spans.xml

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

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<section xml:id="spans">
  <title>Vector Equations and Spans</title>

  <objectives>
    <ol>
      <li>Understand the equivalence between a system of linear equations and a vector equation.</li>
      <li>Learn the definition of <m>\Span\{x_1,x_2,\ldots,x_k\}</m>, and how to draw pictures of spans.</li>
      <li><em>Recipe:</em> solve a vector equation using augmented matrices / decide if a vector is in a span.</li>
      <li><em>Pictures:</em> an inconsistent system of equations, a consistent system of equations, spans in <m>\R^2</m> and <m>\R^3</m>.</li>
      <li><em>Vocabulary word:</em> <term>vector equation</term>.</li>
      <li><em>Essential vocabulary word:</em> <term>span</term>.</li>
    </ol>
  </objectives>

  <subsection>
    <title>Vector Equations</title>

    <p>
      <idx><h>Vector equation</h><h>equivalence with system of equations</h></idx>
      An equation involving vectors with <m>n</m> coordinates is the same as <m>n</m> equations involving only numbers.  For example, the equation
    <me>x\vec{1 2 6} + y \vec{-1 -2 -1} = \vec{8 16 3}</me>
    simplifies to
    <me>\vec{ x 2x 6x} + \vec{-y -2y -y} = \vec{8 16 3}
\sptxt{or}
\spalignsysdelims()\syseq{ x - y; 2x - 2y; 6x - y} = \vec{8 16 3}.
    </me>
    For two vectors to be equal, all of their coordinates must be equal, so this is just the system of linear equations
    <me>\spalignsysdelims\{.\syseq{x - y = 8; 2x - 2y = 16; 6x - y = 3\rlap.}</me>
    </p>

    <definition>
      <idx><h>Vector equation</h><h>definition of</h></idx>
      <statement>
        <p>A <term>vector equation</term> is an equation involving a linear combination of vectors with possibly unknown coefficients.</p>
      </statement>
    </definition>
    
        <bluebox>
      <p>Asking whether or not a vector equation has a solution is the same as asking if a given vector is a linear combination of some other given vectors.  
      </p>
    </bluebox>
    <p>For example the vector equation above is asking if the vector <m>(8,16,3)</m> is a linear combination of the vectors <m>(1,2,6)</m> and <m>(-1,2,-1)</m>.</p>
    
    <p>The thing we really care about is solving systems of linear equations, not solving vector equations.  The whole point of vector equations is that they give us a different, and more geometric, way of viewing systems of linear equations.</p>

    
    

    <note hide-type="true">
      <title>A Picture of a Consistent System</title>
      <idx><h>Vector equation</h><h>consistent</h><see>System of linear equations, consistent</see></idx>
      <idx><h>Vector equation</h><h>inconsistent</h><see>System of linear equations, inconsistent</see></idx>
      <idx><h>System of linear equations</h><h>consistent</h><h>picture of</h></idx>
      <p>Below we will show that the above system of equations  is consistent.  Equivalently, this means that the above vector equation  has a solution.  In other words, there is a linear combination of <m>(1,2,6)</m> and <m>(-1,2,-1)</m> that equals <m>(8,16,3)</m>.  We can visualize the last statement geometrically.  Therefore, the following <xref ref="vectors-figure-consistent"/> gives a <em>picture of a consistent system of equations</em>.  Compare with <xref ref="vectors-fig-inconsistent"/> below, which shows a picture of an inconsistent system.
      </p>
           </note>

    
        <figure xml:id="vectors-figure-consistent">
      <caption>A picture of the above vector equation.  Try to solve the equation geometrically by moving the sliders.</caption>
      <mathbox source="demos/spans.html?v1=1,2,6&amp;v2=-1,-2,-1&amp;target=8,16,3&amp;range=20&amp;camera=3,.5,1.5" height="500px"/>
    </figure>



 


    <p>In order to actually solve the vector equation
    <me>x\textcolor{seq-red}{\vec{1 2 6}}
    + y \textcolor{seq-green}{\vec{-1 -2 -1}}
    = \textcolor{seq-blue}{\vec{8 16 3}},</me>
    one has to solve the system of linear equations
    <me>\spalignsysdelims\{.\syseq{x - y = 8; 2x - 2y = 16; 6x - y = 3\rlap.}</me>
    This means forming the augmented matrix
    <me>\def\r{\color{seq-red}}\def\g{\color{seq-green}}\def\b{\color{seq-blue}}
    \amat{\r1 \g-1 \b8; \r2 \g-2 \b16; \r6 \g-1 \b3}</me>
    and row reducing.  Note that <em>the columns of the augmented matrix are the vectors from the original vector equation</em>, so it is not actually necessary to write the system of equations: one can go directly from the vector equation to the augmented matrix by <q>smooshing the vectors together</q>.  In the following <xref ref="vectors-example-consistent"/> we carry out the row reduction and find the solution.
    </p>
    
        <example xml:id="vectors-example-consistent">
      <statement>
        <p>Is <m>\vec{ 8 16 3}</m> a linear combination of <m>\vec{1 2 6}</m> and <m>\vec{-1 -2 -1}</m>?
        </p>
      </statement>
      <solution>
      <p>As discussed above, this question boils down to a row reduction:
      <me>\amat{1 -1 8; 2 -2 16; 6 -1 3}
      \quad\xrightarrow{\text{RREF}}\quad
      \amat{1 0 -1; 0 1 -9; 0 0 0}.
      </me>
      From this we see that the equation is consistent, and the solution is <m>x=-1</m> and <m>y=-9</m>.  We conclude that  <m>\vec{ 8 16 3}</m> is indeed a linear combination of <m>\vec{1 2 6}</m> and <m>\vec{-1 -2 -1}</m>, with coefficients <m>-1</m> and <m>-9</m>:
      <me>-\vec{1 2 6} -9\vec{-1 -2 -1} = \vec{8 16 3}.
      </me>
      </p>
      </solution>
    </example>



    <bluebox>
      <title>Recipe: Solving a vector equation</title>
      <idx><h>Vector equation</h><h>solving</h></idx>
      <p>In general, the vector equation
      <me>x_1v_1 + x_2v_2 + \cdots + x_kv_k = b</me>
      where <m>v_1,v_2,\ldots,v_k,\,b</m> are vectors in <m>\R^n</m> and <m>x_1,x_2,\ldots,x_k</m> are unknown scalars, has the same solution set as the linear system with augmented matrix
      <me>\amat[c]{| | ,{} , | |; v_1 v_2 \cdots, v_k b; | | ,{} , | |}</me>
      whose columns are the <m>v_i</m><rsq/>s and the <m>b</m><rsq/>s.
      </p>
    </bluebox>

    <p>Now we have <em>three</em> equivalent ways of thinking about a linear system:
      <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>
      </ol>
    The third is geometric in nature: it lends itself to drawing pictures.  
    </p>

  </subsection>

  <subsection>
    <title>Spans</title>

    <p>It will be important to know what are <em>all</em> linear combinations of a set of vectors <m>v_1,v_2,\ldots,v_k</m> in <m>\R^n</m>.  In other words, we would like to understand the set of all vectors <m>b</m> in <m>\R^n</m> such that the vector equation (in the unknowns <m>x_1,x_2,\ldots,x_k</m>)
    <me>x_1v_1 + x_2v_2 + \cdots + x_kv_k = b</me>
    has a solution (i.e. is consistent).
    </p>

    <essential xml:id="vectors-defn-span">
      <idx><h>Span</h><h>definition of</h></idx>
      <statement>
        <p>Let <m>v_1,v_2,\ldots,v_k</m> be vectors in <m>\R^n</m>.  The <term>span</term> of <m>v_1,v_2,\ldots,v_k</m> is the collection of all linear combinations of <m>v_1,v_2,\ldots,v_k</m>, and is denoted <m>\Span\{v_1,v_2,\ldots,v_k\}</m>.  In symbols:
        <notation><usage>\Span\{v_1,v_2,\ldots,v_k\}</usage><description>Span of vectors</description></notation>
        <me>\Span\{v_1,v_2,\ldots,v_k\} =
        \bigl\{x_1v_1 + x_2v_2 + \cdots + x_kv_k \mid x_1,x_2,\ldots,x_k \text{ in }\R\bigr\}</me>
        We also say that <m>\Span\{v_1,v_2,\ldots,v_k\}</m> is the subset <term>spanned by</term> or <term>generated by</term> the vectors <m>v_1,v_2,\ldots,v_k</m>.
        </p>
      </statement>
    </essential>

    <p>The above <xref ref="vectors-defn-span"/> is the first of several <em>essential definitions</em> that we will see in this textbook.  They are essential in that they form the essence of the subject of linear algebra: learning linear algebra means (in part) learning these definitions.  All of the definitions are important, but it is essential that you learn and understand the definitions marked as such.
    </p>

    <note hide-type="true" xml:id="set-builder-notation">
      <title>Set Builder Notation</title>
      <idx><h>Set builder notation</h></idx>
      <idx><h>Subset</h><h>set builder notation</h></idx>
      <p>The notation
      <notation><usage>\{x\mid\text{condition}\}</usage><description>Set builder notation</description></notation>
      <me>\bigl\{x_1v_1 + x_2v_2 + \cdots + x_kv_k \mid x_1,x_2,\ldots,x_k \text{ in }\R\bigr\}</me>
      reads as: <q>the set of all things of the form <m>x_1v_1 + x_2v_2 + \cdots + x_kv_k</m> such that <m>x_1,x_2,\ldots,x_k</m> are in <m>\R</m>.</q>  The vertical line is <q>such that</q>; everything to the left of it is <q>the set of all things of this form</q>, and everything to the right is the condition that those things must satisfy to be in the set.  Specifying a set in this way is called <term>set builder notation</term>.
      </p>

      <p>All mathematical notation is only shorthand: any sequence of symbols must translate into a usual sentence.</p>
    </note>

    <note hide-type="true">
      <title>Three characterizations of consistency</title>
      <idx><h>System of linear equations</h><h>consistent</h><h>span criterion</h></idx>
      <p>Now we have three equivalent ways of making the same statement:
      <ol>
        <li>A vector <m>b</m> is in the span of <m>v_1,v_2,\ldots,v_k</m>.</li>
        <li>The vector equation
        <me>x_1v_1 + x_2v_2 + \cdots + x_kv_k = b</me>
        has a solution.
        </li>
        <li>The linear system with augmented matrix
        <me>\amat[c]{| | ,{} , | |; v_1 v_2 \cdots, v_k b; | | ,{} , | |}</me>
        is consistent.
        </li>
      </ol>
      </p>
    </note>

    <p><term>Equivalent</term> means that, for any given list of vectors <m>v_1,v_2,\ldots,v_k,\,b</m>, either all three statements are true, or all three statements are false.</p>

    <figure xml:id="vectors-fig-inconsistent">
      <caption>This is a picture of an <em>inconsistent</em> linear system: the vector <m>w</m> on the right-hand side of the equation <m>x_1v_1 + x_2v_2 = w</m> is not in the span of <m>v_1,v_2</m>.  Convince yourself of this by trying to solve the equation
      <m>x_1v_1 + x_2v_2 = w</m>
      by moving the sliders, and by row reduction.  Compare this <xref ref="vectors-figure-consistent"/>.</caption>
      <idx><h>System of linear equations</h><h>inconsistent</h><h>picture of</h></idx>
      <mathbox source="demos/spans.html?v1=1,2,6&amp;v2=-1,-2,-1&amp;target=2,-2,0&amp;range=8&amp;camera=3,.5,1.5" height="500px"/>
    </figure>

    <paragraphs>
      <title>Pictures of Spans</title>
      <idx><h>Span</h><h>pictures of</h></idx>

      <p>Drawing a picture of <m>\Span\{v_1,v_2,\ldots,v_k\}</m> is the same as drawing a picture of all linear combinations of <m>v_1,v_2,\ldots,v_k</m>.</p>
      <figure>
        <caption>Pictures of spans in <m>\R^2</m>.</caption>
        <image>
          <latex-image-code>
            <![CDATA[
\begin{tikzpicture}[thin border nodes, scale=.9]
\begin{scope}
  \draw[grid lines] (-3,-2) grid (4, 3);
  \path[clip] (-3,-2) rectangle (4, 3);
  \draw[thin, seq4] ($-2*(2,1)$) -- ($2*(2,1)$);
  \node[seq4,whitebg] at (0,2) {$\Span\{v\}$};
  \draw[vector, seq1] (0,0) to["$v$"] (2,1);
  \point at (0,0);
\end{scope}

\begin{scope}[xshift=8cm]
  \draw[grid lines] (-3,-2) grid (4, 3);
  \fill[seq4!30, nearly transparent] (-3,-2) rectangle (4,3);
  \node[seq4] at (.5,2.5) {$\Span\{v,w\}$};
  \draw[vector, seq1] (0,-1) to["$v$"] (1,1);
  \draw[vector, seq2] (0,-1) to["\strut$w$"'] (1,-1);
  \point at (0,-1);
\end{scope}

\begin{scope}[xshift=4cm, yshift=-7cm]
  \draw[grid lines] (-3,-3) grid (4, 4);
  \path[clip] (-3,-3) rectangle (4, 4);
  \node[seq4] at (.5,2.5) {$\Span\{v,w\}$};
  \draw[thin, seq4] ($-2*(2,2)$) -- ($2*(2,2)$);
  \draw[vector, seq1] (0,0) to["$v$"] (2,2);
  \draw[vector, seq2] (0,0) to["$w$"' pos=.4] (-1,-1);
  \point at (0,0);
\end{scope}
\end{tikzpicture}
            ]]>
          </latex-image-code>
        </image>
      </figure>

      <figure>
        <caption>Pictures of spans in <m>\R^3</m>.  The span of two noncollinear vectors is the plane containing the origin and the heads of the vectors.  Note that three coplanar (but not collinear) vectors span a plane and not a 3-space, just as two collinear vectors span a line and not a plane.</caption>
        <image>
          <latex-image-code>
            <![CDATA[
\begin{tikzpicture}[myxyz, thin border nodes, scale=.8]
\begin{scope}
  \path[clip, resetxy] (-4,-3) rectangle (4,3);

  \draw[seq4] ($-3*(-1,2,1)$) -- (0,0,0);

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

  \draw[seq4] (0,0,0) -- ($3*(-1,2,1)$);
  \node[seq4] at (-1cm, 2cm) {$\Span\{v\}$};

  \draw[vector, seq1] (0,0,0) to["$v$"] (-1,2,1);
  \draw[thin, densely dotted] (-1,2,1) -- (-1,2,0);

  \point at (0,0,0);
\end{scope}

\begin{scope}[xshift=9cm]
  \path[clip, resetxy] (-4,-3) rectangle (4,3);

  \def\v{(-1,2,1)}
  \def\w{(1,2,.3)}

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

  \begin{scope}[x=(X), y=(Y), transformxy]
    \path[clip] (-1.5, 5) -- (1.5, -5) -- (1.5, -7) -- (-1.5, -7) -- 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] (-1.5, 5) -- (1.5, -5) -- (1.5, 7) -- (-1.5, 7) -- 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}

  \node[seq4] at (-1cm, 2cm) {$\Span\{v,w\}$};

  \draw[vector, seq1] (0,0,0) --
    node [midway, above] {$v$} \v;
  \draw[thin, densely dotted] \v -- \projxy\v;
  \draw[vector, seq2] (0,0,0) --
    node [midway, below] {\strut$w$} \w;
  \draw[thin, densely dotted] \w -- \projxy\w;

  \point at (0,0,0);
\end{scope}

\begin{scope}[yshift=-7cm]
  \path[clip, resetxy] (-4,-3) rectangle (4,3);

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

  \draw[vector, seq1] (0,0,0) to["$v$"] (-1,2,1);
  \draw[thin, densely dotted] (-1,2,1) -- (-1,2,0);
  \draw[vector, seq2] (0,0,0) to["\strut$w$"'] (1,2,.3);
  \draw[thin, densely dotted] (1,2,.3) -- (1,2,0);
  \draw[vector, seq3] (0,0,0) to["$u$"] (.5,-.5,1);
  \draw[thin, densely dotted] (.5,-.5,1) -- (.5,-.5,0);

  \point at (0,0,0);

  \fill[seq4!30, nearly transparent, resetxy] (-4,-3) rectangle (4,3);
  \node[seq4] at (-1cm, 2cm) {$\Span\{u,v,w\}$};

\end{scope}

\begin{scope}[xshift=9cm, yshift=-7cm]
  \path[clip, resetxy] (-4,-3) rectangle (4,3);

  \def\v{(-1,2,1)}
  \def\w{(1,2,.3)}
  \def\u{(0,2,.65)}

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

  \begin{scope}[x=(X), y=(Y), transformxy]
    \path[clip] (-1.5, 5) -- (1.5, -5) -- (1.5, -7) -- (-1.5, -7) -- 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] (-1.5, 5) -- (1.5, -5) -- (1.5, 7) -- (-1.5, 7) -- 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}

  \node[seq4] at (-1cm, 2cm) {$\Span\{u,v,w\}$};

  \draw[vector, seq1] (0,0,0) --
    node [midway, above] {$v$} \v;
  \draw[thin, densely dotted] \v -- \projxy\v;
  \draw[vector, seq2] (0,0,0) --
    node [midway, below] {\strut$w$} \w;
  \draw[thin, densely dotted] \w -- \projxy\w;
  \draw[vector, seq3] (0,0,0) --
    \u node [right] {$u$};
  \draw[thin, densely dotted] \u -- \projxy\u;

  \draw[thin, densely dotted] \v -- \w;

  \point at (0,0,0);
\end{scope}

\end{tikzpicture}
            ]]>
          </latex-image-code>
        </image>
      </figure>
      
            <example hide-type="true">
        <title>Interactive: Span of two vectors in <m>\R^2</m></title>
        <figure>
          <caption>Interactive picture of a span of two vectors in <m>\R^2</m>.  Check <q>Show x.v + y.w</q> and move the sliders to see how every point in the violet region is in fact a linear combination of the two vectors.</caption>
          <mathbox source="demos/spans.html?v1=1,2&amp;v2=1,0&amp;showPlane=true&amp;range=5&amp;labels=v,w" height="500px"/>
        </figure>
      </example>



      <example hide-type="true">
        <title>Interactive: Span of two vectors in <m>\R^3</m></title>
        <figure>
          <caption>Interactive picture of a span of two vectors in <m>\R^3</m>.  Check <q>Show x.v + y.w</q> and move the sliders to see how every point in the violet region is in fact a linear combination of the two vectors.</caption>
          <mathbox source="demos/spans.html?v1=3,2,4&amp;v2=-4,2,1&amp;labels=v,w" height="500px"/>
        </figure>
      </example>

      <example hide-type="true">
        <title>Interactive: Span of three vectors in <m>\R^3</m></title>
        <figure>
          <caption>Interactive picture of a span of three vectors in <m>\R^3</m>.  Check <q>Show x.v + y.w + z.u</q> and move the sliders to see how every point in the violet region is in fact a linear combination of the three vectors.</caption>
          <mathbox source="demos/spans.html?labels=v,w,u" height="500px"/>
        </figure>
      </example>

    </paragraphs>

  </subsection>

</section>