matrixeq.xml

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

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<section xml:id="matrix-equations">
  <title>Matrix Equations</title>

  <objectives>
    <ol>
      <li>Understand the equivalence between a system of linear equations, an augmented matrix, a vector equation, and a matrix equation.</li>
      <li>Characterize the vectors <m>b</m> such that <m>Ax=b</m> is consistent, in terms of the span of the columns of <m>A</m>.</li>
      <li>Characterize matrices <m>A</m> such that <m>Ax=b</m> is consistent for all vectors <m>b</m>.</li>
      <li><em>Recipe:</em> multiply a vector by a matrix (two ways).</li>
      <li><em>Picture:</em> the set of all vectors <m>b</m> such that <m>Ax=b</m> is consistent.</li>
      <li><em>Vocabulary word:</em> <term>matrix equation</term>.</li>
    </ol>
  </objectives>

  <subsection>
    <title>The Matrix Equation <m>Ax=b</m>.</title>

    <p>In this section we introduce a very concise way of writing a system of linear equations: <m>Ax=b</m>.  Here <m>A</m> is a matrix and <m>x,b</m> are vectors (generally of different sizes), so first we must explain how to multiply a matrix by a vector.
    </p>

    <bluebox>
      <p>When we say <q><m>A</m> is an <m>m\times n</m> matrix,</q> we mean that <m>A</m> has <m>m</m> rows and <m>n</m> columns.
      <idx><h>Matrix</h><h>size of</h></idx>
      <notation><usage>m\times n\text{ matrix}</usage><description>Size of a matrix</description></notation>
      </p>
    </bluebox>

    <remark>
      <p>In this book, we do <em>not</em> reserve the letters <m>m</m> and <m>n</m> for the numbers of rows and columns of a matrix.  If we write <q><m>A</m> is an <m>n\times m</m> matrix</q>, then <m>n</m> is the number of rows of <m>A</m> and <m>m</m> is the number of columns.</p>
    </remark>

    <definition xml:id="matrixeq-defn-Ax1">
      <idx><h>Matrix-vector product</h><h>definition of</h></idx>
      <idx><h>Matrix</h><h>product with vector</h><see>Matrix-vector product</see></idx>
      <idx><h>Vector</h><h>product with matrix</h><see>Matrix-vector product</see></idx>
      <statement>
        <p>Let <m>A</m> be an <m>m\times n</m> matrix with columns <m>v_1,v_2,\ldots,v_n</m>:
        <me>A = \mat{| | ,{}, |; v_1 v_2 \cdots, v_n;| | ,{}, | }</me>
        The <term>product</term> of <m>A</m> with a vector <m>x</m> in <m>\R^n</m> is the linear combination
        <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 a vector in <m>\R^m</m>.
        </p>
      </statement>
    </definition>

    <example>
      <p><me>
        \mat{
        4 5 6;
        7 8 9}
        \vec{1 2 3}
        = 1\vec{4 7} + 2\vec{5 8} + 3\vec{6 9} = \vec{32 50}.
      </me></p>
    </example>

    <p>In order for <m>Ax</m> to make sense, the number of entries of <m>x</m> has to be the same as the number of columns of <m>A</m>: we are using the entries of <m>x</m> as the coefficients of the columns of <m>A</m> in a linear combination.  The resulting vector has the same number of entries as the number of <em>rows</em> of <m>A</m>, since each column of <m>A</m> has that number of entries.
    </p>

    <bluebox>
      <p>If <m>A</m> is an <m>m\times n</m> matrix (<m>m</m> rows, <m>n</m> columns), then <m>Ax</m> makes sense when <m>x</m> has <m>n</m> entries.  The product <m>Ax</m> has <m>m</m> entries.
      </p>
    </bluebox>


    <proposition hide-type="true" xml:id="matrix-linearity">
      <title>Properties of the Matrix-Vector Product</title>
      <statement>
        <p>
          Let <m>A</m> be an <m>m\times n</m> matrix, let <m>u,v</m> be vectors in <m>\R^n</m>, and let <m>c</m> be a scalar.  Then:
          <ul>
            <li><m>A(u+v) = Au + Av</m></li>
            <li><m>A(cu) = cAu</m></li>
          </ul>
        </p>
      </statement>
    </proposition>

    <definition>
      <idx><h>Matrix equation</h><h>definition of</h></idx>
      <statement>
        <p>A <term>matrix equation</term> is an equation of the form <m>Ax=b</m>, where <m>A</m> is an <m>m\times n</m> matrix, <m>b</m> is a vector in <m>\R^m</m>, and <m>x</m> is a vector whose coefficients <m>x_1,x_2,\ldots,x_n</m> are unknown.</p>
      </statement>
    </definition>

	<p>
	In this book we will study two complementary questions about a matrix equation <m>Ax=b</m>:
	<ol>
	<li>Given a specific choice of <m>b</m>, what are all of the solutions to <m>Ax=b</m>?</li>
	<li>What are all of the choices of <m>b</m> so that <m>Ax=b</m> is consistent?</li>
	</ol>
	The first question is more like the questions you might be used to from your earlier courses in algebra; you have a lot of practice solving equations like <m>x^2-1=0</m> for <m>x</m>.  The second question is perhaps a new concept for you.  The <xref ref="rank-theorem"/>, which is the culmination of this chapter, tells us that the two questions are intimately related.  
	</p>

    <note hide-type="true">
      <title>Matrix Equations and Vector Equations</title>
      <idx><h>Matrix equation</h><h>equivalence with vector equation</h></idx>
      <idx><h>Vector equation</h><h>equivalence with matrix equation</h></idx>
      <p>Let <m>v_1,v_2,\ldots,v_n</m> and <m>b</m> be vectors in <m>\R^m</m>.  Consider the vector equation
      <me>x_1v_1 + x_2v_2 + \cdots + x_nv_n = b.</me>
      This is equivalent to the matrix equation <m>Ax=b</m>, where
      <me>A = \mat{ | | {} |; v_1 v_2 \cdots, v_n; | | {} |}
      \sptxt{and}
      x = \vec{x_1 x_2 \vdots, x_n}.
      </me>
      Conversely, if <m>A</m> is any <m>m\times n</m> matrix, then
      <m>Ax=b</m> is equivalent to the vector equation
      <me>x_1v_1 + x_2v_2 + \cdots + x_nv_n = b,</me>
      where <m>v_1,v_2,\ldots,v_n</m> are the columns of <m>A</m>, and <m>x_1,x_2,\ldots,x_n</m> are the entries of <m>x</m>.
      </p>
    </note>

    <example>
      <statement>
        <p>Write the vector equation
        <me>2v_1 + 3v_2 - 4v_3 = \vec{7 2 1}</me>
        as a matrix equation, where <m>v_1,v_2,v_3</m> are vectors in <m>\R^3</m>.
        </p>
      </statement>

      <solution>
        <p>Let <m>A</m> be the matrix with columns <m>v_1,v_2,v_3</m>, and let <m>x</m> be the vector with entries <m>2,3,-4</m>.  Then
        <me>Ax =  \mat{ | | |; v_1 v_2 v_3; | | |}\vec{2 3 -4}
        = 2v_1 + 3v_2 - 4v_3, </me>
        so the vector equation is equivalent to the matrix equation
        <m>Ax=\vec{7 2 1}.</m>
        </p>
      </solution>
    </example>

    <note hide-type="true">
      <title>Four Ways of Writing a Linear System</title>
      <p>We now have <em>four</em> equivalent ways of writing (and thinking about) a system of linear equations:
      <idx><h>System of linear equations</h><h>four ways of writing</h></idx>
      <ol>
        <li>As a system of equations:
        <me>\syseq{2x_1 + 3x_2 - 2x_3 = 7; x_1 - x_2 - 3x_3 = 5}
        </me>
        </li>
        <li>As an augmented matrix:
        <me>\amat{2 3 -2 7; 1 -1 -3 5}</me>
        </li>
        <li>As a vector equation (<m>x_1v_1 + x_2v_2 + \cdots + x_nv_n = b</m>):
        <me>x_1\vec{2 1} + x_2\vec{3 -1} + x_3\vec{-2 -3} = \vec{7 5}</me>
        </li>
        <li>As a matrix equation (<m>Ax=b</m>):
        <me>\mat{2 3 -2; 1 -1 -3}\vec{x_1 x_2 x_3} = \vec{7 5}.
        </me>
        </li>
      </ol>
      In particular, <em>all four have the same solution set</em>.
      </p>
    </note>

    <bluebox>
      <p>We will move back and forth freely between the four ways of writing a linear system, over and over again, for the rest of the book.</p>
    </bluebox>

    <paragraphs>
      <title>Another Way to Compute <m>Ax</m></title>

      <p>The above <xref ref="matrixeq-defn-Ax1"/> is a useful way of defining the product of a matrix with a vector when it comes to understanding the relationship between matrix equations and vector equations.  Here we give a definition that is better-adapted to computations by hand.
      </p>

      <definition xml:id="matrixeq-row-column-prod">
        <idx><h>Row vector</h><see>Vector</see></idx>
        <idx><h>Vector</h><h>row vector</h></idx>
        <idx><h>Vector</h><h>row vector</h><h>product with column vector</h></idx>
        <statement>
          <p>A <term>row vector</term> is a matrix with one row.  The <term>product</term> of a row vector of length <m>n</m> and a (column) vector of length <m>n</m> is
          <me>\mat{a_1 a_2 \cdots, a_n} \vec{x_1 x_2 \vdots, x_n}
          = a_1x_1 + a_2x_2 + \cdots + a_nx_n.</me>
          This is a scalar.
          </p>
        </statement>
      </definition>

      <bluebox xml:id="matrixeq-row-column">
        <title>Recipe: The row-column rule for matrix-vector multiplication</title>
        <idx><h>Matrix-vector product</h><h>row-column rule</h></idx>

        <p>
          If <m>A</m> is an <m>m\times n</m> matrix with rows <m>r_1,r_2,\ldots,r_m</m>, and <m>x</m> is a vector in <m>\R^n</m>, then
          <me>Ax = \mat[c]{ \matrow{r_1};
          \matrow{r_2};
          \vdots ;
          \matrow{r_m}}
          x
          = \vec{r_1x r_2x \vdots, r_mx}.
          </me>
        </p>
      </bluebox>

      <example>
        <p><me>
          \mat{
          4 5 6;
          7 8 9}
          \vec{1 2 3} =
          \mat{\mat{4, 5, 6}\vec{1 2 3};
          \mat{7 8 9}\vec{1 2 3}}
          = \vec{4\cdot1+5\cdot2+6\cdot3 7\cdot1+8\cdot2+9\cdot3}
          = \vec{32 50}.
        </me>
        This is the same answer as before:
        <me>
        \mat{
        4 5 6;
        7 8 9}
        \vec{1 2 3}
        = 1\vec{4 7} + 2\vec{5 8} + 3\vec{6 9}
        = \vec{1\cdot4+2\cdot5+3\cdot6 1\cdot7+2\cdot8+3\cdot9}
        = \vec{32 50}.
        </me>
        </p>
      </example>
    </paragraphs>

  </subsection>

  <subsection>
    <title>Spans and Consistency</title>

    <p>Let <m>A</m> be a matrix with columns <m>v_1,v_2,\ldots,v_n</m>:
    <me>A = \mat{ | | {} |; v_1 v_2 \cdots, v_n; | | {} |}.</me>
    Then
    <me>\begin{split}
    Ax=b&amp;\text{ has a solution} \\
    &amp;\iff \text{there exist $x_1,x_2,\ldots,x_n$ such that }
    A\vec{x_1 x_2 \vdots, x_n} = b \\
    &amp;\iff \text{there exist $x_1,x_2,\ldots,x_n$ such that }
    x_1v_1 + x_2v_2 + \cdots + x_nv_n = b \\
    &amp;\iff \text{$b$ is a linear combination of } v_1,v_2,\ldots,v_n \\
    &amp;\iff \text{$b$ is in the span of the columns of $A$}.
    \end{split}
    </me>
    </p>

    <bluebox xml:id="matrixeq-spans-consistency" type-name="Note">
      <title>Spans and Consistency</title>
      <idx><h>Matrix equation</h><h>spans and consistency</h></idx>
      <idx><h>System of linear equations</h><h>consistent</h><h>span criterion</h></idx>
      <idx><h>Column span</h><see>Column space</see></idx>
      <p>The matrix equation <m>Ax=b</m> has a solution if and only if <m>b</m> is in the span of the columns of <m>A</m>.</p>
    </bluebox>

    <p>This gives an equivalence between an <em>algebraic</em> statement (<m>Ax=b</m> is consistent), and a <em>geometric</em> statement (<m>b</m> is in the span of the columns of <m>A</m>).
    </p>

    <example>
      <title>An Inconsistent System</title>
      <statement><p>
        Let <m>A=\mat[r]{2 1; -1 0; 1 -1}</m>.  Does the equation <m>Ax=\vec{0 2 2}</m> have a solution?
      </p></statement>

      <solution>
        <p>First we answer the question geometrically.  The columns of <m>A</m> are
        <me>\textcolor{seq-red}{v_1 = \vec{2 -1 1}}
        \sptxt{and}
        \textcolor{seq-blue}{v_2 = \vec{1 0 -1}},</me>
        and the target vector (on the right-hand side of the equation) is <m>\textcolor{seq-green}{w = \vec{0 2 2}}.</m>  The equation <m>Ax=w</m> is consistent if and only if <m>w</m> is contained in the span of the columns of <m>A</m>.  So we draw a picture:
        <latex-code>
              <![CDATA[
\begin{tikzpicture}[myxyz, z={(0,0,1.2)}, thin border nodes]
  \path[clip, resetxy] (-4,-4) rectangle (4,4);

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

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

  \begin{scope}[x=(X), y=(Y), transformxy]
    \path[clip] (-5, -5) -- (5, 5) -- (5, 10) -- (-5, 10) -- cycle;
    \fill[seq4!30, nearly opaque] (-1.5,-1) rectangle (1.5,2);
    \draw[step=.5cm, seq4, very thin] (-1.5,-1) grid (1.5,2);
  \end{scope}

  \begin{scope}[transformxy]
    \fill[white, nearly opaque] (-2, -2) rectangle (3, 3);
    \draw[grid lines] (-2, -2) grid (3, 3);
  \end{scope}

  \begin{scope}[x=(X), y=(Y), transformxy]
    \path[clip] (-5, -5) -- (5, 5) -- (5, -10) -- (-5, -10) -- cycle;
    \fill[seq4!30, nearly opaque] (-1.5,-1) rectangle (1.5,2);
    \draw[step=.5cm, seq4, very thin] (-1.5,-1) grid (1.5,2);
  \end{scope}

  \draw[vector, seq-blue] (0,0,0) --
    node [midway, above right] {$v_1$} \v;
  \draw[thin, densely dotted] \v -- \projxy\v;

  \begin{scope}
    \path[clip, transformxy] (-2, -2) rectangle (3, 3);
    \draw[vector, seq-green!50!white] (0,0,0) --
      node [pos=1.0, below left] {$v_2$} \w;
  \end{scope}
  \begin{scope}
    \path[clip, transformxy] (5, -2) rectangle (3, 3);
    \draw[vector, seq-green] (0,0,0) --
      node [pos=1.0, below left] {$v_2$} \w;
  \end{scope}
  \draw[thin, densely dotted, black!50!white] \w -- \projxy\w;

  \draw[vector, black] (0,0,0) --
    node [pos=0.8, above left] {$w$} \b;
  \draw[thin, densely dotted] \b -- \projxy\b;

  \node[seq4] at (-1.5cm, 2.3cm) {$\Span\{v_1,v_2\}$};

  \point at (0,0,0);

\end{tikzpicture}
              ]]>
        </latex-code>
        It does not appear that <m>w</m> lies in <m>\Span\{v_1,v_2\},</m> so the equation is inconsistent.</p>
        <figure>
          <caption>The vector <m>w</m> is not contained in <m>\Span\{v_1,v_2\}</m>, so the equation <m>Ax=b</m> is inconsistent.  (Try moving the sliders to solve the equation.)</caption>
          <mathbox source="demos/spans.html?v1=2,-1,1&amp;v2=1,0,-1&amp;target=0,2,2&amp;range=5" height="500px"/>
        </figure>

        <p>Let us check our geometric answer by solving the matrix equation using row reduction.  We put the system into an augmented matrix and row reduce:
        <me>\amat{2 1 0; -1 0 2; 1 -1 2}
        \quad\xrightarrow{\text{RREF}}\quad
        \amat{1 0 0; 0 1 0; 0 0 1}.
        </me>
        The last equation is <m>0=1</m>, so the system is indeed inconsistent, and the matrix equation
        <me>\mat[r]{2 1; -1 0; 1 -1}x = \vec{0 2 2}</me>
        has no solution.
        </p>

      </solution>
    </example>

    <example>
      <title>A Consistent System</title>

      <statement><p>
        Let <m>A=\mat[r]{2 1; -1 0; 1 -1}</m>.  Does the equation <m>Ax=\vec{1 -1 2}</m> have a solution?
      </p></statement>

      <solution>
        <p>First we answer the question geometrically.  The columns of <m>A</m> are
        <me>\textcolor{seq-red}{v_1 = \vec{2 -1 1}}
        \sptxt{and}
        \textcolor{seq-blue}{v_2 = \vec{1 0 -1}},</me>
        and the target vector (on the right-hand side of the equation) is <m>\textcolor{seq-green}{w = \vec{1 -1 2}}.</m>  The equation <m>Ax=w</m> is consistent if and only if <m>w</m> is contained in the span of the columns of <m>A</m>.  So we draw a picture:
        <latex-code>
              <![CDATA[
\begin{tikzpicture}[myxyz, z={(0,0,1.2)}, thin border nodes]
  \path[clip, resetxy] (-4,-4) rectangle (4,4);

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

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

  \begin{scope}[x=(X), y=(Y), transformxy]
    \path[clip] (-5, -5) -- (5, 5) -- (5, 10) -- (-5, 10) -- cycle;
    \fill[seq4!30, nearly opaque] (-1.5,-1.5) rectangle (1.5,2);
    \draw[step=.5cm, seq4, very thin] (-1.5,-1.5) grid (1.5,2);
  \end{scope}

  \begin{scope}[transformxy]
    \fill[white, nearly opaque] (-2, -2) rectangle (3, 3);
    \draw[help lines] (-2, -2) grid (3, 3);
  \end{scope}

  \begin{scope}[x=(X), y=(Y), transformxy]
    \path[clip] (-5, -5) -- (5, 5) -- (5, -10) -- (-5, -10) -- cycle;
    \fill[seq4!30, nearly opaque] (-1.5,-1.5) rectangle (1.5,2);
    \draw[step=.5cm, seq4, very thin] (-1.5,-1.5) grid (1.5,2);
  \end{scope}

  \draw[vector, seq-blue] (0,0,0) --
    node [midway, above right] {$v_1$} \v;
  \draw[thin, densely dotted] \v -- \projxy\v;

  \begin{scope}
    \path[clip, transformxy] (-2, -2) rectangle (3, 3);
    \draw[vector, seq-green!50!white] (0,0,0) --
      node [pos=1.0, below left] {$v_2$} \w;
  \end{scope}
  \begin{scope}
    \path[clip, transformxy] (5, -2) rectangle (3, 3);
    \draw[vector, seq-green] (0,0,0) --
      node [pos=1.0, below left] {$v_2$} \w;
  \end{scope}
  \draw[thin, densely dotted, black!50!white] \w -- \projxy\w;

  \draw[vector, black] (0,0,0) --
    node [midway, above right] {$w$} \b;
  \draw[thin, densely dotted] \b -- \projxy\b;

  \node[seq4] at (1.75cm, 2.1cm) {$\Span\{v_1,v_2\}$};

  \point at (0,0,0);
\end{tikzpicture}
              ]]>
        </latex-code>
        It appears that <m>w</m> is indeed contained in the span of the columns of <m>A</m>; in fact, we can see
        <me>w = v_1 - v_2 \implies x = \vec{1 -1}.</me>
        </p>
        <figure>
          <caption>The vector <m>w</m> is contained in <m>\Span\{v_1,v_2\}</m>, so the equation <m>Ax=b</m> is consistent.  (Move the sliders to solve the equation.)</caption>
          <mathbox source="demos/spans.html?v1=2,-1,1&amp;v2=1,0,-1&amp;target=1,-1,2&amp;range=5" height="500px"/>
        </figure>

        <p>Let us check our geometric answer by solving the matrix equation using row reduction.  We put the system into an augmented matrix and row reduce:
        <me>\amat{2 1 1; -1 0 -1; 1 -1 2}
        \quad\xrightarrow{\text{RREF}}\quad
        \amat{1 0 1; 0 1 -1; 0 0 0}.
        </me>
        This gives us <m>x=1</m> and <m>y=-1</m>, which is consistent with the picture:
        <me>1\vec{2 -1 1} -1\vec{1 0 -1} = \vec{1 -1 2}
        \sptxt{or}
        A\vec{1 -1} = \vec{1 -1 2}.</me>
        </p>

      </solution>
    </example>

    <paragraphs>
      <title>When Solutions Always Exist</title>

      <p>Building on this <xref ref="matrixeq-spans-consistency">note</xref>, we have the following criterion for when <m>Ax=b</m> is consistent for <em>every</em> choice of <m>b</m>.</p>

      <theorem xml:id="matrixeq-thm-full-span">
        <idx><h>Matrix equation</h><h>always consistent</h></idx>
        <statement>
          <p>Let <m>A</m> be an <m>m\times n</m> (non-augmented) matrix.  The following are equivalent:
          <ol>
            <li><m>Ax=b</m> has a solution for all <m>b</m> in <m>\R^m</m>.</li>
            <li>The span of the columns of <m>A</m> is all of <m>\R^m</m>.</li>
            <li><m>A</m> has a <xref ref="defn-pivot-pos" text="title">pivot position</xref> in every row.</li>
          </ol>
          </p>
        </statement>

        <proof>
          <p>The equivalence of 1 and 2 is established by this <xref ref="matrixeq-spans-consistency"/> as applied to every <m>b</m> in <m>\R^m</m>.
          </p>

          <p>Now we show that 1 and 3 are equivalent.  (Since we know 1 and 2 are equivalent, this implies 2 and 3 are equivalent as well.)  If <m>A</m> has a pivot in every row, then its reduced row echelon form looks like this:
          <me>\mat{
          1   0   \star,   0   \star ;
          0   1   \star , 0   \star ;
          0   0   0   1   \star
          },</me>
          and therefore <m>\amat{A b}</m> reduces to this:
          <me>\amat[c]{
          1   0   \star,   0   \star, \star ;
          0   1   \star , 0   \star, \star ;
          0   0   0   1   \star, \star
          }.</me>
          There is no <m>b</m> that makes it inconsistent, so there is always a solution.  Conversely, if <m>A</m> does not have a pivot in each row, then its reduced row echelon form looks like this:
          <me>\mat{
          1   0   \star,   0   \star ;
          0   1   \star , 0   \star ;
          0   0   0   0  0
          },</me>
          which can give rise to an inconsistent system after augmenting with <m>b</m>:
          <me>\amat{
          1   0   \star,   0   \star, 0 ;
          0   1   \star , 0   \star, 0 ;
          0   0   0   0   0 16
          }.
          </me>
          </p>
        </proof>
      </theorem>

      <p>Recall that <term>equivalent</term> means that, for any given matrix <m>A</m>, either <em>all</em> of the conditions of the above <xref ref="matrixeq-thm-full-span"/> are true, or they are all false.
      </p>

      <bluebox>
        <p>Be careful when reading the statement of the above <xref ref="matrixeq-thm-full-span"/>.  The first two conditions look very much like this <xref ref="matrixeq-spans-consistency"/>, but they are logically quite different because of the quantifier <q><em>for all</em> <m>b</m></q>.
        </p>
      </bluebox>

      <example hide-type="true">
        <title>Interactive: The criteria of the theorem are satisfied</title>
        <figure>
          <caption>An example where the criteria of the above <xref ref="matrixeq-thm-full-span"/> are satisfied.  The violet region is the span of the columns <m>v_1,v_2,v_3</m> of <m>A</m>, which is the same as the set of all <m>b</m> such that <m>Ax=b</m> has a solution.  If you drag <m>b</m>, the demo will solve <m>Ax=b</m> for you and move <m>x</m>.</caption>
          <mathbox source="demos/Axequalsb.html?mat=2,1,-1:1,0,2&amp;range2=5&amp;show=false&amp;closed=true" height="600px"/>
        </figure>
      </example>

      <example hide-type="true">
        <title>Interactive: The critera of the theorem are not satisfied</title>
        <figure>
          <caption>An example where the criteria of the above <xref ref="matrixeq-thm-full-span"/> are <em>not</em> satisfied.  The violet line is the span of the columns <m>v_1,v_2,v_3</m> of <m>A</m>, which is the same as the set of all <m>b</m> such that <m>Ax=b</m> has a solution.  Try dragging <m>b</m> in and out of the column span.</caption>
          <mathbox source="demos/Axequalsb.html?show=false&amp;closed=true" height="600px"/>
        </figure>
      </example>

      <remark>
        <p>
          We will see in this <xref ref="dimension-basis-colspace-dim"/> that the dimension of the span of the columns is equal to the number of pivots of <m>A</m>.  That is, the columns of <m>A</m> span a line if <m>A</m> has one pivot, they span a plane if <m>A</m> has two pivots, etc.  The whole space <m>\R^m</m> has dimension <m>m</m>, so this generalizes the fact that the columns of <m>A</m> span <m>\R^m</m> when <m>A</m> has <m>m</m> pivots.
        </p>
      </remark>

    </paragraphs>

  </subsection>

</section>