b-coordinates.xml

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<restrict-version versions="1554 default">
<section xml:id="bases-as-coord-systems">
  <title>Bases as Coordinate Systems</title>

  <objectives>
    <ol>
      <li>Learn to view a basis as a coordinate system on a subspace.</li>
      <li><em>Recipes:</em> compute the <m>\cB</m>-coordinates of a vector, compute the usual coordinates of a vector from its <m>\cB</m>-coordinates.</li>
      <li><em>Picture:</em> the <m>\cB</m>-coordinates of a vector using its location on a nonstandard coordinate grid.</li>
      <li><em>Vocabulary word:</em> <term><m>\cB</m>-coordinates</term>.</li>
    </ol>
  </objectives>

  <p>
    In this section, we interpret a basis of a subspace <m>V</m> as a <em>coordinate system</em> on <m>V</m>, and we learn how to write a vector in <m>V</m> in that coordinate system.
  </p>

  <fact xml:id="dimension-unique-coeff">
    <idx><h>Basis</h><h>uniqueness with respect to</h></idx>
    <idx><h>Basis</h><h>coordinates with respect to</h><see><m>\cB</m>-coordinates</see></idx>
    <statement>
      <p>
        If <m>\cB = \{v_1,v_2,\ldots,v_m\}</m> is a basis for a subspace <m>V</m>, then any vector <m>x</m> in <m>V</m> can be written as a linear combination
        <me>x = c_1v_1 + c_2v_2 + \cdots + c_mv_m</me>
        in exactly one way.
      </p>
    </statement>
    <proof>
      <p>
        Recall that to say <m>\cB</m> is a <em>basis</em> for <m>V</m> means that <m>\cB</m> spans <m>V</m> and <m>\cB</m> is linearly independent.  Since <m>\cB</m> spans <m>V</m>, we can write any <m>x</m> in <m>V</m> as a linear combination of <m>v_1,v_2,\ldots,v_m</m>.  For uniqueness, suppose that we had two such expressions:
        <md>
          <mrow>x &amp;= c_1v_1 + c_2v_2 + \cdots + c_mv_m</mrow>
          <mrow>x &amp;= c_1'v_1 + c_2'v_2 + \cdots + c_m'v_m.</mrow>
        </md>
        Subtracting the first equation from the second yields
        <me>0 = x-x = (c_1-c_1')v_1 + (c_2-c_2')v_2 + \cdots + (c_m-c_m')v_m.</me>
        Since <m>\cB</m> is linearly independent, the only solution to the above equation is the trivial solution: all the coefficients must be zero.  It follows that <m>c_i-c_i'</m> for all <m>i</m>, which proves that <m>c_1=c_1',\,c_2=c_2',\ldots,\,c_m=c_m'</m>.
      </p>
    </proof>
  </fact>

  <specialcase>
    <p>
      Consider the standard basis of <m>\R^3</m> from this <xref ref="dimension-eg-Rn"/>:
      <me>e_1=\vec{1 0 0},\quad e_2=\vec{0 1 0}, \quad e_3=\vec{0 0 1}.</me>
      According to the above <xref ref="dimension-unique-coeff"/>, every vector in <m>\R^3</m> can be written as a linear combination of <m>e_1,e_2,e_3</m>, with unique coefficients.  For example,
      <me>v = \vec{\color{seq1}3 \color{seq2}5 \color{seq3}{-2}}
      = \textcolor{seq1}{3}\vec{1 0 0} + \textcolor{seq2}{5}\vec{0 1 0}
      \textcolor{seq3}{{}- 2}\vec{0 0 1}
      = \textcolor{seq1}{3}e_1 + \textcolor{seq2}{5}e_2
      \textcolor{seq3}{{}- 2}e_3.</me>
      In this case, <em>the coordinates of <m>v</m> are exactly the coefficients of <m>e_1,e_2,e_3</m>.</em>
    </p>
  </specialcase>

  <p>
    What exactly are coordinates, anyway?  One way to think of coordinates is that they give directions for how to get to a certain point from the origin.  In the above example, the linear combination <m>3e_1+5e_2-2e_3</m> can be thought of as the following list of instructions: start at the origin, travel <m>3</m> units north, then travel <m>5</m> units east, then <m>2</m> units down.
  </p>

  <definition xml:id="dimension-defn-Bcoords">
    <idx><h><m>\cB</m>-coordinates</h><h>definition of</h></idx>
    <statement>
      <p>
        Let <m>\cB=\{v_1,v_2,\ldots,v_m\}</m> be a basis of a subspace <m>V</m>, and let
        <me>x = c_1v_1 + c_2v_2 + \cdots + c_mv_m</me>
        be a vector in <m>V</m>.  The coefficients <m>c_1,c_2,\ldots,c_m</m> are the <term>coordinates of <m>x</m> with respect to <m>\cB</m></term>.  The <term><m>\cB</m>-coordinate vector of <m>x</m></term> is the vector
        <me>[x]_\cB = \vec{c_1 c_2 \vdots, c_m} \quad\text{in }\R^m.</me>
      </p>
    </statement>
  </definition>

  <p>
    <idx><h><m>\cB</m>-coordinates</h><h>informally</h></idx>
    If we change the basis, then we can still give instructions for how to get to the point <m>(3,5,-2)</m>, but the instructions will be different. Say for example we take the basis
    <me>v_1=e_1+e_2 = \vec{1,1,0}, \quad v_2=e_2 = \vec{0 1 0},\quad v_3 =e_3 = \vec{0 0 1}.</me>
    We can write <m>(3,5,-2)</m> in this basis as <m>3v_1+2v_2-2v_3</m>. In other words: start at the origin, travel northeast <m>3</m> times as far as <m>v_1</m>, then <m>2</m> units east, then <m>2</m> units down.
    In this situation, we can say that <q><m>3</m> is the <m>v_1</m>-coordinate of <m>(3,5,-2)</m>, <m>2</m> is the <m>v_2</m>-coordinate of <m>(3,5,-2)</m>, and <m>-2</m> is the <m>v_3</m>-coordinate of <m>(3,5,-2)</m>.</q>
  </p>

  <bluebox>
    <idx><h><m>\cB</m>-coordinates</h><h>labeling points</h></idx>
    <p>
      The above <xref ref="dimension-defn-Bcoords"/> gives a way of using <m>\R^m</m> to <em>label</em> the points of a subspace of dimension <m>m</m>: a point is simply labeled by its <m>\cB</m>-coordinate vector.  For instance, if we choose a basis for a plane, we can label the points of that plane with the points of <m>\R^2</m>.
    </p>
  </bluebox>

  <example>
    <title>A nonstandard coordinate system on <m>\R^2</m></title>
    <statement>
      <p>
        Define
        <me>v_1 = \vec{1 1},\; v_2 = \vec{1 -1},\quad \cB = \{v_1,\,v_2\}.</me>
        <ol>
          <li>Verify that <m>\cB</m> is a basis for <m>\R^2</m>.</li>
          <li>If <m>[w]_\cB = {1\choose 2}</m>, then what is <m>w</m>?</li>
          <li>Find the <m>\cB</m>-coordinates of <m>v = {5\choose 3}.</m>
          </li>
        </ol>
      </p>
    </statement>
    <solution>
      <p>
        <ol>
          <li>
            By the <xref ref="basis-theorem"/>, any two linearly independent vectors form a basis for <m>\R^2</m>.  Clearly <m>v_1,v_2</m> are not multiples of each other, so they are linearly independent.
          </li>
          <li>
            To say <m>[w]_\cB={1\choose 2}</m> means that <m>1</m> is the <m>v_1</m>-coordinate of <m>w</m>, and that <m>2</m> is the <m>v_2</m>-coordinate:
            <me>w = v_1 + 2v_2 = \vec{1 1} + 2\vec{1 -1} = \vec{3 -1}.</me>
          </li>
          <li>
            We have to solve the vector equation <m>v=c_1v_1 + c_2v_2</m> in the unknowns <m>c_1,c_2</m>.  We form an augmented matrix and row reduce:
            <me>\amat{1 1 5; 1 -1 3}
            \quad\xrightarrow{\text{RREF}}\quad
            \amat{1 0 4; 0 1 1}.</me>
            We have <m>c_1=4</m> and <m>c_2=1</m>, so <m>v = 4v_1+v_2</m> and
            <m>[v]_\cB = {4\choose1}.</m>
          </li>
        </ol>
        In the following picture, we indicate the coordinate system defined by <m>\cB</m> by drawing lines parallel to the <q><m>v_1</m>-axis</q> and <q><m>v_2</m>-axis</q>.  Using this grid it is easy to see that the <m>\cB</m>-coordinates of <m>v</m> are <m>{5\choose 1}</m>, and that the <m>\cB</m>-coordinates of <m>w</m> are <m>{1\choose 2}</m>.
        <latex-code>
\begin{tikzpicture}[scale=.7, thin border nodes]

  \draw (-6, -6) rectangle (6, 6);
  \clip (-6, -6) rectangle (6, 6);

  \draw[scale={sqrt(2)}, rotate=45, help lines] (-7, -7) grid (7, 7);
  \draw[vector, seq-blue] (0, 0) to["$v_1$" above left] (1, 1);
  \draw[vector, seq-green] (0, 0) to["$v_2$" below left] (1, -1);

  \point[seq-red, "$w$" {above,text=seq-red}] at (3, -1);
  \point[seq-orange, "$v$" {above,text=seq-orange}] at (5, 3);

  \point at (0, 0);

\end{tikzpicture}
        </latex-code>
        This picture could be the grid of streets in <url href="https://www.google.com/maps/place/Palo+Alto,+CA/@37.4387098,-122.1548738,15.75z/data=!4m5!3m4!1s0x808fb07b9dba1c39:0xe1ff55235f576cf!8m2!3d37.4418834!4d-122.1430195?hl=en">Palo Alto, California</url>.  Residents of Palo Alto refer to northwest as <q>north</q> and to northeast as <q>east</q>.  There is a reason for this: the old road to San Francisco is called El Camino Real, and that road runs from the southeast to the northwest in Palo Alto.  So when a Palo Alto resident says <q>go south two blocks and east one block</q>, they are giving directions from the origin to the Whole Foods at <m>w</m>.
      </p>
      <figure>
        <caption>A picture of the basis <m>\cB = \{v_1,v_2\}</m> of <m>\R^2</m>.  The grid indicates the coordinate system defined by the basis <m>\cB</m>; one set of lines measures the <m>v_1</m>-coordinate, and the other set measures the <m>v_2</m>-coordinate.  Use the sliders to find the <m>\cB</m>-coordinates of <m>w</m>.</caption>
        <mathbox source="demos/spans.html?v1=1,1&amp;v2=1,-1&amp;target=3,-1&amp;range=6&amp;&amp;tlabel=w" height="500px"/>
      </figure>
    </solution>
  </example>

  <specialcase>
    <idx><h><m>\cB</m>-coordinates</h><h>nonstandard grid</h></idx>
    <p>
      Let
      <me>v_1 = \vec[r]{2 -1 1} \quad v_2 = \vec[r]{1 0 -1}.</me>
      These form a basis <m>\cB</m> for a plane <m>V = \Span\{v_1,v_2\}</m> in <m>\R^3</m>.  We indicate the coordinate system defined by <m>\cB</m> by drawing lines parallel to the <q><m>v_1</m>-axis</q> and <q><m>v_2</m>-axis</q>:
      <latex-code>
  \begin{tikzpicture}[myxyz, z={(0cm,1.2cm)}, thin border nodes]
    \useasboundingbox[resetxy] (-3,-3) rectangle (3,2.3);

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

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

    \begin{scope}[x=(X), y=(Y), transformxy]
      \fill[seq4!10, nearly opaque] (-1,-1) rectangle (1,1);
      \draw[step=.5cm, seq4!80, thin] (-1,-1) grid (1,1);
    \end{scope}

    \begin{scope}[x=($.5*(X)$), y=($.5*(Y)$), every label/.append style={fill=none}]
      \point[minimum size=1.5mm, seq1, "$u_1$" {seq1, xshift=-2mm}] at (1,1);
      \point[minimum size=1.5mm, seq2, "$u_2$" {seq2, right}] at (-1,.5);
      \point[minimum size=1.5mm, seq3, "$u_3$" seq3] at (1.5,-.5);
      \point[minimum size=1.5mm, seq5, "$u_4$" {seq5, right}] at (0,1.5);
    \end{scope}

    \draw[vector] (0,0,0) to["$v_1$"'] ($.5*(X)$);
    \draw[vector] (0,0,0) to["$v_2$"] ($.5*(Y)$);

    \node[x=(X),y=(Y),seq-violet] at (0,-1.2) {$V$};

    \point at (0,0,0);
  \end{tikzpicture}
      </latex-code>
      We can see from the picture that the <m>v_1</m>-coordinate of <m>\color{seq-red}u_1</m> is equal to <m>1</m>, as is the <m>v_2</m>-coordinate, so
      <m>[\textcolor{seq1}{u_1}]_\cB = {1\choose 1}.</m>
      Similarly, we have
      <me>
        [\textcolor{seq2}{u_2}]_\cB
        = \vec[r]{-1 \frac 12}\qquad
        [\textcolor{seq3}{u_3}]_\cB
        = \vec[r]{\frac 32 -\frac 12}\qquad
        [\textcolor{seq5}{u_4}]_\cB
        = \vec{0 \frac 32}.
      </me>
    </p>
    <figure xml:id="dimension-coordinates">
      <caption>Left: the <m>\cB</m>-coordinates of a vector <m>x</m>.  Right: the vector <m>x</m>.  The violet grid on the right is a picture of the coordinate system defined by the basis <m>\cB</m>; one set of lines measures the <m>v_1</m>-coordinate, and the other set measures the <m>v_2</m>-coordinate.  Drag the heads of the vectors <m>x</m> and <m>[x]_\cB</m> to understand the correspondence between <m>x</m> and its <m>\cB</m>-coordinate vector.
      </caption>
      <mathbox source="demos/Axequalsb.html?captions=basis&amp;mat=2,1:-1,0:1,-1&amp;range2=5" height="500px"/>
    </figure>
  </specialcase>

  <example>
    <title>A coordinate system on a plane</title>
    <statement>
      <p>
        Define
        <me>v_1 = \vec{1 0 1},\; v_2 = \vec{1 1 1},\quad
        \cB = \{v_1,\,v_2\}, \quad V = \Span\{v_1,\,v_2\}.</me>
        <ol>
          <li>Verify that <m>\cB</m> is a basis for <m>V</m>.</li>
          <li>If <m>[w]_\cB = {5\choose 2}</m>, then what is <m>w</m>?</li>
          <li>Find the <m>\cB</m>-coordinates of
          <m>v = \vec{5 3 5}.</m>
          </li>
        </ol>
      </p>
    </statement>
    <solution>
      <p>
        <ol>
          <li>
            We need to verify that <m>\cB</m> spans <m>V</m>, and that it is linearly independent.  By definition, <m>V</m> is the span of <m>\cB</m>; since <m>v_1</m> and <m>v_2</m> are not multiples of each other, they are linearly independent.  This shows in particular that <m>V</m> is a <em>plane</em>.
          </li>
          <li>
            To say <m>[w]_\cB={5\choose 2}</m> means that <m>5</m> is the <m>v_1</m>-coordinate of <m>w</m>, and that <m>2</m> is the <m>v_2</m>-coordinate:
            <me>w = 5v_1 + 2v_2 = 5\vec{1 0 1} + 2\vec{1 1 1} = \vec{7 2 7}.</me>
          </li>
          <li>
            We have to solve the vector equation <m>v=c_1v_1 + c_2v_2</m> in the unknowns <m>c_1,c_2</m>.  We form an augmented matrix and row reduce:
            <me>\amat{1 1 5; 0 1 3; 1 1 5}
            \quad\xrightarrow{\text{RREF}}\quad
            \amat{1 0 2; 0 1 3; 0 0 0}.</me>
            We have <m>c_1=2</m> and <m>c_2=3</m>, so <m>v = 2v_1+3v_2</m> and
            <m>[v]_\cB = {2\choose 3}.</m>
          </li>
        </ol>
      </p>
      <figure xml:id="dimension-coordinates-1">
        <caption>A picture of the plane <m>V</m> and the basis <m>\cB = \{v_1,v_2\}</m>.  The violet grid is a picture of the coordinate system defined by the basis <m>\cB</m>; one set of lines measures the <m>v_1</m>-coordinate, and the other set measures the <m>v_2</m>-coordinate.  Use the sliders to find the <m>\cB</m>-coordinates of <m>v</m>.</caption>
        <mathbox source="demos/spans.html?v1=1,0,1&amp;v2=1,1,1&amp;target=5,3,5&amp;range=6&amp;camera=-3,0,2&amp;tlabel=v" height="500px"/>
      </figure>
    </solution>
  </example>

  <example>
    <title>A coordinate system on another plane</title>
    <statement>
      <p>Define
      <me>v_1 = \vec{2 3 2},\; v_2 = \vec[r]{-1 1 1},\; v_3 = \vec{2 8 6},\quad
      V = \Span\{v_1,\,v_2,\,v_3\}.
      </me>
      <ol>
        <li>Find a basis <m>\cB</m> for <m>V</m>.</li>
        <li>Find the <m>\cB</m>-coordinates of <m>x = \vec{4 11 8}</m>.</li>
      </ol>
      </p>
    </statement>
    <solution>
      <p>
        <ol>
          <li>
            We write <m>V</m> as the column space of a matrix <m>A</m>, then row reduce to find the pivot columns, as in this <xref ref="dimension-eg-basis-span"/>.
            <me>A = \mat{2 -1 2; 3 1 8; 2 1 6} \quad\xrightarrow{\text{RREF}}\quad
            \mat{1 0 2; 0 1 2; 0 0 0}.
            </me>
            The first two columns are pivot columns, so we can take <m>\cB=\{v_1,v_2\}</m> as our basis for <m>V</m>.
          </li>
          <li>
            We have to solve the vector equation <m>x = c_1v_1 + c_2v_2</m>.  We form an augmented matrix and row reduce:
            <me>\amat{2 -1 4; 3 1 11; 2 1 8}
            \quad\xrightarrow{\text{RREF}}\quad
            \amat{1 0 3; 0 1 2; 0 0 0}.</me>
            We have <m>c_1 = 3</m> and <m>c_2 = 2</m>, so
            <m>x=3v_1 + 2v_2</m>, and thus <m>[x]_\cB = {3\choose 2}.</m>
          </li>
        </ol>
      </p>
      <figure xml:id="dimension-coordinates-2">
        <caption>A picture of the plane <m>V</m> and the basis <m>\cB = \{v_1,v_2\}</m>.  The violet grid is a picture of the coordinate system defined by the basis <m>\cB</m>; one set of lines measures the <m>v_1</m>-coordinate, and the other set measures the <m>v_2</m>-coordinate.  Use the sliders to find the <m>\cB</m>-coordinates of <m>x</m>.</caption>
        <mathbox source="demos/spans.html?v1=2,3,2&amp;v2=-1,1,1&amp;target=4,11,8&amp;tlabel=x&amp;range=12&amp;camera=.5,-3,2&amp;tlabel=x" height="500px"/>
      </figure>
    </solution>
  </example>

  <bluebox>
    <title>Recipes: <m>\cB</m>-coordinates</title>
    <idx><h><m>\cB</m>-coordinates</h><h>computing</h><h>row reduction</h></idx>
    <p>
      If <m>\cB = \{v_1,v_2,\ldots,v_m\}</m> is a basis for a subspace <m>V</m> and <m>x</m> is in <m>V</m>, then
      <me>[x]_\cB = \vec{c_1 c_2 \vdots, c_m} \sptxt{means}
      x = c_1v_1 + c_2v_2 + \cdots + c_mv_m.
      </me>
      Finding the <m>\cB</m>-coordinates of <m>x</m> means solving the vector equation
      <me>x = c_1v_1 + c_2v_2 + \cdots + c_mv_m</me>
      in the unknowns <m>c_1,c_2,\ldots,c_m</m>.  This generally means row reducing the augmented matrix
      <me>\amat[c]{| | ,, | |; v_1 v_2 \cdots, v_m x; | | ,, | |}.</me>
    </p>
  </bluebox>

  <remark>
    <p>
      Let <m>\cB = \{v_1,v_2,\ldots,v_m\}</m> be a basis of a subspace <m>V</m>.
      Finding the <m>\cB</m>-coordinates of a vector <m>x</m> means solving the vector equation
      <me>x = c_1v_1 + c_2v_2 + \cdots + c_mv_m.</me>
      If <m>x</m> is <em>not</em> in <m>V</m>, then this equation has no solution, as <m>x</m> is not in <m>V = \Span\{v_1,v_2,\ldots,v_m\}</m>.  In other words, the above equation is <em>inconsistent</em> when <m>x</m> is not in <m>V</m>.
    </p>
  </remark>

</section>
</restrict-version>