\right\},</me>
        since all three columns of <m>A</m> are collinear.  Therefore, any vector not on the line through <m>1\choose-2</m> is not in the range of <m>T</m>.  For instance, if <m>b={1\choose 1}</m> then <m>T(x)=b</m> has no solution.
      </p>
      <figure>
        <caption>A picture of the matrix transformation <m>T</m>.  The range of <m>T</m> is the violet line on the right; this is smaller than the codomain <m>\R^2</m>.  If you drag <m>b</m> off of the violet line, then the equation <m>Ax=b</m> becomes inconsistent; this means <m>T(x)=b</m> has no solution.</caption>
        <mathbox source="demos/Axequalsb.html?show=false&amp;closed=true" height="500px"/>
      </figure>
    </solution>
  </example>

  <p>
    <idx><h>Onto</h><h>finding a vector not in the range</h></idx>
    The previous two examples illustrate the following observation.  Suppose that <m>T(x)=Ax</m> is a matrix transformation that is <em>not</em> onto.  This means that <m>\text{range}(T) = \Col(A)</m> is a subspace of <m>\R^m</m> of dimension less than <m>m</m>: perhaps it is a line in the plane, or a line in <m>3</m>-space, or a plane in <m>3</m>-space, etc.  Whatever the case, the range of <m>T</m> is <em>very small</em> compared to the codomain.  To find a vector not in the range of <m>T</m>, choose a random nonzero vector <m>b</m> in <m>\R^m</m>; you have to be extremely unlucky to choose a vector that is in the range of <m>T</m>.  Of course, to check whether a given vector <m>b</m> is in the range of <m>T</m>, you have to solve the matrix equation <m>Ax=b</m> to see whether it is consistent.
  </p>

  <note hide-type="true" xml:id="one-to-one-tall-matrices">
    <title>Tall matrices do not have onto transformations</title>
    <idx><h>Onto</h><h>tall matrices</h></idx>
    <idx><h>Matrix transformation</h><h>tall matrices</h></idx>
    <idx><h>Tall matrix</h><see>Matrix transformation</see></idx>
    <p>
      If <m>T\colon\R^n\to\R^m</m> is an onto matrix transformation, what can we say about the relative sizes of <m>n</m> and <m>m</m>?
    </p>
    <p>
      The matrix associated to <m>T</m> has <m>n</m> columns and <m>m</m> rows.  Each row and each column can only contain one pivot, so in order for <m>A</m> to have a pivot in every row, it must have <em>at least as many columns</em> as rows: <m>m\leq n</m>.
    </p>
    <p>
      This says that, for instance, <m>\R^2</m> is <q>too small</q> to admit an onto linear transformation to <m>\R^3</m>.
    </p>
    <p>
      Note that there exist wide matrices that are not onto: for example,
      <me>\mat{1 -1 2; -2 2 -4}</me>
      does not have a pivot in every row.
    </p>
  </note>

</subsection>


<subsection>
  <title>Comparison</title>

  <p>
    <idx><h>One-to-one</h><h>versus onto</h></idx>
    <idx><h>Onto</h><h>versus one-to-one</h></idx>
    The above expositions of one-to-one and onto transformations were written to mirror each other.  However, <q>one-to-one</q> and <q>onto</q> are complementary notions: neither one implies the other.  Below we have provided a chart for comparing the two.  In the chart, <m>A</m> is an <m>m\times n</m> matrix, and <m>T\colon\R^n\to\R^m</m> is the matrix transformation <m>T(x)=Ax</m>.
    <latex-code>
\parbox{2.5in}{\centering%
  \textbf{$T$ is one-to-one}\\[3mm]
  $T(x)=b$ has \emph{at most one solution} for every $b$.\\[3mm]
  The columns of $A$ are linearly independent.\\[3mm]
  $A$ has a pivot in every column.\\[3mm]
  The range of $T$ has dimension $n$.
}\qquad\parbox{2.5in}{\centering%
  \textbf{$T$ is onto}\\[3mm]
  $T(x)=b$ has \emph{at least one solution} for every $b$.\\[3mm]
  The columns of $A$ span $\R^m$.\\~\\[3mm]
  $A$ has a pivot in every row.\\[3mm]
  The range of $T$ has dimension $m$.
}
    </latex-code>
  </p>

  <example>
    <title>Functions of one variable</title>
    <p>
      The function <m>\sin\colon\R\to\R</m> is neither one-to-one nor onto.
    </p>

    <p>
      The function <m>\exp\colon\R\to\R</m> defined by <m>\exp(x) = e^x</m> is one-to-one but not onto.
    </p>

    <p>
      The function <m>f\colon\R\to\R</m> defined by <m>f(x)=x^3</m> is one-to-one and onto.
    </p>

    <p>
      The function <m>f\colon\R\to\R</m> defined by <m>f(x)=x^3-x</m> is onto but not one-to-one.
    </p>
  </example>

  <example>
    <title>A matrix transformation that is neither one-to-one nor onto</title>
    <p>
      Let
      <me>A = \mat{1 -1 2; -2 2 -4},</me>
      and define <m>T\colon\R^3\to\R^2</m> by <m>T(x) = Ax</m>.  This transformation is neither one-to-one nor onto, as we saw in this <xref ref="one-to-one-eg-neither1"/> and this <xref ref="one-to-one-eg-neither2"/>.
    </p>
    <figure>
      <caption>A picture of the matrix transformation <m>T</m>.  The violet plane is the solution set of <m>T(x)=0</m>.  If you drag <m>x</m> along the violet plane, the output <m>T(x) = Ax</m> does not change.  This demonstrates that <m>T(x)=0</m> has more than one solution, so <m>T</m> is not one-to-one.   The range of <m>T</m> is the violet line on the right; this is smaller than the codomain <m>\R^2</m>.  If you drag <m>b</m> off of the violet line, then the equation <m>Ax=b</m> becomes inconsistent; this means <m>T(x)=b</m> has no solution.</caption>
      <mathbox source="demos/Axequalsb.html?show=true&amp;closed=true" height="500px"/>
    </figure>
  </example>

  <example>
    <title>A matrix transformation that is one-to-one but not onto</title>
    <p>
      Let <m>A</m> be the matrix
      <me>A = \mat{1 0; 0 1; 1 0},</me>
      and define <m>T\colon\R^2\to\R^3</m> by <m>T(x) = Ax</m>.  This transformation is one-to-one but not onto, as we saw in this <xref ref="matrix-trans-11-notonto1"/> and this <xref ref="matrix-trans-11-notonto2"/>.
    </p>
    <figure>
      <caption>A picture of the matrix transformation <m>T</m>.  The range of <m>T</m> is the violet plane on the right; this is smaller than the codomain <m>\R^3</m>.  If you drag <m>b</m> off of the violet plane, then the equation <m>Ax=b</m> becomes inconsistent; this means <m>T(x)=b</m> has no solution.  However, for <m>b</m> lying on the violet plane, there is a unique vector <m>x</m> such that <m>T(x)=b</m>.</caption>
      <mathbox source="demos/Axequalsb.html?mat=1,0:0,1:1,0&amp;range2=5&amp;closed=true&amp;show=" height="500px"/>
    </figure>
  </example>

  <example>
    <title>A matrix transformation that is onto but not one-to-one</title>
    <p>
      Let <m>A</m> be the matrix
      <me>A = \mat{1 1 0; 0 1 1},</me>
      and define <m>T\colon\R^3\to\R^2</m> by <m>T(x) = Ax</m>.  This transformation is onto but not one-to-one, as we saw in this <xref ref="one-to-one-not11-onto2"/> and this <xref ref="one-to-one-not11-onto2"/>.
    </p>
    <figure>
      <caption>A picture of the matrix transformation <m>T</m>.  Every vector on the right side is the output of <m>T</m> for a suitable input.  If you drag <m>b</m>, the demo will find an input vector <m>x</m> with output <m>b</m>.  The violet line is the null space of <m>A</m>, i.e., solution set of <m>T(x)=0</m>.  If you drag <m>x</m> along the violet line, the output <m>T(x) = Ax</m> does not change.  This demonstrates that <m>T(x)=0</m> has more than one solution, so <m>T</m> is not one-to-one.</caption>
      <mathbox source="demos/Axequalsb.html?mat=1,1,0:0,1,1&amp;range2=5&amp;closed=true&amp;show=true&amp;x=1,2,.5" height="500px"/>
    </figure>
  </example>

  <example xml:id="one-to-one-both-11-onto">
    <title>Matrix transformations that are both one-to-one and onto</title>
    <p>
      In this <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> is both one-to-one and onto.  A <m>2\times 2</m> matrix <m>A</m> has a pivot in every row if and only if it has a pivot in every column (if and only if it has two pivots), so in this case, the transformation <m>T</m> is one-to-one if and only if it is onto.  One can see geometrically that they are onto (what is the input for a given output?), or that they are one-to-one using the fact that the columns of <m>A</m> are not collinear.
    </p>
    <figure>
      <caption>Counterclockwise rotation by <m>90^\circ</m> is a matrix transformation.  This transformation is onto (if <m>b</m> is a vector in <m>\R^2</m>, then it is the output vector for the input vector which is <m>b</m> rotated <em>clockwise</em> by <m>90^\circ</m>), and it is one-to-one (different vectors rotate to different vectors).</caption>
      <mathbox source="demos/twobytwo.html?mat=0,-1,1,0&amp;closed=true&amp;pic=theo8.jpg" height="500px"/>
    </figure>
  </example>

  <note hide-type="true">
    <title>One-to-one is the same as onto for square matrices</title>
    <idx><h>One-to-one</h><h>square matrices</h></idx>
    <idx><h>Onto</h><h>square matrices</h></idx>
    <p>
      We observed in the previous <xref ref="one-to-one-both-11-onto"/> that a square matrix has a pivot in every row if and only if it has a pivot in every column.  Therefore, a matrix transformation <m>T</m> from <m>\R^n</m> to itself is one-to-one if and only if it is onto: in this case, the two notions are equivalent.
    </p>
    <p>
      Conversely, by this <xref ref="one-to-one-wide-matrices"/> and this <xref ref="one-to-one-tall-matrices"/>, if a matrix transformation <m>T\colon\R^m\to\R^n</m> is both one-to-one and onto, then <m>m=n</m>.</p>
  </note>

  <p>
    Note that in general, a transformation <m>T</m> is both one-to-one and onto if and only if <m>T(x)=b</m> has <em>exactly one</em> solution for all <m>b</m> in <m>\R^m</m>.
  </p>

</subsection>

</section>