dimension.xml

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

<!--********************************************************************
Copyright 2017 Georgia Institute of Technology

Permission is granted to copy, distribute and/or modify this document
under the terms of the GNU Free Documentation License, Version 1.3 or
any later version published by the Free Software Foundation.  A copy of
the license is included in gfdl.xml.
*********************************************************************-->

<section xml:id="dimension">
  <title>Basis and Dimension</title>

  <objectives>
    <ol>
      <li>Understand the definition of a basis of a subspace.</li>
      <li>Understand the basis theorem.</li>
      <li><em>Recipes:</em> basis for a column space, basis for a null space, basis of a span.</li>
      <li><em>Picture:</em> basis of a subspace of <m>\R^2</m> or <m>\R^3</m>.</li>
      <li><em>Theorem:</em> basis theorem.</li>
      <li><em>Essential vocabulary words:</em> <term>basis</term>, <term>dimension</term>.</li>
    </ol>
  </objectives>

  <subsection>
    <title>Basis of a Subspace</title>

    <p>
      As we discussed in <xref ref="subspaces"/>, a subspace is the same as a span, except we do not have a set of spanning vectors in mind.  There are infinitely many choices of spanning sets for a nonzero subspace; to avoid reduncancy, usually it is most convenient to choose a spanning set with the <em>minimal</em> number of vectors in it.  This is the idea behind the notion of a basis.
    </p>

    <essential xml:id="dimension-defn-basis">
      <idx><h>Basis</h><h>definition of</h></idx>
      <statement>
        <p>Let <m>V</m> be a subspace of <m>\R^n</m>.  A <term>basis</term> of <m>V</m> is a set of vectors <m>\{v_1,v_2,\ldots,v_m\}</m> in <m>V</m> such that:
        <ol>
          <li><m>V = \Span\{v_1,v_2,\ldots,v_m\}</m>, and</li>
          <li>the set <m>\{v_1,v_2,\ldots,v_m\}</m> is linearly independent.</li>
        </ol>
        </p>
      </statement>
    </essential>

    <p>
      Recall that a set of vectors is <em>linearly independent</em> if and only if, when you remove any vector from the set, the span shrinks (<xref ref="linindep-criterion-inspan" text="type-global"/>).  In other words, if <m>\{v_1,v_2,\ldots,v_m\}</m> is a basis of a subspace <m>V</m>, then no proper subset of <m>\{v_1,v_2,\ldots,v_m\}</m> will span <m>V</m>: it is a <em>minimal</em> spanning set.  Any subspace admits a basis by this <xref ref="subspaces-spans-are-subspaces"/>.
    </p>

    <bluebox>
      <idx><h>Basis</h><h>infinitely many</h></idx>
      <p>A nonzero subspace has <em>infinitely many</em> different bases, but they all contain the same number of vectors.</p>
    </bluebox>

    <p>We leave it as an exercise to prove that any two bases have the same number of vectors; one might want to wait until after learning the invertible matrix theorem in <xref ref="matrix-inverses"/>.</p>

    <essential xml:id="dimension-defn-dimension">
      <idx><h>Dimension</h><h>definition of</h></idx>
      <notation><usage>\dim V</usage><description>Dimension of a subspace</description></notation>
      <statement>
        <p>
          Let <m>V</m> be a subspace of <m>\R^n</m>.  The number of vectors in any basis of <m>V</m> is called the <term>dimension</term> of <m>V</m>, and is written <m>\dim V</m>.
        </p>
      </statement>
    </essential>

    <example xml:id="dimension-eg-R2">
      <title>A basis of <m>\R^2</m></title>
      <statement>
        <p>Find a basis of <m>\R^2</m>.</p>
      </statement>

      <solution>
        <p>
          We need to find two vectors in <m>\R^2</m> that span <m>\R^2</m> and are linearly independent.  One such basis is <m>\bigl\{{1\choose 0},{0\choose 1}\bigr\}</m>:
          <ol>
            <li>They span because any vector <m>a\choose b</m> can be written as a linear combination of <m>{1\choose 0},{0\choose 1}</m>:
            <me>\vec{a b} = a\vec{1 0} + b\vec{0 1}.</me>
            </li>
            <li>They are linearly independent: if
            <me>x\vec{1 0} + y\vec{0 1} = \vec{x y} = \vec{0 0}</me>
            then <m>x=y=0</m>.
            </li>
          </ol>
          This shows that the plane <m>\R^2</m> has dimension 2.
          <latex-code>
\begin{tikzpicture}
  \draw[grid lines] (-2,-2) grid (2,2);
  \draw[thick vector] (0,0) -- (1,0) node[right,whitebg] {$1\choose 0$};
  \draw[thick vector] (0,0) -- (0,1) node[above,whitebg] {$0\choose 1$};
  \point at (0,0);
\end{tikzpicture}
          </latex-code>
        </p>
      </solution>
    </example>

    <example>
      <title>All bases of <m>\R^2</m></title>
      <statement>
        <p>Find all bases of <m>\R^2</m>.</p>
      </statement>

      <solution>
        <p>
          We know from the previous <xref ref="dimension-eg-R2"/> that <m>\R^2</m> has dimension 2, so any basis of <m>\R^2</m> has two vectors in it.  Let <m>v_1,v_2</m> be vectors in <m>\R^2</m>, and let <m>A</m> be the matrix with columns <m>v_1,v_2</m>.
          <ol>
            <li>To say that <m>\{v_1,v_2\}</m> spans <m>\R^2</m> means that <m>A</m> has a pivot in every <em>row</em>: see this <xref ref="matrixeq-thm-full-span"/>.</li>
            <li>To say that <m>\{v_1,v_2\}</m> is linearly independent means that <m>A</m> has a pivot in every <em>column</em>: see this <xref ref="linindep-matrix-cols"/>.
            </li>
          </ol>
          Since <m>A</m> is a <m>2\times 2</m> matrix, it has a pivot in every row exactly when it has a pivot in every column.  Hence any two noncollinear vectors form a basis of <m>\R^2</m>.  For example,
          <me>\left\{\vec{1 0},\;\vec{1 1}\right\}</me>
          is a basis.
          <latex-code>
\begin{tikzpicture}
  \draw[grid lines] (-2,-2) grid (2,2);
  \draw[thick vector] (0,0) -- (1,0) node[anchor=west,whitebg] {$1\choose 0$};
  \draw[thick vector] (0,0) -- (1,1) node[anchor=south,whitebg] {$1\choose 1$};
  \point at (0,0);
\end{tikzpicture}
          </latex-code>
        </p>
      </solution>
    </example>

    <example xml:id="dimension-eg-Rn">
      <title>The standard basis of <m>\R^n</m></title>
      <idx><h>Basis</h><h>of <m>\R^n</m></h></idx>
      <p>
        One shows exactly as in the above <xref ref="dimension-eg-R2"/> that the standard coordinate vectors
        <me>e_1=\vec{1 0 \vdots, 0 0},\quad e_2=\vec{0 1 \vdots, 0 0}, \quad\ldots,\quad
        e_{n-1}=\vec{0 0 \vdots, 1 0},\quad e_n=\vec{0 0 \vdots, 0 1}
        </me>
        form a basis for <m>\R^n</m>.  This is sometimes known as the <term>standard basis</term>.
      </p>

      <p>In particular, <m>\R^n</m> has dimension <m>n</m>.</p>
    </example>

    <specialcase xml:id="dimension-eg-Rn-any">
      <idx><h>Basis</h><h>of <m>\R^n</m></h></idx>
      <p>
        The previous <xref ref="dimension-eg-Rn"/> implies that any basis for <m>\R^n</m> has <m>n</m> vectors in it.  Let <m>v_1,v_2,\ldots,v_n</m> be vectors in <m>\R^n</m>, and let <m>A</m> be the <m>n\times n</m> matrix with columns <m>v_1,v_2,\ldots,v_n</m>.
          <ol>
            <li>To say that <m>\{v_1,v_2,\ldots,v_n\}</m> spans <m>\R^n</m> means that <m>A</m> has a <xref ref="defn-pivot-pos" text="title">pivot position</xref> in every <em>row</em>: see this <xref ref="matrixeq-thm-full-span"/>.</li>
            <li>To say that <m>\{v_1,v_2,\ldots,v_n\}</m> is linearly independent means that <m>A</m> has a pivot position in every <em>column</em>: see this <xref ref="linindep-matrix-cols"/>.
            </li>
          </ol>
          Since <m>A</m> is a square matrix, it has a pivot in every row if and only if it has a pivot in every column.
          We will see in <xref ref="matrix-inverses"/> that the above two conditions are equivalent to the <em>invertibility</em> of the matrix <m>A</m>.
      </p>
    </specialcase>

    <example>
      <statement>
        <p>Let
        <me>V = \left\{ \vec{x y z} \text{ in } \R^3\bigm| x + 3y + z = 0 \right\} \qquad
        \cB = \left\{ \vec{-3 1 0},\;\vec{0 1 -3} \right\}.</me>
        Verify that <m>V</m> is a subspace, and show directly that <m>\cB</m> is a basis for <m>V</m>.
        </p>
      </statement>
      <solution>
        <p>
          First we observe that <m>V</m> is the solution set of the homogeneous equation <m>x + 3y + z = 0</m>, so it is a subspace: see this <xref ref="subspace-is-col-or-nul"/>.  To show that <m>\cB</m> is a basis, we really need to verify three things:
          <ol start="0">
            <li>Both vectors are in <m>V</m> because
            <me>\spalignsysdelims..
            \syseq{
            (-3) + 3(1) + (0) = 0;
            (0)  + 3(1) + (-3) = 0\rlap.}
            </me>
            </li>
            <li>
              <em>Span:</em> suppose that <m>\vec{x y z}</m> is in <m>V</m>.  Since <m>x + 3y + z = 0</m> we have <m>y = -\frac 13(x+z)</m>, so
              <me>
                \vec{x y z} = \vec{x -\frac 13(x+z) z}
                = -\frac x3\vec{-3 1 0} - \frac z3\vec{0 1 -3}.
              </me>
              Hence <m>\cB</m> spans <m>V</m>.
            </li>
            <li>
              <em>Linearly independent:</em>
              <me>c_1\vec{-3 1 0} + c_2\vec{0 1 -3} = 0
    \implies \vec{-3c_1 c_1+c_2 -3c_2} = \vec{0 0 0}
              \implies c_1 = c_2 = 0.</me>
              Alternatively, one can observe that the two vectors are not collinear.
            </li>
          </ol>
          Since <m>V</m> has a basis with two vectors, it has dimension two: it is a <em>plane</em>.
        </p>
        <figure xml:id="dimension-figure-basis">
          <caption>A picture of the plane <m>V</m> and its basis <m>\cB = \{v_1,v_2\}</m>.  Note that <m>\cB</m> spans <m>V</m> and is linearly independent.</caption>
          <mathbox source="demos/spans.html?v1=-3,1,0&amp;v2=0,1,-3&amp;range=5&amp;captions=indep" height="500px"/>
        </figure>
        <p>
          This example is somewhat contrived, in that we will learn systematic methods for verifying that a subset is a basis.  The intention is to illustrate the defining properties of a basis.
        </p>
      </solution>
    </example>

  </subsection>

  <subsection>
    <title>Computing a Basis for a Subspace</title>

    <p>
      Now we show how to find bases for the column space of a matrix and the null space of a matrix.  In order to find a basis for a given subspace, it is usually best to rewrite the subspace as a column space or a null space first: see this <xref ref="subspace-is-col-or-nul"/>.
    </p>

    <paragraphs>
      <title>A basis for the column space</title>

      <p>First we show how to compute a basis for the column space of a matrix.</p>

    <theorem xml:id="dimension-basis-colspace">
      <idx><h>Basis</h><h>of a column space</h></idx>
      <idx><h>Column space</h><h>basis of</h><see>Basis</see></idx>
      <statement>
        <p>The pivot columns of a matrix <m>A</m> form a basis for <m>\Col(A)</m>.</p>
      </statement>
      <proof>
        <p>This is a restatement of a <xref ref="linindep-pivot-cols"/>.</p>
      </proof>
    </theorem>

    <bluebox>
      <p>
        The above theorem is referring to the pivot columns in the <em>original</em> matrix, not its reduced row echelon form.  Indeed, a matrix and its reduced row echelon form generally have different column spaces.  For example, in the matrix <m>A</m> below:
        <latex-code>
            <![CDATA[
\begin{tikzpicture}
  \tikzset{
    my matrix/.style={
      matrix, math matrix,
      column sep={2.2em,between origins},
      every node/.append style={anchor=base east}},
  }
  \node (symb) {$A=\;\;\null$};
  \path (symb.east) node[my matrix, right] (A) {
    1  \&  2 \& 0 \& -1 \\
    -2 \& -3 \& 4 \&  5 \\
    2  \&  4 \& 0 \& -2 \\
  } (A.east) ++(2cm,0) node {$\xrightarrow{\text{RREF}}$}
    ++(2cm, 0) node[my matrix, right] (B) {
    1 \& 0 \& -8 \& -7 \\
    0 \& 1 \&  4 \&  3 \\
    0 \& 0 \&  0 \&  0 \\
  };
  \node[fit=(A-1-1) (A-3-1), draw=orange, rounded corners] (box1) {};
  \node[fit=(A-1-2) (A-3-2), draw=orange, rounded corners] (box2) {};
  \path ($(box1.south)!.5!(box2.south)$) ++(0,-1cm)
      node[blue!50, font=\small,anchor=base] (orig) {pivot columns $=$ basis};
  \draw[->,blue!50, shorten >=1pt] (orig.north) to[out=90,in=-90] (box1.south);
  \draw[->,blue!50, shorten >=1pt] (orig.north) to[out=90,in=-90] (box2.south);
  \path let \p1=($(B-3-1.south)!.5!(B-3-2.south)$) in (\p1 |- orig.base)
      node[blue!50, font=\small,anchor=base] (rref) {pivot columns in RREF};
  \draw[->,blue!50, shorten >=1pt] (rref.north) to[out=90,in=-90] (B-3-1.south);
  \draw[->,blue!50, shorten >=1pt] (rref.north) to[out=90,in=-90] (B-3-2.south);
  \ifpdfvers
    \draw[->] (rref.west) -- (orig.east);
  \else
    \draw[->,
        decoration={snake, amplitude=.4mm, segment length=1mm, post length=.5mm},
        decorate] (rref.west) -- (orig.east);
  \fi

\end{tikzpicture}
            ]]>
        </latex-code>
        the pivot columns are the first two columns, so a basis for <m>\Col(A)</m> is
        <me>\left\{\vec{1 -2 2},\;\vec{2 -3 4}\right\}.</me>
        The first two columns of the reduced row echelon form certainly span a different subspace, as
        <me>\Span\left\{\vec{1 0 0},\;\vec{0 1 0}\right\}
        = \left\{\vec{a b 0}\Bigm|a, b \text{ in } \R\right\}
        = \text{($xy$-plane)},
        </me>
        but <m>\Col(A)</m> contains vectors whose last coordinate is nonzero.
      </p>
    </bluebox>

    <corollary xml:id="dimension-basis-colspace-dim">
      <statement>
        <p>The dimension of <m>\Col(A)</m> is the number of pivots of <m>A</m>.</p>
      </statement>
    </corollary>

    </paragraphs>

    <paragraphs>
      <title>A basis of a span</title>

    <p>
      <idx><h>Basis</h><h>of a span</h></idx>
      <idx><h>Span</h><h>basis of</h><see>Basis</see></idx>
      Computing a basis for a span is the same as computing a basis for a column space.  Indeed, the span of finitely many vectors <m>v_1,v_2,\ldots,v_m</m> <em>is</em> the column space of a matrix, namely, the matrix <m>A</m> whose columns are <m>v_1,v_2,\ldots,v_m</m>:
      <me>A = \mat{| | ,, |; v_1 v_2 \cdots, v_m; | | ,, |}.</me>
    </p>

    <example xml:id="dimension-eg-basis-span">
      <title>A basis of a span</title>
      <statement>
        <p>
          Find a basis of the subspace
          <me>V = \Span\left\{\vec{1 -2 2},\;\vec{2 -3 4},\;\vec{0 4 0},\;\vec{-1 5 -2}\right\}.</me>
        </p>
      </statement>
      <solution>
        <p>
          The subspace <m>V</m> is the column space of the matrix
          <me>A = \mat[r]{1 2 0 -1; -2 -3 4 5; 2 4 0 -2}.</me>
          The reduced row echelon form of this matrix is
          <me>\mat[r]{1 0 -8 -7; 0 1 4 3; 0 0 0 0}.</me>
          The first two columns are pivot columns, so a basis for <m>V</m> is
          <me>\left\{\vec{1 -2 2},\;\vec{2 -3 4}\right\}.</me>
        </p>
        <figure xml:id="dimension-figure-basis-2">
          <caption>A picture of the plane <m>V</m> and its basis <m>\cB = \{v_1,v_2\}</m>.</caption>
          <mathbox source="demos/spans.html?v1=1,-2,2&amp;v2=2,-3,4&amp;range=6&amp;captions=indep" height="500px"/>
        </figure>
      </solution>
    </example>

    <example xml:id="dimension-eg-basis-span2">
      <title>Another basis of the same span</title>
      <statement>
        <p>
          Find a basis of the subspace
          <me>V = \Span\left\{\vec{1 -2 2},\;\vec{2 -3 4},\;\vec{0 4 0},\;\vec{-1 5 -2}\right\}</me>
          which does not consist of the first two vectors, as in the previous <xref ref="dimension-eg-basis-span"/>.
        </p>
      </statement>
      <solution>
        <p>
          The point of this example is that the above <xref ref="dimension-basis-colspace"/> gives <em>one</em> basis for <m>V</m>; as always, there are infinitely more.
        </p>
        <p>
          Reordering the vectors, we can express <m>V</m> as the column space of
          <me>A' = \mat[r]{0 -1 1 2; 4 5 -2 -3; 0 -2 2 4}.</me>
          The reduced row echelon form of this matrix is
          <me>\mat{1 0 3/4 7/4; 0 1 -1 -2; 0 0 0 0}.</me>
          The first two columns are pivot columns, so a basis for <m>V</m> is
          <me>\left\{\vec{0 4 0},\;\vec{-1 5 -2}\right\}.</me>
          These are the <em>last</em> two vectors in the given spanning set.
        </p>
        <figure xml:id="dimension-figure-basis-3">
          <caption>A picture of the plane <m>V</m> and its basis <m>\cB = \{v_1,v_2\}</m>.</caption>
          <mathbox source="demos/spans.html?v1=0,4,0&amp;v2=-1,5,-2&amp;range=6&amp;captions=indep" height="500px"/>
        </figure>
      </solution>
    </example>

    </paragraphs>

    <paragraphs>
      <title>A basis for the null space </title>

    <p>In order to compute a basis for the null space of a matrix, one has to find the parametric vector form of the solutions of the homogeneous equation <m>Ax=0</m>.</p>

    <theorem xml:id="dimension-basis-nulspace">
      <idx><h>Basis</h><h>of a null space</h></idx>
      <idx><h>Null space</h><h>basis of</h><see>Basis</see></idx>
      <statement>
        <p>The vectors attached to the free variables in the parametric vector form of the solution set of <m>Ax=0</m> form a basis of <m>\Nul(A)</m>.</p>
      </statement>
    </theorem>

    <p>The proof of the theorem has two parts.  The first part is that every solution lies in the span of the given vectors.  This is automatic: the vectors are exactly chosen so that every solution is a linear combination of those vectors.  The second part is that the vectors are linearly independent.  This part was discussed in this <xref ref="param-vect-form-lin-ind"/>.</p>

    </paragraphs>

    <paragraphs>
      <title>A basis for a general subspace</title>

      <p>
        As mentioned at the beginning of this subsection, when given a subspace written in a different form, in order to compute a basis it is usually best to rewrite it as a column space or null space of a matrix.
      </p>

      <example>
        <title>A basis of a subspace</title>
        <statement>
          <p>
            Let <m>V</m> be the subspace defined by
            <me>
              V = \left\{\vec{x y z}\bigm| x + 2y = z\right\}.
            </me>
            Find a basis for <m>V</m>.  What is <m>\dim(V)</m>?
          </p>
        </statement>
        <solution>
          <p>
            First we notice that <m>V</m> is exactly the solution set of the homogeneous linear equation <m>x + 2y - z = 0</m>.  Hence <m>V = \Nul\mat{1 2 -1}.</m>  This matrix is in reduced row echelon form; the parametric form of the general solution is <m>x = -2y + z</m>, so the parametric vector form is
            <me>\vec{x y z} = y\vec{-2 1 0} + z\vec{1 0 1}.</me>
            It follows that a basis is
            <me>\left\{\vec{-2 1 0},\;\vec{1 0 1}\right\}.</me>
            Since <m>V</m> has a basis with two vectors, its dimension is <m>2</m>: it is a plane.
          </p>
        </solution>
      </example>

    </paragraphs>

  </subsection>

  <subsection>
    <title>The Basis Theorem</title>

    <p>
      Recall that <m>\{v_1,v_2,\ldots,v_n\}</m> forms a basis for <m>\R^n</m> if and only if the matrix <m>A</m> with columns <m>v_1,v_2,\ldots,v_n</m> has a pivot in every row and column (see this <xref ref="dimension-eg-Rn-any"/>).  Since <m>A</m> is an <m>n\times n</m> matrix, these two conditions are equivalent: the vectors span if and only if they are linearly independent.
      The basis theorem is an abstract version of the preceding statement, that applies to any subspace.
    </p>

    <theorem type-name="Basis Theorem" xml:id="basis-theorem">
      <idx><h>Basis</h><h>basis theorem</h></idx>
      <statement>
        <p>
          Let <m>V</m> be a subspace of dimension <m>m</m>.  Then:
          <ul>
            <li>Any <m>m</m> linearly independent vectors in <m>V</m> form a basis for <m>V</m>.</li>
            <li>Any <m>m</m> vectors that span <m>V</m> form a basis for <m>V</m>.</li>
          </ul>
        </p>
      </statement>
      <proof>
        <p>
          Suppose that <m>\cB = \{v_1,v_2,\ldots,v_m\}</m> is a set of linearly independent vectors in <m>V</m>.  In order to show that <m>\cB</m> is a basis for <m>V</m>, we must prove that <m>V = \Span\{v_1,v_2,\ldots,v_m\}.</m>  If not, then there exists some vector <m>v_{m+1}</m> in <m>V</m> that is not contained in <m>\Span\{v_1,v_2,\ldots,v_m\}.</m>  By the <xref ref="linindep-increasing-span">increasing span criterion</xref>, the set <m>\{v_1,v_2,\ldots,v_m,v_{m+1}\}</m> is also linearly independent.  Continuing in this way, we keep choosing vectors until we eventually do have a linearly independent spanning set: say <m>V = \Span\{v_1,v_2,\ldots,v_m,\ldots,v_{m+k}\}</m>.  Then <m>\{v_1,v_2,\ldots,v_{m+k}\}</m> is a basis for <m>V</m>, which implies that <m>\dim(V) = m+k > m</m>.  But we were assuming that <m>V</m> has dimension <m>m</m>, so <m>\cB</m> must have already been a basis.
        </p>

        <p>
          Now suppose that <m>\cB = \{v_1,v_2,\ldots,v_m\}</m> spans <m>V</m>.  If <m>\cB</m> is not linearly independent, then by this <xref ref="linindep-criterion-inspan"/>, we can remove some number of vectors from <m>\cB</m> without shrinking its span.  After reordering, we can assume that we removed the last <m>k</m> vectors without shrinking the span, and that we cannot remove any more. Now <m>V = \Span\{v_1,v_2,\ldots,v_{m-k}\}</m>, and <m>\{v_1,v_2,\ldots,v_{m-k}\}</m> is a basis for <m>V</m> because it is linearly independent.  This implies that <m>\dim V=m-k &lt; m</m>.  But we were assuming that <m>\dim V = m</m>, so <m>\cB</m> must have already been a basis.
        </p>
      </proof>
    </theorem>

    <p>
      In other words, if you <em>already</em> know that <m>\dim V = m</m>, and if you have a set of <m>m</m> vectors <m>\cB = \{v_1,v_2,\ldots,v_m\}</m> in <m>V</m>, then you only have to check <em>one</em> of:
      <ol>
        <li><m>\cB</m> is linearly independent, <em>or</em></li>
        <li><m>\cB</m> spans <m>V</m>,</li>
      </ol>
      in order for <m>\cB</m> to be a basis of <m>V</m>.  If you did not already know that <m>\dim V = m</m>, then you would have to check <em>both</em> properties.
    </p>

	<p>
	To put it yet another way, suppose we have a set of vectors <m>\cB  = \{v_1,v_2,\ldots,v_m\}</m> in a subspace <m>V</m>.  Then if any two of the following statements is true, the third must also be true:
       <ol>
        <li><m>\cB</m> is linearly independent,</li>
        <li><m>\cB</m> spans <m>V</m>, and</li>
	<li><m>\dim V = m.</m></li>
      </ol>
	</p>



    <p>For example, if <m>V</m> is a plane, then any two noncollinear vectors in <m>V</m> form a basis.</p>

    <example>
      <title>Two noncollinear vectors form a basis of a plane</title>
      <statement>
        <p>
          Find a basis of the subspace
          <me>V = \Span\left\{\vec{1 -2 2},\;\vec{2 -3 4},\;\vec{0 4 0},\;\vec{-1 5 -2}\right\}</me>
          which is different from the bases in this <xref ref="dimension-eg-basis-span"/> and this <xref ref="dimension-eg-basis-span2"/>.
        </p>
      </statement>
      <solution>
        <p>
          We know from the previous examples that <m>\dim V = 2</m>.  By the <xref ref="basis-theorem"/>, it suffices to find any two noncollinear vectors in <m>V</m>.  We write two linear combinations of the four given spanning vectors, chosen at random:
          <me>
            w_1 = \vec{1 -2 2} + \vec{2 -3 4} = \vec{3 -5 6} \qquad
            w_2 = -\vec{2 -3 4} + \frac 12\vec{0 4 0} = \vec{-2 5 -4}.
          </me>
          Since <m>w_1,w_2</m> are not collinear, <m>\cB = \{w_1,w_2\}</m> is a basis for <m>V</m>.
        </p>
        <figure xml:id="dimension-figure-basis-4">
          <caption>A picture of the plane <m>V</m> and its basis <m>\cB = \{w_1,w_2\}</m>.</caption>
          <mathbox source="demos/spans.html?v1=3,-5,6&amp;v2=-2,5,-4&amp;range=6&amp;captions=indep&amp;labels=w1,w2" height="500px"/>
        </figure>
      </solution>
    </example>

    <example>
      <title>Finding a basis by inspection</title>
      <statement>
        <p>
          Find a basis for the plane
          <me>V = \left\{\vec{x_1 x_2 x_3}\bigm| x_1 + x_2 = x_3\right\}</me>
          by inspection.  (This plane is expressed in <xref ref="set-builder-notation" text="title">set builder notation</xref>.)
        </p>
      </statement>
      <solution>
        <p>
          First note that <m>V</m> is the null space of the matrix <m>\mat{1 1 -1}</m>; this matrix is in reduced row echelon form and has two free variables, so <m>V</m> is indeed a plane.  We write down two vectors satisfying <m>x_1 + x_2 = x_3</m>:
          <me>
            v_1 = \vec{1 0 1} \qquad v_2 = \vec{0 1 1}.
          </me>
          Since <m>v_1</m> and <m>v_2</m> are not collinear, they are linearly independent; since <m>\dim(V) = 2</m>, the basis theorem implies that <m>\{v_1,v_2\}</m> is a basis for <m>V</m>.
        </p>
      </solution>
    </example>

  </subsection>


</section>