similarity.xml

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<restrict-version versions="1554 default">
<section xml:id="similarity">
  <title>Similarity</title>

  <objectives>
    <ol>
      <li>Learn to interpret similar matrices geoemetrically.</li>
      <li>Understand the relationship between the eigenvalues, eigenvectors, and characteristic polynomials of similar matrices.</li>
      <li><em>Recipe:</em> compute <m>Ax</m> in terms of <m>B,C</m> for <m>A=CBC\inv</m>.</li>
      <li><em>Picture:</em> the geometry of similar matrices.</li>
      <li><em>Vocabulary word:</em> <term>similarity</term>.</li>
    </ol>
  </objectives>

  <introduction>
    <p>
      Some matrices are easy to understand.  For instance, a diagonal matrix
      <me>
        D = \mat{2 0 ; 0 1/2}
      </me>
      just scales the coordinates of a vector: <m>D{x\choose y} = {2x\choose y/2}</m>. The purpose of most of the rest of this chapter is to understand complicated-looking matrices by analyzing to what extent they <q>behave like</q> simple matrices.  For instance, the matrix
      <me>
        A = \frac 1{10}\mat{11 6; 9 14}
      </me>
      has eigenvalues <m>2</m> and <m>1/2</m>, with corresponding eigenvectors <m>v_1 = {2/3\choose 1}</m> and <m>v_2 = {-1\choose 1}</m>.  Notice that
      <me>
        \spalignsysdelims..\syseq{
        D(xe_1+ye_2) = xDe_1+yDe_2 = 2xe_1-\frac12ye_2;
        A(xv_1+yv_2) = xAv_1+yAv_2\hfil, = 2xv_1-\frac12yv_2\rlap.}
      </me>
      Using <m>v_1,v_2</m> instead of the usual coordinates makes <m>A</m> <q>behave</q> like a diagonal matrix.
    </p>

    <figure>
      <caption>The matrices <m>A</m> and <m>D</m> behave similarly.  Click <q>multiply</q> to multiply the colored points by <m>D</m> on the left and <m>A</m> on the right.  (We will see in <xref ref="diagonalization"/> why the points follow hyperbolic paths.)</caption>
      <mathbox source="demos/dynamics2.html?mat=2,0:0,1/2&amp;v1=2/3,1&amp;v2=-1,1&amp;y=1,5" height="600px"/>
    </figure>

    <p>
      The other case of particular importance will be matrices that <q>behave</q> like a rotation matrix: indeed, this will be crucial for understanding <xref ref="complex-eigenvalues"/> geometrically.  See this <xref ref="similarity-to-rotation"/>.
    </p>

    <p>
      In this section, we study in detail the situation when two matrices behave similarly with respect to different coordinate systems.  In <xref ref="diagonalization"/> and <xref ref="complex-eigenvalues"/>, we will show how to use eigenvalues and eigenvectors to find a simpler matrix that behaves like a given matrix.
    </p>
  </introduction>

  <subsection>
    <title>Similar Matrices</title>

    <p>
      We begin with the algebraic definition of similarity.
    </p>

    <definition>
      <idx><h>Similarity</h><h>definition of</h></idx>
      <idx><h>Matrix</h><h>similar</h><see>Similarity</see></idx>
      <statement>
        <p>
          Two <m>n\times n</m> matrices <m>A</m> and <m>B</m> are <term>similar</term> if there exists an invertible <m>n\times n</m> matrix <m>C</m> such that <m>A = CBC\inv</m>.
        </p>
      </statement>
    </definition>

    <example xml:id="similarity-eg1">
      <p>
        The matrices
        <me>
          \mat{-12 15; -10 13} \sptxt{and} \mat{3 0; 0 -2}
        </me>
        are similar because
        <me>
          \mat{-12 15; -10 13} = \mat{-2 3; 1 -1} \mat{3 0; 0 -2} \mat{-2 3; 1 -1}\inv,
        </me>
        as the reader can verify.
      </p>
    </example>

    <example>
      <p>
        The matrices
        <me>
          \mat{3 0 ; 0 -2} \sptxt{and} \mat{1 0; 0 1}
        </me>
        are not similar.  Indeed, the second matrix is the identity matrix <m>I_2</m>, so if <m>C</m> is any invertible <m>2\times 2</m> matrix, then
        <me>
          CI_2C\inv = CC\inv = I_2 \neq \mat{3 0; 0 -2}.
        </me>
      </p>
    </example>

    <p>
      <idx><h>Similarity</h><h>identity matrix</h></idx>
      <idx><h>Identity matrix</h><h>similarity</h></idx>
      As in the above example, one can show that <m>I_n</m> is the only matrix that is similar to <m>I_n</m>, and likewise for any scalar multiple of <m>I_n</m>.
    </p>

    <bluebox>
      <p>
        Similarity is unrelated to row equivalence.  Any invertible matrix is row equivalent to <m>I_n</m>, but <m>I_n</m> is the only matrix similar to <m>I_n</m>.  For instance,
        <me>\mat{2 1; 0 2} \sptxt{and} \mat{1 0; 0 1}</me>
        are row equivalent but not similar.
      </p>
    </bluebox>

    <p>
      As suggested by its name, similarity is what is called an <em>equivalence relation</em>.  This means that it satisfies the following properties.
    </p>

    <proposition xml:id="similarity-eq-reln">
      <idx><h>Similarity</h><h>equivalence relation</h></idx>
      <statement>
        <p>
          Let <m>A,B</m>, and <m>C</m> be <m>n\times n</m> matrices.
          <ol>
            <li>
              <alert>Reflexivity:</alert> <m>A</m> is similar to itself.
            </li>
            <li>
              <alert>Symmetry:</alert> if <m>A</m> is similar to <m>B</m>, then <m>B</m> is similar to <m>A</m>.
            </li>
            <li>
              <alert>Transitivity:</alert> if <m>A</m> is similar to <m>B</m> and <m>B</m> is similar to <m>C</m>, then <m>A</m> is similar to <m>C</m>.
            </li>
          </ol>
        </p>
      </statement>
      <proof>
        <p>
          <ol>
            <li>
              Taking <m>C = I_n = I_n\inv</m>, we have <m>A = I_n A I_n\inv</m>.
            </li>
            <li>
              Suppose that <m>A = CBC\inv.</m>  Multiplying both sides on the left by <m>C\inv</m> and on the right by <m>C</m> gives
              <me>
                C\inv A C = C\inv(CBC\inv)C = B.
              </me>
              Since <m>(C\inv)\inv = C</m>, we have <m>B = C\inv A (C\inv)\inv</m>, so that <m>B</m> is similar to <m>A</m>.
            </li>
            <li>
              Suppose that <m>A = DBD\inv</m> and <m>B = ECE\inv</m>.  Subsituting for <m>B</m> and remembering that <m>(DE)\inv = E\inv D\inv</m>, we have
              <me>
                A = D(ECE\inv)D\inv = (DE)C(DE)\inv,
              </me>
              which shows that <m>A</m> is similar to <m>C</m>.
            </li>
          </ol>
        </p>
      </proof>
    </proposition>

    <example>
      <p>
        The matrices
        <me>\mat{-12 15; -10 13} \sptxt{and} \mat{3 0; 0 -2}</me>
        are similar, as we saw in this <xref ref="similarity-eg1"/>.  Likewise, the matrices
        <me>\mat{-12 15; -10 13} \sptxt{and} \mat{-12 5; -30 13}</me>
        are similar because
        <me>
          \mat{-12 5; -30 13} = \mat{2 -1; 2 1}\mat{-12 15; -10 13}\mat{2 -1; 2 1}\inv.
        </me>
        It follows that
        <me>\mat{-12 5; -30 13} \sptxt{and} \mat{3 0; 0 -2}</me>
        are similar to each other.
      </p>
    </example>

    <p>
      We conclude with an observation about similarity and powers of matrices.
    </p>

    <fact xml:id="similarity-powers">
      <idx><h>Similarity</h><h>and powers</h></idx>
      <idx><h>Matrix multiplication</h><h>powers</h><h>and similarity</h></idx>
      <statement>
        <p>
          Let <m>A = CBC\inv</m>.  Then for any <m>n\geq 1</m>, we have
          <me>A^n = CB^n C\inv.</me>
        </p>
      </statement>
      <proof visible="true">
        <p>
          First note that
          <me>
            A^2 = AA = (CBC\inv)(CBC\inv) = CB(C\inv C)BC\inv = CBI_nBC\inv = CB^2C\inv.
          </me>
          Next we have
          <me>
            A^3 = A^2A = (CB^2C\inv)(CBC\inv) = CB^2(C\inv C)BC\inv = CB^3C\inv.
          </me>
          The pattern is clear.
        </p>
      </proof>
    </fact>

    <example>
      <statement>
        <p>
          Compute <m>A^{100}</m>, where
          <me>A = \mat{5 13; -2 -5} = \mat{-2 3; 1 -1}\mat{0 -1; 1 0}\mat{-2 3; 1 -1}\inv.</me>
        </p>
      </statement>
      <solution>
        <p>
          By the <xref ref="similarity-powers"/>, we have
          <me>
            A^{100} = \mat{-2 3; 1 -1}\mat{0 -1; 1 0}^{100}\mat{-2 3; 1 -1}\inv.
          </me>
          The matrix <m>\smallmat 0{-1}10</m> is a counterclockwise rotation by <m>90^\circ</m>.  If we rotate by <m>90^\circ</m> four times, then we end up where we started.  Hence rotating by <m>90^\circ</m> one hundred times is the identity transformation, so
          <me>
            A^{100} = \mat{-2 3; 1 -1}\mat{1 0; 0 1}\mat{-2 3; 1 -1}\inv
            = \mat{1 0; 0 1}.
          </me>
        </p>
      </solution>
    </example>

  </subsection>

  <subsection>
    <title>Geometry of Similar Matrices</title>
    <idx><h>Similarity</h><h>geometry of</h></idx>

    <p>
      Similarity is a very interesting construction when viewed geometrically.  We will see that, roughly, <em>similar matrices do the same thing in different coordinate systems.</em>  The reader might want to review <m>\cB</m>-coordinates and nonstandard coordinate grids in <xref ref="bases-as-coord-systems"/> before reading this subsection.
    </p>

    <p>
      By conditions 4 and 5 of the <xref ref="imt-2"/>, an <m>n\times n</m> matrix <m>C</m> is invertible if and only if its columns <m>v_1,v_2,\ldots,v_n</m> form a basis for <m>\R^n</m>.  This means we can speak of the <m>\cB</m>-coordinates of a vector in <m>\R^n</m>, where <m>\cB</m> is the basis of columns of <m>C</m>.  Recall that
      <me>
        [x]_\cB = \vec{c_1 c_2 \vdots, c_n} \sptxt{means}
        x = c_1v_1 + c_2v_2 + \cdots + c_nv_n
        = \mat{| | ,, |; v_1 v_2 \cdots, v_n; | | ,, |}\vec{c_1 c_2 \vdots, c_n}.
      </me>
      Since <m>C</m> is the matrix with columns <m>v_1,v_2,\ldots,v_n</m>, this says that <m>x = C[x]_\cB</m>.  Multiplying both sides by <m>C\inv</m> gives <m>[x]_\cB = C\inv x.</m>  To summarize:
    </p>

    <bluebox>
      <idx><h><m>\cB</m>-coordinates</h><h>change of basis matrix</h></idx>
      <p>
        Let <m>C</m> be an invertible <m>n\times n</m> matrix with columns <m>v_1,v_2,\ldots,v_n</m>, and let <m>\cB = \{v_1,v_2,\ldots,v_n\}</m>, a basis for <m>\R^n</m>.  Then for any <m>x</m> in <m>\R^n</m>, we have
        <me>
          C[x]_\cB = x \sptxt{and} C\inv x = [x]_\cB.
        </me>
        This says that <em><m>C</m> changes from the <m>\cB</m>-coordinates to the usual coordinates</em>, and <em><m>C\inv</m> changes from the usual coordinates to the <m>\cB</m>-coordinates.</em>
      </p>
    </bluebox>

    <p>
      Suppose that <m>A = CBC\inv.</m>  The above observation gives us another way of computing <m>Ax</m> for a vector <m>x</m> in <m>\R^n</m>.  Recall that
      <m>CBC\inv x = C(B(C\inv x)),</m>
      so that multiplying <m>CBC\inv</m> by <m>x</m> means <em>first</em> multiplying by <m>C\inv</m>, <em>then</em> by <m>B</m>, <em>then</em> by <m>C</m>.  See this <xref ref="matrix-mult-order-ops"/>.
    </p>

    <bluebox>
      <title>Recipe: Computing <m>Ax</m> in terms of <m>B</m></title>
      <idx><h>Similarity</h><h>action on a vector</h></idx>
    <p>
        Suppose that <m>A = CBC\inv,</m> where <m>C</m> is an invertible matrix with columns <m>v_1,v_2,\ldots,v_n.</m>  Let <m>\cB = \{v_1,v_2,\ldots,v_n\}</m>, a basis for <m>\R^n</m>.  Let <m>x</m> be a vector in <m>\R^n</m>.  To compute <m>Ax</m>, one does the following:
        <ol>
          <li>
            Multiply <m>x</m> by <m>C\inv</m>, which changes to the <m>\cB</m>-coordinates: <m>[x]_\cB = C\inv x</m>.
          </li>
          <li>
            Multiply this by <m>B</m>: <m>B[x]_\cB = BC\inv x</m>.
          </li>
          <li>
            Interpreting this vector as a <m>\cB</m>-coordinate vector, we multiply it by <m>C</m> to change back to the usual coordinates:
            <m>Ax = CBC\inv x = CB[x]_\cB.</m>
          </li>
        </ol>
        <latex-code>
\begin{tikzpicture}[scale=.8, thin border nodes]
  \draw[help lines] (-3,-3) grid (3,3);
  \node at (0, 3.5) {$\cB$-coordinates};

  \draw[seq-green, vector] (0, 0) to["{$[x]_\cB$}" above right] (-2, 1);
  \draw[seq-red, vector] (0, 0) to["{$B[x]_\cB$}"] (-2, -1);

  \draw[->, shorten=2mm, thick] (8, 1) to[bend right]
     node[above=1mm] {multiply by $C\inv$} (3, 1);
  \draw[->, shorten=2mm, thick] (3, -1) to[bend right]
     node[below=1mm] {multiply by $C$} (8, -1);

  \point at (0, 0);

  \begin{scope}[xshift=11cm]
    \draw[help lines] (-3,-3) grid (3,3);
    \node at (0, 3.5) {usual coordinates};

    \draw[seq-green, vector] (0, 0) to["$x$" below right] (1, 2);
    \draw[seq-red, vector] (0, 0) to["$Ax$"] (-1, 2);

    \point at (0, 0);

  \end{scope}

\end{tikzpicture}
        </latex-code>
      </p>
      <p>
        To summarize: if <m>A = CBC\inv</m>, then <m>A</m> and <m>B</m> do the same thing, only in different coordinate systems.
      </p>
    </bluebox>

    <p>
      The following example is the heart of this section.
    </p>

    <specialcase xml:id="similarity-worked-eg">
      <idx><h>Similarity</h><h>worked example</h></idx>
      <p>
        Consider the matrices
        <me>
          A = \mat{1/2 3/2; 3/2 1/2} \qquad
          B = \mat{2 0; 0 -1} \qquad
          C = \mat{1 1; 1 -1}.
        </me>
        One can verify that <m>A = CBC\inv</m>: see this <xref ref="diagonal-eg-22"/>.  Let <m>v_1 = {1\choose 1}</m> and <m>v_2 = {1\choose -1}</m>, the columns of <m>C</m>, and let <m>\cB = \{v_1,v_2\}</m>, a basis of <m>\R^2</m>.
      </p>
      <p>
        The matrix <m>B</m> is diagonal: it scales the <m>x</m>-direction by a factor of <m>2</m> and the <m>y</m>-direction by a factor of <m>-1</m>.
        <latex-code>
\def\theo{\includegraphics[width=6cm]{theo11.jpg}}

\begin{tikzpicture}[scale=.8, thin border nodes]

  \node[transform shape] at (0, 0) {\theo};
  \draw[seq-violet, opacity=1, vector] (0,0) -- (1,0) node[right] {$e_1$};
  \draw[seq-blue,   opacity=1, vector] (0,0) -- (0,1) node[above] {$e_2$};

  \point at (0, 0);

  \begin{scope}[xshift=10cm, xscale=2, yscale=-1]
    \clip (-1.5, -3) rectangle (1.5, 3);

    \node[transform shape] at (0, 0) {\theo};

    \draw[seq-violet, opacity=1, vector] (0,0) -- (1, 0) node[right] {$Be_1$};
    \draw[seq-blue,   opacity=1, vector] (0,0) -- (0, 1) node[below] {$Be_2$};

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

  \draw (-3, -3) rectangle (3, 3);
  \draw[xshift=10cm] (-3, -3) rectangle (3, 3);

  \draw[->, shorten=2mm, thick] (3, 1) to[bend left, "$B$" above=.5mm] (7, 1);

\end{tikzpicture}
        </latex-code>
        To compute <m>Ax</m>, first we multiply by <m>C\inv</m> to find the <m>\cB</m>-coordinates of <m>x</m>, then we multiply by <m>B</m>, then we multiply by <m>C</m> again.  For instance, let <m>\textcolor{seq-green}x = {0\choose -2}</m>.
        <ol>
          <li>
            We see from the <m>\cB</m>-coordinate grid below that
            <m>\textcolor{seq-green}x = -\textcolor{seq-violet}{v_1} + \textcolor{seq-blue}{v_2}</m>.  Therefore, <m>C\inv\textcolor{seq-green}x = \textcolor{seq-green}{[x]_\cB} = {-1\choose 1}.</m>
          </li>
          <li>
            Multiplying by <m>B</m> scales the coordinates: <m>\textcolor{seq-red}{B[x]_\cB} = {-2\choose -1}</m>.
          </li>
          <li>
            Interpreting <m>{-2\choose -1}</m> as a <m>\cB</m>-coordinate vector, we multiply by <m>C</m> to get
            <me>
              \textcolor{seq-red}{Ax} = C\vec{-2 -1} = -2\textcolor{seq-violet}{v_1} - \textcolor{seq-blue}{v_2} = \vec{-3 -1}.
            </me>
            Of course, this vector lies at <m>(-2, -1)</m> on the <m>\cB</m>-coordinate grid.
          </li>
        </ol>
        <latex-code>
\begin{tikzpicture}[scale=.85, thin border nodes]
  \draw[help lines] (-3,-3) grid (3,3);
  \node at (0, 3.5) {$\cB$-coordinates};

  \draw[seq-violet, vector, opacity=.5] (0,0) -- (1,0);
  \draw[seq-blue,   vector, opacity=.5] (0,0) -- (0,1);

  \draw[seq-green, vector] (0, 0) to["{$[x]_\cB$}" {above left, pos=1}] (-1, 1);
  \draw[seq-red, vector] (0, 0) to["{$B[x]_\cB$}"] (-2, -1);

  \draw[->, shorten=2mm, thick] (9, 1) to[bend right]
     node[above=1mm] {multiply by $C\inv$} (3, 1);
  \node[align=center] (A) at (6, 0)
     {scale $x$ by $2$\\scale $y$ by $-1$};
  \draw[->, shorten >=2mm, thick] (A) -- (3, 0);
  \draw[->, shorten=2mm, thick] (3, -1) to[bend right]
     node[below=1mm] {multiply by $C$} (9, -1);

  \point at (0, 0);

  \begin{scope}[xshift=12cm]
    \node at (0, 3.5) {usual coordinates};
    \draw[help lines] (-3, -3) rectangle (3, 3);
    \path[clip] (-3, -3) rectangle (3, 3);
    \begin{scope}[cm={(1,1,1,-1,(0,0))}, scale=.7]
      \draw[help lines] (-5,-5) grid (5,5);

      \draw[seq-violet, vector, opacity=.5] (0,0) -- (1,0);
      \draw[seq-blue,   vector, opacity=.5] (0,0) -- (0,1);

      \draw[seq-green, vector] (0, 0) to["$x$" left] (-1, 1);
      \draw[seq-red, vector] (0, 0) to["$Ax$" above left] (-2, -1);

    \end{scope}

    \point at (0, 0);

  \end{scope}

\end{tikzpicture}
        </latex-code>
        Now let <m>\textcolor{seq-green}x = \frac 12{5\choose -3}</m>.
        <ol>
          <li>
            We see from the <m>\cB</m>-coordinate grid that
            <m>\textcolor{seq-green}x = \frac 12\textcolor{seq-violet}{v_1} + 2\textcolor{seq-blue}{v_2}</m>.  Therefore, <m>C\inv\textcolor{seq-green}x = \textcolor{seq-green}{[x]_\cB} = {1/2\choose 2}.</m>
          </li>
          <li>
            Multiplying by <m>B</m> scales the coordinates: <m>\textcolor{seq-red}{B[x]_\cB} = {1\choose -2}</m>.
          </li>
          <li>
            Interpreting <m>{1\choose -2}</m> as a <m>\cB</m>-coordinate vector, we multiply by <m>C</m> to get
            <me>
              \textcolor{seq-red}{Ax} = C\vec{1 -2} = \textcolor{seq-violet}{v_1} - 2\textcolor{seq-blue}{v_2} = \vec{-1 3}.
            </me>
            This vector lies at <m>(1, -2)</m> on the <m>\cB</m>-coordinate grid.
          </li>
        </ol>
        <latex-code>
\begin{tikzpicture}[scale=.85, thin border nodes]
  \draw[help lines] (-3,-3) grid (3,3);
  \node at (0, 3.5) {$\cB$-coordinates};

  \draw[seq-violet, vector, opacity=.5] (0,0) -- (1,0);
  \draw[seq-blue,   vector, opacity=.5] (0,0) -- (0,1);

  \draw[seq-green, vector] (0, 0) to["{$[x]_\cB$}" {above left, pos=1}] (1/2, 2);
  \draw[seq-red, vector] (0, 0) to["{$B[x]_\cB$}" pos=.7] (1, -2);

  \draw[->, shorten=2mm, thick] (9, 1) to[bend right]
     node[above=1mm] {multiply by $C\inv$} (3, 1);
  \node[align=center] (A) at (6, 0)
     {scale $x$ by $2$\\scale $y$ by $-1$};
  \draw[->, shorten >=2mm, thick] (A) -- (3, 0);
  \draw[->, shorten=2mm, thick] (3, -1) to[bend right]
     node[below=1mm] {multiply by $C$} (9, -1);

  \point at (0, 0);

  \begin{scope}[xshift=12cm]
    \node at (0, 3.5) {usual coordinates};
    \draw[help lines] (-3, -3) rectangle (3, 3);
    \path[clip] (-3, -3) rectangle (3, 3);
    \begin{scope}[cm={(1,1,1,-1,(0,0))}, scale=.7]
      \draw[help lines] (-5,-5) grid (5,5);

      \draw[seq-violet, vector, opacity=.5] (0,0) -- (1,0);
      \draw[seq-blue,   vector, opacity=.5] (0,0) -- (0,1);

      \draw[seq-green, vector] (0, 0) to["$x$" {above right, pos=.7}] (1/2, 2);
      \draw[seq-red, vector] (0, 0) to["$Ax$" left] (1, -2);

    \end{scope}

    \point at (0, 0);

  \end{scope}

\end{tikzpicture}
        </latex-code>
        To summarize:
        <ul>
          <li>
            <m>B</m> scales the <m>e_1</m>-direction by <m>2</m> and the <m>e_2</m>-direction by <m>-1</m>.
          </li>
          <li>
            <m>A</m> scales the <m>v_1</m>-direction by <m>2</m> and the <m>v_2</m>-direction by <m>-1</m>.
          </li>
        </ul>
        <latex-code>
\def\theo{\includegraphics[width=6cm]{theo11.jpg}}

\begin{tikzpicture}[scale=.8, thin border nodes]

  \node[transform shape] at (0, 0) {\theo};
  \draw[seq-violet, opacity=1, vector] (0,0) -- (1,0) node[right] {$e_1$};
  \draw[seq-blue,   opacity=1, vector] (0,0) -- (0,1) node[above] {$e_2$};

  \point at (0, 0);

  \begin{scope}[xshift=9cm, xscale=2, yscale=-1]
    \clip (-1.5, -3) rectangle (1.5, 3);

    \node[transform shape] at (0, 0) {\theo};

    \draw[seq-violet, opacity=1, vector] (0,0) -- (1, 0) node[right] {$Be_1$};
    \draw[seq-blue,   opacity=1, vector] (0,0) -- (0, 1) node[below] {$Be_2$};

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

  \begin{scope}[yshift=-9cm]
    \clip (-3, -3) rectangle (3, 3);
    \begin{scope}[cm={(1,1,1,-1,(0,0))}]
      \node[transform shape] at (0, 0) {\theo};
      \draw[seq-violet, opacity=1, vector] (0,0) -- (1,0) node[right] {$v_1$};
      \draw[seq-blue,   opacity=1, vector] (0,0) -- (0,1) node[right] {$v_2$};
    \end{scope}

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

  \begin{scope}[yshift=-9cm, xshift=9cm]
    \clip (-3, -3) rectangle (3, 3);
    \begin{scope}[cm={(1,1,1,-1,(0,0))}]
    \begin{scope}[xscale=2, yscale=-1]
      \node[transform shape] at (0, 0) {\theo};
      \draw[seq-violet, opacity=1, vector] (0,0) -- (1,0) node[right] {$Av_1$};
      \draw[seq-blue,   opacity=1, vector] (0,0) -- (0,1) node[left] {$Av_2$};
    \end{scope}
    \end{scope}

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

  \draw (-3, -3) rectangle (3, 3);
  \draw[xshift=9cm] (-3, -3) rectangle (3, 3);
  \draw[yshift=-9cm] (-3, -3) rectangle (3, 3);
  \draw[xshift=9cm, yshift=-9cm] (-3, -3) rectangle (3, 3);

  \draw[->, shorten=2mm, thick] (3, 1) to[bend left, "$B$" above=.5mm] (6, 1);
  \draw[->, shorten=2mm, thick, yshift=-9cm]
    (3, -1) to[bend right, "$A$" above=.5mm] (6, -1);
  \draw[->, shorten=2mm, thick] (-1, -6) to[bend left, "$C\inv$" right=.5mm] (-1, -3);
  \draw[->, shorten=2mm, thick, xshift=9cm]
    (1, -3) to[bend left, "$C$" left=.5mm] (1, -6);

\end{tikzpicture}
        </latex-code>
      </p>
      <figure>
        <caption>The geometric relationship between the similar matrices <m>A</m> and <m>B</m> acting on <m>\R^2</m>.  Click and drag the heads of <m>x</m> and <m>[x]_\cB</m>.  Study this picture until you can reliably predict where the other three vectors will be after moving one of them: this is the essence of the geometry of similar matrices.</caption>
        <mathbox source="demos/similarity.html?B=2,0:0,-1&amp;C=1,1:1,-1&amp;closed" height="600px"/>
      </figure>
    </specialcase>

    <example hide-type="true" xml:id="similarity-another-matrix">
      <title>Interactive: Another matrix similar to <m>B</m></title>
      <p>
        Consider the matrices
        <me>
          A' = \frac 15\mat{-8 -9; 6 13} \qquad
          B = \mat{2 0; 0 -1} \qquad
          C' = \frac 12\mat{-1 -3; 2 1}.
        </me>
        Then <m>A' = C'B(C')\inv</m>, as one can verify.  Let <m>v_1'=\frac 12{-1\choose 2}</m> and <m>v_2' = \frac 12{-3\choose 1}</m>, the columns of <m>C'</m>, and let <m>\cB' = \{v_1',v_2'\}</m>.  Then <m>A'</m> does the same thing as <m>B</m>, as in the previous <xref ref="similarity-worked-eg"/>, except <m>A'</m> uses the <m>\cB'</m>-coordinate system.  In other words:
        <ul>
          <li>
            <m>B</m> scales the <m>e_1</m>-direction by <m>2</m> and the <m>e_2</m>-direction by <m>-1</m>.
          </li>
          <li>
            <m>A'</m> scales the <m>v_1'</m>-direction by <m>2</m> and the <m>v_2'</m>-direction by <m>-1</m>.
          </li>
        </ul>
        <latex-code>
\def\theo{\includegraphics[width=6cm]{theo6.jpg}}

\begin{tikzpicture}[scale=.8, thin border nodes]

  \node[transform shape] at (0, 0) {\theo};
  \draw[seq-violet, opacity=1, vector] (0,0) -- (1,0) node[right] {$e_1$};
  \draw[seq-blue,   opacity=1, vector] (0,0) -- (0,1) node[above] {$e_2$};

  \point at (0, 0);

  \begin{scope}[xshift=9cm, xscale=2, yscale=-1]
    \clip (-1.5, -3) rectangle (1.5, 3);

    \node[transform shape] at (0, 0) {\theo};

    \draw[seq-violet, opacity=1, vector] (0,0) -- (1, 0) node[right] {$Be_1$};
    \draw[seq-blue,   opacity=1, vector] (0,0) -- (0, 1) node[below] {$Be_2$};

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

  \begin{scope}[yshift=-9cm]
    \clip (-3, -3) rectangle (3, 3);
    \begin{scope}[cm={(-1/2,1,-3/2,1/2,(0,0))}]
      \node[transform shape] at (0, 0) {\theo};
      \draw[seq-violet, opacity=1, vector] (0,0) -- (1,0) node[above] {$v_1'$};
      \draw[seq-blue,   opacity=1, vector] (0,0) -- (0,1) node[left] {$v_2'$};
    \end{scope}

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

  \begin{scope}[yshift=-9cm, xshift=9cm]
    \clip (-3, -3) rectangle (3, 3);
    \begin{scope}[cm={(-1/2,1,-3/2,1/2,(0,0))}]
    \begin{scope}[xscale=2, yscale=-1]
      \node[transform shape] at (0, 0) {\theo};
      \draw[seq-violet, opacity=1, vector] (0,0) -- (1,0) node[above=-1mm] {$A'v_1'$};
      \draw[seq-blue,   opacity=1, vector] (0,0) -- (0,1) node[right] {$A'v_2'$};
    \end{scope}
    \end{scope}

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

  \draw (-3, -3) rectangle (3, 3);
  \draw[xshift=9cm] (-3, -3) rectangle (3, 3);
  \draw[yshift=-9cm] (-3, -3) rectangle (3, 3);
  \draw[xshift=9cm, yshift=-9cm] (-3, -3) rectangle (3, 3);

  \draw[->, shorten=2mm, thick] (3, 1) to[bend left, "$B$" above=.5mm] (6, 1);
  \draw[->, shorten=2mm, thick, yshift=-9cm]
    (3, -1) to[bend right, "$A'$" above=.5mm] (6, -1);
  \draw[->, shorten=2mm, thick] (-1, -6) to[bend left, "$(C')\inv$" right=.5mm] (-1, -3);
  \draw[->, shorten=2mm, thick, xshift=9cm]
    (1, -3) to[bend left, "$C'$" left=.5mm] (1, -6);

\end{tikzpicture}
        </latex-code>
      </p>
      <figure>
        <caption>The geometric relationship between the similar matrices <m>A'</m> and <m>B</m> acting on <m>\R^2</m>.  Click and drag the heads of <m>x</m> and <m>[x]_{\cB'}</m>.</caption>
        <mathbox source="demos/similarity.html?B=2,0:0,-1&amp;C=-1/2,-3/2:1,1/2&amp;closed" height="600px"/>
      </figure>
    </example>

    <example xml:id="similarity-eg-rotation">
      <title>A matrix similar to a rotation matrix</title>
      <p>
        Consider the matrices
        <me>
          A = \frac 16\mat{7 -17; 5 -7} \qquad
          B = \mat{0 -1; 1 0} \qquad
          C = \mat{2 -1/2; 1 1/2}.
        </me>
        One can verify that <m>A = CBC\inv.</m>  Let <m>v_1 = {2\choose 1}</m> and <m>v_2 = \frac 12{-1\choose 1}</m>, the columns of <m>C</m>, and let <m>\cB = \{v_1,v_2\}</m>, a basis of <m>\R^2</m>.
      </p>
      <p>
        The matrix <m>B</m> rotates the plane counterclockwise by <m>90^\circ</m>.
        <latex-code>
\def\theo{\includegraphics[width=6cm]{theo10.jpg}}

\begin{tikzpicture}[scale=.8, thin border nodes]

  \node[transform shape] at (0, 0) {\theo};
  \draw[seq-violet, opacity=1, vector] (0,0) -- (1,0) node[right] {$e_1$};
  \draw[seq-blue,   opacity=1, vector] (0,0) -- (0,1) node[above] {$e_2$};

  \point at (0, 0);

  \begin{scope}[xshift=10cm]
    \clip (-3, -3) rectangle (3, 3);

    \begin{scope}[rotate=90]
      \node[transform shape] at (0, 0) {\theo};

      \draw[seq-violet, opacity=1, vector] (0,0) -- (1, 0) node[above] {$Be_1$};
      \draw[seq-blue,   opacity=1, vector] (0,0) -- (0, 1) node[left] {$Be_2$};
    \end{scope}

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

  \draw (-3, -3) rectangle (3, 3);
  \draw[xshift=10cm] (-3, -3) rectangle (3, 3);

  \draw[->, shorten=2mm, thick] (3, 1) to[bend left, "$B$" above=.5mm] (7, 1);

\end{tikzpicture}
        </latex-code>
        To compute <m>Ax</m>, first we multiply by <m>C\inv</m> to find the <m>\cB</m>-coordinates of <m>x</m>, then we multiply by <m>B</m>, then we multiply by <m>C</m> again.  For instance, let <m>\textcolor{seq-green}x = \frac 32{1\choose 1}</m>.
        <ol>
          <li>
            We see from the <m>\cB</m>-coordinate grid below that
            <m>\textcolor{seq-green}x = \textcolor{seq-violet}{v_1} + \textcolor{seq-blue}{v_2}</m>.  Therefore, <m>C\inv\textcolor{seq-green}x = \textcolor{seq-green}{[x]_\cB} = {1\choose 1}.</m>
          </li>
          <li>
            Multiplying by <m>B</m> rotates by <m>90^\circ</m>: <m>\textcolor{seq-red}{B[x]_\cB} = {-1\choose 1}</m>.
          </li>
          <li>
            Interpreting <m>{-1\choose 1}</m> as a <m>\cB</m>-coordinate vector, we multiply by <m>C</m> to get
            <me>
              \textcolor{seq-red}{Ax} = C\vec{-1 1} = -\textcolor{seq-violet}{v_1} + \textcolor{seq-blue}{v_2} = \frac 12\vec{-5 -1}.
            </me>
            Of course, this vector lies at <m>(-1, 1)</m> on the <m>\cB</m>-coordinate grid.
          </li>
        </ol>
        <latex-code>
\begin{tikzpicture}[scale=.75, thin border nodes]
  \node at (0, 3.5) {$\cB$-coordinates};

  \begin{scope}[scale=3/2, arrows={[line width=.6pt]}]
    \draw[help lines] (-2,-2) grid (2, 2);
    \draw[help lines, seq-yellow] (0, 0) circle[radius=1];
    \draw[seq-violet, vector, opacity=.5] (0,0) -- (1,0);
    \draw[seq-blue,   vector, opacity=.5] (0,0) -- (0,1);

    \draw[|->, shorten=1.5mm, dashed]
        (1,1) arc[radius=sqrt(2), start angle=45, delta angle=90];
    \draw[seq-green, vector] (0, 0) -- (1, 1) node[above right] {$[x]_\cB$};
    \draw[seq-red, vector] (0, 0) -- (-1, 1) node[above left] {$B[x]_\cB$};
  \end{scope}

  \draw[->, shorten=2mm, thick] (9, 1) to[bend right]
     node[above=1mm] {multiply by $C\inv$} (3, 1);
  \node (A) at (6, 0) {rotate by $90^\circ$};
  \draw[->, shorten >=2mm, thick] (A) -- (3, 0);
  \draw[->, shorten=2mm, thick] (3, -1) to[bend right]
     node[below=1mm] {multiply by $C$} (9, -1);

  \point at (0, 0);

  \begin{scope}[xshift=12cm]
    \node at (0, 3.5) {usual coordinates};
    \draw[help lines] (-3, -3) rectangle (3, 3);
    \path[clip] (-3, -3) rectangle (3, 3);
    \begin{scope}[cm={(2,1,-1/2,1/2,(0,0))}, arrows={[line width=.6pt]}]
      \draw[help lines] (-10,-10) grid (10, 10);
      \draw[help lines, seq-yellow] (0, 0) circle[radius=1];

      \draw[seq-violet, vector, opacity=.5] (0,0) -- (1,0);
      \draw[seq-blue,   vector, opacity=.5] (0,0) -- (0,1);
      \draw[|->, dashed, shorten=1.5mm]
          (1,1) arc[radius=sqrt(2), start angle=45, delta angle=90];

      \draw[seq-green, vector] (0, 0) -- (1, 1) node[above right] {$x$};
      \draw[seq-red, vector] (0, 0) -- (-1, 1) node[below=1mm, xshift=1mm] {$Ax$};

    \end{scope}

    \point at (0, 0);

  \end{scope}

\end{tikzpicture}
        </latex-code>
        Now let <m>\textcolor{seq-green}x = \frac 12{-1\choose -2}</m>.
        <ol>
          <li>
            We see from the <m>\cB</m>-coordinate grid that
            <m>\textcolor{seq-green}x = -\frac 12\textcolor{seq-violet}{v_1} - \textcolor{seq-blue}{v_2}</m>.  Therefore, <m>C\inv\textcolor{seq-green}x = \textcolor{seq-green}{[x]_\cB} = {-1/2\choose -1}.</m>
          </li>
          <li>
            Multiplying by <m>B</m> rotates by <m>90^\circ</m>: <m>\textcolor{seq-red}{B[x]_\cB} = {1\choose -1/2}</m>.
          </li>
          <li>
            Interpreting <m>{1\choose -1/2}</m> as a <m>\cB</m>-coordinate vector, we multiply by <m>C</m> to get
            <me>
              \textcolor{seq-red}{Ax} = C\vec{1 -1/2} = \textcolor{seq-violet}{v_1} - \frac 12\textcolor{seq-blue}{v_2} = \frac 14\vec{9 3}.
            </me>
            This vector lies at <m>(1, -\frac 12)</m> on the <m>\cB</m>-coordinate grid.
          </li>
        </ol>
        <latex-code>
\begin{tikzpicture}[scale=.75, thin border nodes]
  \node at (0, 3.5) {$\cB$-coordinates};

  \begin{scope}[scale=3/2, arrows={[line width=.6pt]}]
    \draw[help lines] (-2,-2) grid (2, 2);
    \draw[help lines, seq-yellow] (0, 0) circle[radius=1];
    \draw[seq-violet, vector, opacity=.5] (0,0) -- (1,0);
    \draw[seq-blue,   vector, opacity=.5] (0,0) -- (0,1);

    \draw[|->, shorten=1.5mm, dashed]
        (-1/2,-1) arc[radius=sqrt(5/4), start angle={atan2(-1,-1/2)}, delta angle=90];
    \draw[seq-green, vector] (0, 0) -- (-1/2, -1) node[below left=-1mm] {$[x]_\cB$};
    \draw[seq-red, vector] (0, 0) -- (1, -1/2) node[right] {$B[x]_\cB$};
  \end{scope}

  \draw[->, shorten=2mm, thick] (9, 1) to[bend right]
     node[above=1mm] {multiply by $C\inv$} (3, 1);
  \node (A) at (6, 0) {rotate by $90^\circ$};
  \draw[->, shorten >=2mm, thick] (A) -- (3, 0);
  \draw[->, shorten=2mm, thick] (3, -1) to[bend right]
     node[below=1mm] {multiply by $C$} (9, -1);

  \point at (0, 0);

  \begin{scope}[xshift=12cm]
    \node at (0, 3.5) {usual coordinates};
    \draw[help lines] (-3, -3) rectangle (3, 3);
    \path[clip] (-3, -3) rectangle (3, 3);
    \begin{scope}[cm={(2,1,-1/2,1/2,(0,0))}, arrows={[line width=.6pt]}]
      \draw[help lines] (-10,-10) grid (10, 10);
      \draw[help lines, seq-yellow] (0, 0) circle[radius=1];

      \draw[seq-violet, vector, opacity=.5] (0,0) -- (1,0);
      \draw[seq-blue,   vector, opacity=.5] (0,0) -- (0,1);

      \draw[|->, shorten=1.5mm, dashed]
          (-1/2,-1) arc[radius=sqrt(5/4), start angle={atan2(-1,-1/2)}, delta angle=90];
      \draw[seq-green, vector] (0, 0) -- (-1/2,-1) node[below] {$x$};
      \draw[seq-red, vector] (0, 0) -- (1, -1/2) node[right] {$Ax$};

    \end{scope}

    \point at (0, 0);

  \end{scope}

\end{tikzpicture}
        </latex-code>
        To summarize:
        <ul>
          <li>
            <m>B</m> rotates counterclockwise around the circle centered at the origin and passing through <m>e_1</m> and <m>e_2</m>.
          </li>
          <li>
            <m>A</m> rotates counterclockwise around the ellipse centered at the origin and passing through <m>v_1</m> and <m>v_2</m>.
          </li>
        </ul>
        <latex-code>
\def\theo{\includegraphics[width=6cm]{theo10.jpg}}

\begin{tikzpicture}[scale=.75, thin border nodes]

  \node[transform shape] at (0, 0) {\theo};
  \draw[seq-yellow, opacity=.5] (0,0) circle[radius=1];
  \draw[seq-violet, opacity=1, vector] (0,0) -- (1,0) node[right] {$e_1$};
  \draw[seq-blue,   opacity=1, vector] (0,0) -- (0,1) node[above] {$e_2$};

  \point at (0, 0);

  \begin{scope}[xshift=9cm, rotate=90]
    \clip (-3, -3) rectangle (3, 3);

    \node[transform shape] at (0, 0) {\theo};

    \draw[seq-yellow, opacity=.5] (0,0) circle[radius=1];
    \draw[seq-violet, opacity=1, vector] (0,0) -- (1, 0) node[right] {$Be_1$};
    \draw[seq-blue,   opacity=1, vector] (0,0) -- (0, 1) node[below] {$Be_2$};

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

  \begin{scope}[yshift=-9cm]
    \clip (-3, -3) rectangle (3, 3);
    \begin{scope}[cm={(2,1,-1/2,1/2,(0,0))}]
      \node[transform shape] at (0, 0) {\theo};
      \draw[seq-yellow, opacity=.5] (0,0) circle[radius=1];
      \draw[seq-violet, opacity=1, vector] (0,0) -- (1,0) node[above] {$v_1$};
      \draw[seq-blue,   opacity=1, vector] (0,0) -- (0,1) node[left] {$v_2$};
    \end{scope}
    \point at (0, 0);

  \end{scope}

  \begin{scope}[yshift=-9cm, xshift=9cm]
    \clip (-3, -3) rectangle (3, 3);
    \begin{scope}[cm={(2,1,-1/2,1/2,(0,0))}, rotate=90]
      \node[transform shape] at (0, 0) {\theo};
      \draw[seq-yellow, opacity=.5] (0,0) circle[radius=1];
      \draw[seq-violet, opacity=1, vector] (0,0) -- (1,0) node[above] {$Av_1$};
      \draw[seq-blue,   opacity=1, vector] (0,0) -- (0,1) node[below] {$Av_2$};
    \end{scope}
    \point at (0, 0);

  \end{scope}

  \draw (-3, -3) rectangle (3, 3);
  \draw[xshift=9cm] (-3, -3) rectangle (3, 3);
  \draw[yshift=-9cm] (-3, -3) rectangle (3, 3);
  \draw[xshift=9cm, yshift=-9cm] (-3, -3) rectangle (3, 3);

  \draw[->, shorten=2mm, thick] (3, 1) to[bend left, "$B$" above=.5mm] (6, 1);
  \draw[->, shorten=2mm, thick, yshift=-9cm]
    (3, -1) to[bend right, "$A'$" above=.5mm] (6, -1);
  \draw[->, shorten=2mm, thick] (-1, -6) to[bend left, "$C\inv$" right=.5mm] (-1, -3);
  \draw[->, shorten=2mm, thick, xshift=9cm]
    (1, -3) to[bend left, "$C$" left=.5mm] (1, -6);

\end{tikzpicture}
        </latex-code>
      </p>
      <figure>
        <caption>The geometric relationship between the similar matrices <m>A</m> and <m>B</m> acting on <m>\R^2</m>.  Click and drag the heads of <m>x</m> and <m>[x]_\cB</m>.</caption>
        <mathbox source="demos/similarity.html?B=0,-1:1,0&amp;C=2,-1/2:1,1/2&amp;snap=off" height="600px"/>
      </figure>
    </example>

    <p>
      To summarize and generalize the previous example:
    </p>

    <bluebox xml:id="similarity-to-rotation">
      <title>A Matrix Similar to a Rotation Matrix</title>
      <p>
        Let
        <me>
          B = \mat{\cos\theta,-\sin\theta; \sin\theta,\cos\theta}
          \qquad
          C = \mat{| |; v_1 v_2; | |}
          \qquad
          A = CBC\inv,
        </me>
        where <m>C</m> is assumed invertible.  Then:
        <ul>
          <li>
            <m>B</m> rotates the plane by an angle of <m>\theta</m> around the circle centered at the origin and passing through <m>e_1</m> and <m>e_2</m>, in the direction from <m>e_1</m> to <m>e_2</m>.
          </li>
          <li>
            <m>A</m> rotates the plane by an angle of <m>\theta</m> around the ellipse centered at the origin and passing through <m>v_1</m> and <m>v_2</m>, in the direction from <m>v_1</m> to <m>v_2</m>.
          </li>
        </ul>
        <latex-code>
\def\theo{\includegraphics[width=6cm]{theo10.jpg}}

\begin{tikzpicture}[scale=.75, thin border nodes]

  \node[transform shape] at (0, 0) {\theo};
  \draw[seq-yellow, opacity=.5] (0,0) circle[radius=1];
  \draw[seq-violet, opacity=1, vector] (0,0) -- (1,0) node[right] {$e_1$};
  \draw[seq-blue,   opacity=1, vector] (0,0) -- (0,1) node[above] {$e_2$};

  \point at (0, 0);

  \begin{scope}[xshift=9cm, rotate=90]
    \clip (-3, -3) rectangle (3, 3);

    \node[transform shape] at (0, 0) {\theo};

    \draw[seq-yellow, opacity=.5] (0,0) circle[radius=1];
    \draw[seq-violet, opacity=1, vector] (0,0) -- (1, 0) node[right] {$Be_1$};
    \draw[seq-blue,   opacity=1, vector] (0,0) -- (0, 1) node[below] {$Be_2$};

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

  \begin{scope}[yshift=-9cm]
    \clip (-3, -3) rectangle (3, 3);
    \begin{scope}[cm={(2,1,-1/2,1/2,(0,0))}]
      \node[transform shape] at (0, 0) {\theo};
      \draw[seq-yellow, opacity=.5] (0,0) circle[radius=1];
      \draw[seq-violet, opacity=1, vector] (0,0) -- (1,0) node[above] {$v_1$};
      \draw[seq-blue,   opacity=1, vector] (0,0) -- (0,1) node[left] {$v_2$};
    \end{scope}
    \point at (0, 0);

  \end{scope}

  \begin{scope}[yshift=-9cm, xshift=9cm]
    \clip (-3, -3) rectangle (3, 3);
    \begin{scope}[cm={(2,1,-1/2,1/2,(0,0))}, rotate=90]
      \node[transform shape] at (0, 0) {\theo};
      \draw[seq-yellow, opacity=.5] (0,0) circle[radius=1];
      \draw[seq-violet, opacity=1, vector] (0,0) -- (1,0) node[above] {$Av_1$};
      \draw[seq-blue,   opacity=1, vector] (0,0) -- (0,1) node[below] {$Av_2$};
    \end{scope}
    \point at (0, 0);

  \end{scope}

  \draw (-3, -3) rectangle (3, 3);
  \draw[xshift=9cm] (-3, -3) rectangle (3, 3);
  \draw[yshift=-9cm] (-3, -3) rectangle (3, 3);
  \draw[xshift=9cm, yshift=-9cm] (-3, -3) rectangle (3, 3);

  \draw[->, shorten=2mm, thick] (3, 1) to[bend left, "$B$" above=.5mm] (6, 1);
  \draw[->, shorten=2mm, thick, yshift=-9cm]
    (3, -1) to[bend right, "$A'$" above=.5mm] (6, -1);
  \draw[->, shorten=2mm, thick] (-1, -6) to[bend left, "$C\inv$" right=.5mm] (-1, -3);
  \draw[->, shorten=2mm, thick, xshift=9cm]
    (1, -3) to[bend left, "$C$" left=.5mm] (1, -6);

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

    <example hide-type="true" xml:id="similarity-33-eg">
      <title>Interactive: Similar <m>3\times 3</m> matrices</title>
      <p>
        Consider the matrices
        <me>
          A = \mat{-1 0 0; -1 0 2; -1 1 1} \qquad
          B = \mat{-1 0 0; 0 -1 0; 0 0 2} \qquad
          C = \mat{-1 1 0; 1 1 1; -1 0 1}.
        </me>
        Then <m>A = CBC\inv</m>, as one can verify.  Let <m>v_1,v_2,v_3</m> be the columns of <m>C</m>, and let <m>\cB = \{v_1,v_2,v_3\}</m>, a basis of <m>\R^3</m>.  Then <m>A</m> does the same thing as <m>B</m>, except <m>A</m> uses the <m>\cB</m>-coordinate system.  In other words:
        <ul>
          <li>
            <m>B</m> scales the <m>e_1,e_2</m>-plane by <m>-1</m> and the <m>e_3</m>-direction by <m>2</m>.
          </li>
          <li>
            <m>A</m> scales the <m>v_1,v_2</m>-plane by <m>-1</m> and the <m>v_3</m>-direction by <m>2</m>.
          </li>
        </ul>
      </p>
      <figure>
        <caption>The geometric relationship between the similar matrices <m>A</m> and <m>B</m> acting on <m>\R^3</m>.  Click and drag the heads of <m>x</m> and <m>[x]_\cB</m>.</caption>
        <mathbox source="demos/similarity.html?B=-1,0,0:0,-1,0:0,0,2&amp;C=-1,1,0:1,1,1:-1,0,1&amp;closed" height="600px"/>
      </figure>
    </example>
  </subsection>

  <subsection>
    <title>Eigenvalues of Similar Matrices</title>

    <p>
      Since similar matrices behave in the same way with respect to different coordinate systems, we should expect their eigenvalues and eigenvectors to be closely related.
    </p>

    <theorem type-name="Fact" xml:id="similarity-charpoly">
      <idx><h>Similarity</h><h>and the characteristic polynomial</h></idx>
      <idx><h>Characteristic polynomial</h><h>of similar matrices</h></idx>
      <statement>
        <p>Similar matrices have the same characteristic polynomial.</p>
      </statement>
      <proof visible="true">
        <p>
          Suppose that <m>A = CBC\inv</m>, where <m>A,B,C</m> are <m>n\times n</m> matrices.  We calculate
          <me>
            \begin{split}
            A - \lambda I_n
            \amp= CBC\inv - \lambda CC\inv
            = CBC\inv - C\lambda C\inv \\
            \amp= CBC\inv - C\lambda I_n C\inv
            = C(B-\lambda I_n)C\inv.
            \end{split}
          </me>
          Therefore,
          <me>
            \det(A-\lambda I_n) = \det(C(B-\lambda I_n)C\inv)
            = \det(C)\det(B-\lambda I_n)\det(C)\inv
            = \det(B-\lambda I_n).
          </me>
          Here we have used the <xref ref="det-defn-mult-prop">multiplicativity property</xref> and its <xref ref="det-defn-power-prop">corollary</xref>.
        </p>
      </proof>
    </theorem>

    <p>
      Since the eigenvalues of a matrix are the roots of its characteristic polynomial, we have shown:
    </p>

    <bluebox>
      <idx><h>Similarity</h><h>and eigenvalues</h></idx>
      <idx><h>Eigenvalue</h><h>of similar matrices</h></idx>
      <p>
        Similar matrices have the same eigenvalues.
      </p>
    </bluebox>

    <p>
      <idx><h>Similarity</h><h>and the trace</h></idx>
      <idx><h>Similarity</h><h>and the determinant</h></idx>
      <idx><h>Determinant</h><h>of similar matrices</h></idx>
      <idx><h>Matrix</h><h>trace of</h><h>similar matrices</h></idx>
      By this <xref ref="charpoly-is-poly"/>, similar matrices also have the same trace and determinant.
    </p>

    <note>
      <p>
        The converse of the <xref ref="similarity-charpoly"/> is false.  Indeed, the matrices
        <me>\mat{1 1; 0 1} \sptxt{and} \mat{1 0 ; 0 1}</me>
        both have characteristic polynomial <m>f(\lambda) = (\lambda-1)^2</m>, but they are not similar, because the only matrix that is similar to <m>I_2</m> is <m>I_2</m> itself.
      </p>
    </note>

    <p>
      Given that similar matrices have the same eigenvalues, one might guess that they have the same eigen<em>vectors</em> as well.  Upon reflection, this is not what one should expect: indeed, the eigenvectors should only match up after changing from one coordinate system to another.  This is the content of the next fact, remembering that <m>C</m> and <m>C\inv</m> change between the usual coordinates and the <m>\cB</m>-coordinates.
    </p>

    <theorem type-name="Fact" xml:id="similarity-evecs">
      <idx><h>Eigenvector</h><h>of similar matrices</h></idx>
      <idx><h>Similarity</h><h>and eigenvectors</h></idx>
      <statement>
        <p>
          Suppose that <m>A = CBC\inv</m>.  Then
          <me>
            \begin{split}
            v \text{ is an eigenvector of } A
            \amp\quad\implies\quad C\inv v \text{ is an eigenvector of } B \\
            v \text{ is an eigenvector of } B
            \amp\quad\implies\quad C v \text{ is an eigenvector of } A. \\
            \end{split}
          </me>
          The eigenvalues of <m>v</m> / <m>C\inv v</m> or <m>v</m> / <m>Cv</m> are the same.
        </p>
      </statement>
      <proof visible="true">
        <p>
          Suppose that <m>v</m> is an eigenvector of <m>A</m> with eigenvalue <m>\lambda</m>, so that <m>Av = \lambda v</m>.  Then
          <me>B(C\inv v) = C\inv(CBC\inv v) = C\inv(Av) = C\inv\lambda v = \lambda(C\inv v),</me>
          so that <m>C\inv v</m> is an eigenvector of <m>B</m> with eigenvalue <m>\lambda</m>.  Likewise if <m>v</m> is an eigenvector of <m>B</m> with eigenvalue <m>\lambda</m>, then <m>Bv = \lambda v</m>, and we have
          <me>A(Cv) = (CBC\inv)Cv = CBv = C(\lambda v) = \lambda(Cv),</me>
          so that <m>Cv</m> is an eigenvalue of <m>A</m> with eigenvalue <m>\lambda</m>.
        </p>
      </proof>
    </theorem>

    <bluebox>
      <idx><h>Eigenspace</h><h>of similar matrices</h></idx>
      <idx><h>Similarity</h><h>and eigenspaces</h></idx>
      <p>
        If <m>A = CBC\inv</m>, then <m>C\inv</m> takes the <m>\lambda</m>-eigenspace of <m>A</m> to the <m>\lambda</m>-eigenspace of <m>B</m>, and <m>C</m> takes the <m>\lambda</m>-eigenspace of <m>B</m> to the <m>\lambda</m>-eigenspace of <m>A</m>.
      </p>
    </bluebox>

    <specialcase>
      <p>
        We continue with the above <xref ref="similarity-worked-eg"/>: let
        <me>
          A = \mat{1/2 3/2; 3/2 1/2} \qquad
          B = \mat{2 0; 0 -1} \qquad
          C = \mat{1 1; 1 -1},
        </me>
        so <m>A = CBC\inv.</m>  Let <m>v_1 = {1\choose 1}</m> and <m>v_2 = {1\choose -1}</m>, the columns of <m>C</m>.  Recall that:
        <ul>
          <li>
            <m>B</m> scales the <m>e_1</m>-direction by <m>2</m> and the <m>e_2</m>-direction by <m>-1</m>.
          </li>
          <li>
            <m>A</m> scales the <m>v_1</m>-direction by <m>2</m> and the <m>v_2</m>-direction by <m>-1</m>.
          </li>
        </ul>
        This means that the <m>x</m>-axis is the <m>2</m>-eigenspace of <m>B</m>, and the <m>y</m>-axis is the <m>-1</m>-eigenspace of <m>B</m>; likewise, the <q><m>v_1</m>-axis</q> is the <m>2</m>-eigenspace of <m>A</m>, and the <q><m>v_2</m>-axis</q> is the <m>-1</m>-eigenspace of <m>A</m>.  This is consistent with the <xref ref="similarity-evecs"/>, as multiplication by <m>C</m> changes <m>e_1</m> into <m>Ce_1 = v_1</m> and <m>e_2</m> into <m>Ce_2 = v_2</m>.
        <latex-code>
\def\theo{\includegraphics[width=6cm]{theo11.jpg}}

\begin{tikzpicture}[scale=.8, thin border nodes]

  \node[transform shape] at (0, 0) {\theo};
  \draw[seq-violet, opacity=1] (-3,0) -- (3,0)
      node[below,pos=.75,font=\small] {$2$-eigenspace};
  \draw[seq-blue,   opacity=1] (0,-3) -- (0,3)
      node[above,pos=.25, rotate=90, font=\small] {$-1$-eigenspace};

  \point at (0, 0);

  \begin{scope}[xshift=9cm]
    \clip (-3, -3) rectangle (3, 3);
    \begin{scope}[cm={(1,1,1,-1,(0,0))}]
      \node[transform shape] at (0, 0) {\theo};
      \draw[seq-violet, opacity=1] (-3,0) -- (3,0)
          node[above,pos=.75, rotate=45, font=\small] {$2$-eigenspace};
      \draw[seq-blue,   opacity=1] (0,-3) -- (0,3)
          node[above,pos=.25, rotate=-45, font=\small] {$-1$-eigenspace};
    \end{scope}

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

  \draw (-3, -3) rectangle (3, 3);
  \draw[xshift=9cm] (-3, -3) rectangle (3, 3);

  \draw[->, shorten=2mm, thick] (3, 1) to[bend left, "$C$" above=.5mm] (6, 1);

\end{tikzpicture}
        </latex-code>
      </p>
      <figure>
        <caption>The eigenspaces of <m>A</m> are the lines through <m>v_1</m> and <m>v_2</m>.  These are the images under <m>C</m> of the coordinate axes, which are the eigenspaces of <m>B</m>.</caption>
        <mathbox source="demos/eigenspace.html?mat=1/2,3/2:3/2,1/2&amp;nomult" height="600px"/>
      </figure>
    </specialcase>

    <example hide-type="true">
      <title>Interactive: Another matrix similar to <m>B</m></title>
      <p>
        Continuing with this <xref ref="similarity-another-matrix"/>, let
        <me>
          A' = \frac 15\mat{-8 -9; 6 13} \qquad
          B = \mat{2 0; 0 -1} \qquad
          C' = \frac 12\mat{-1 -3; 2 1},
        </me>
        so <m>A' = C'B(C')\inv</m>.  Let <m>v_1'=\frac 12{-1\choose 2}</m> and <m>v_2' = \frac 12{-3\choose 1}</m>, the columns of <m>C'</m>.  Then:
        <ul>
          <li>
            <m>B</m> scales the <m>e_1</m>-direction by <m>2</m> and the <m>e_2</m>-direction by <m>-1</m>.
          </li>
          <li>
            <m>A'</m> scales the <m>v_1'</m>-direction by <m>2</m> and the <m>v_2'</m>-direction by <m>-1</m>.
          </li>
        </ul>
        As before, the <m>x</m>-axis is the <m>2</m>-eigenspace of <m>B</m>, and the <m>y</m>-axis is the <m>-1</m>-eigenspace of <m>B</m>; likewise, the <q><m>v_1'</m>-axis</q> is the <m>2</m>-eigenspace of <m>A'</m>, and the <q><m>v_2'</m>-axis</q> is the <m>-1</m>-eigenspace of <m>A'</m>.  This is consistent with the <xref ref="similarity-evecs"/>, as multiplication by <m>C'</m> changes <m>e_1</m> into <m>C'e_1 = v_1'</m> and <m>e_2</m> into <m>C'e_2 = v_2'</m>.
        <latex-code>
\def\theo{\includegraphics[width=6cm]{theo6.jpg}}

\begin{tikzpicture}[scale=.8, thin border nodes]

  \node[transform shape] at (0, 0) {\theo};
  \draw[seq-violet, opacity=1] (-3,0) -- (3,0)
      node[below,pos=.75,font=\small] {$2$-eigenspace};
  \draw[seq-blue,   opacity=1] (0,-3) -- (0,3)
      node[above,pos=.25, rotate=90, font=\small] {$-1$-eigenspace};

  \point at (0, 0);

  \begin{scope}[xshift=9cm]
    \clip (-3, -3) rectangle (3, 3);
    \begin{scope}[cm={(-1/2,1,-3/2,1/2,(0,0))}]
      \node[transform shape] at (0, 0) {\theo};
      \draw[seq-violet, opacity=1] (-3,0) -- (3,0)
          node[above,pos=.72, rotate={atan2(-1,1/2)}, font=\small] {$2$-eigenspace};
      \draw[seq-blue,   opacity=1] (0,-3) -- (0,3)
          node[below,pos=.315, rotate={atan2(-1/2,3/2)}, font=\small] {$-1$-eigenspace};
    \end{scope}

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

  \draw (-3, -3) rectangle (3, 3);
  \draw[xshift=9cm] (-3, -3) rectangle (3, 3);

  \draw[->, shorten=2mm, thick] (3, 1) to[bend left, "$C$" above=.5mm] (6, 1);

\end{tikzpicture}
        </latex-code>
      </p>
      <figure>
        <caption>The eigenspaces of <m>A'</m> are the lines through <m>v_1'</m> and <m>v_2'</m>.  These are the images under <m>C'</m> of the coordinate axes, which are the eigenspaces of <m>B</m>.</caption>
        <mathbox source="demos/eigenspace.html?mat=-8/5,-9/5:6/5,13/5&amp;nomult" height="600px"/>
      </figure>
    </example>

    <example hide-type="true">
      <title>Interactive: Similar <m>3\times 3</m> matrices</title>
      <p>
        Continuing with this <xref ref="similarity-33-eg"/>, let
        <me>
          A = \mat{-1 0 0; -1 0 2; -1 1 1} \qquad
          B = \mat{-1 0 0; 0 -1 0; 0 0 2} \qquad
          C = \mat{-1 1 0; 1 1 1; -1 0 1},
        </me>
        so <m>A = CBC\inv</m>.  Let <m>v_1,v_2,v_3</m> be the columns of <m>C</m>.  Then:
        <ul>
          <li>
            <m>B</m> scales the <m>e_1,e_2</m>-plane by <m>-1</m> and the <m>e_3</m>-direction by <m>2</m>.
          </li>
          <li>
            <m>A</m> scales the <m>v_1,v_2</m>-plane by <m>-1</m> and the <m>v_3</m>-direction by <m>2</m>.
          </li>
        </ul>
        In other words, the <m>xy</m>-plane is the <m>-1</m>-eigenspace of <m>B</m>, and the <m>z</m>-axis is the <m>2</m>-eigenspace of <m>B</m>; likewise, the <q><m>v_1,v_2</m>-plane</q> is the <m>-1</m>-eigenspace of <m>A</m>, and the <q><m>v_3</m>-axis</q> is the <m>2</m>-eigenspace of <m>A</m>.  This is consistent with the <xref ref="similarity-evecs"/>, as multiplication by <m>C</m> changes <m>e_1</m> into <m>Ce_1 = v_1</m>, <m>e_2</m> into <m>Ce_2 = v_2</m>, and <m>e_3</m> into <m>Ce_3 = v_3</m>.
      </p>
      <figure>
        <caption>The <m>-1</m>-eigenspace of <m>A</m> is the green plane, and the <m>2</m>-eigenspace of <m>A</m> is the violet line.  These are the images under <m>C</m> of the <m>xy</m>-plane and the <m>z</m>-axis, respectively, which are the eigenspaces of <m>B</m>.</caption>
        <mathbox source="demos/eigenspace.html?mat=-1,0,0:-1,0,2:-1,1,1&amp;nomult" height="600px"/>
      </figure>
    </example>

  </subsection>

</section>
</restrict-version>