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  <section xml:id="characteristic-polynomial">
    <title>The Characteristic Polynomial</title>
  
    <objectives>
      <ol>
        <li>Learn that the eigenvalues of a triangular matrix are the diagonal entries.</li>
        <li>Find all eigenvalues of a matrix using the characteristic polynomial.</li>
        <li>Learn some strategies for finding the zeros of a polynomial.</li>
        <li><em>Recipe:</em> the characteristic polynomial of a <m>2\times 2</m> matrix.</li>
        <li><em>Vocabulary words:</em> <term>characteristic polynomial</term>, <term>trace</term>.</li>
      </ol>
    </objectives>
  
    <p>
      In <xref ref="eigenvectors"/> we discussed how to decide whether a given number <m>\lambda</m> is an eigenvalue of a matrix, and if so, how to find all of the associated eigenvectors.  In this section, we will give a method for computing all of the <em>eigenvalues</em> of a matrix.  This does not reduce to solving a system of linear equations: indeed, it requires solving a <em>nonlinear</em> equation in one variable, namely, finding the roots of the characteristic polynomial.
    </p>
  
    <definition>
      <idx><h>Characteristic polynomial</h><h>definition of</h></idx>
      <idx><h>Polynomial</h><h>characteristic</h><see>Characteristic polynomial</see></idx>
      <statement>
        <p>
          Let <m>A</m> be an <m>n\times n</m> matrix.  The <term>characteristic polynomial</term> of <m>A</m> is the function <m>f(\lambda)</m> given by
          <me>f(\lambda) = \det(A-\lambda I_n).</me>
        </p>
      </statement>
    </definition>
  
    <p>
      We will see <xref ref="charpoly-is-poly">below</xref> that the characteristic polynomial is in fact a polynomial.  Finding the characterestic polynomial means computing the determinant of the matrix <m>A-\lambda I_n</m>, whose entries contain the unknown <m>\lambda</m>.
    </p>
  
    <example xml:id="charpoly-eg-22">
      <statement>
        <p>
          Find the characteristic polynomial of the matrix
          <me>A = \mat{5 2; 2 1}.</me>
        </p>
      </statement>
      <solution>
        <p>
          We have
          <me>
            \begin{split}
            f(\lambda) = \det(A - \lambda I_2)
            \amp= \det\left(\mat{5 2; 2 1} - \mat{\lambda, 0; 0 \lambda}\right) \\
            \amp= \det\mat{5-\lambda, 2; 2 1-\lambda} \\
            \amp= (5-\lambda)(1-\lambda) - 2\cdot 2
            = \lambda^2 - 6\lambda + 1.
            \end{split}
          </me>
        </p>
      </solution>
    </example>
  
    <example xml:id="charpoly-eg-33">
      <statement>
        <p>
          Find the characteristic polynomial of the matrix
          <me>A = \mat{0 6 8; \frac 12 0 0; 0 \frac 12 0}.</me>
        </p>
      </statement>
      <solution>
        <p>
          We compute the determinant by expanding cofactors along the third column:
          <me>
            \begin{split}
            f(\lambda) = \det(A - \lambda I_3)
            \amp= \det\mat{-\lambda, 6 8; \frac 12 -\lambda, 0; 0 \frac 12 -\lambda} \\
            \amp= 8\left(\frac 14 - 0\cdot-\lambda\right)
            - \lambda\left(\lambda^2 - 6\cdot\frac 12\right) \\
            \amp= -\lambda^3 + 3\lambda + 2.
            \end{split}
          </me>
        </p>
      </solution>
    </example>
  
    <p>
      The point of the characteristic polynomial is that we can use it to compute eigenvalues.
    </p>
  
    <theorem>
      <title>Eigenvalues are roots of the characteristic polynomial</title>
      <idx><h>Characteristic polynomial</h><h>and eigenvalues</h></idx>
      <idx><h>Eigenvalue</h><h>and the characteristic polynomial</h><see>Characteristic polynomial</see></idx>
      <statement>
        <p>
          Let <m>A</m> be an <m>n\times n</m> matrix, and let <m>f(\lambda) = \det(A-\lambda I_n)</m> be its characteristic polynomial.  Then a number <m>\lambda_0</m> is an eigenvalue of <m>A</m> if and only if <m>f(\lambda_0) = 0</m>.
        </p>
      </statement>
      <proof visible="true">
        <p>
          By the <xref ref="imt-2"/>, the matrix equation <m>(A-\lambda_0 I_n)x=0</m> has a nontrivial solution if and only if <m>\det(A-\lambda_0 I_n) = 0</m>.  Therefore,
          <me>
            \begin{split}
            \lambda_0 \text{ is an eigenvalue of } A
            \amp\iff Ax = \lambda_0 x \text{ has a nontrivial solution} \\
            \amp\iff (A-\lambda_0 I_n)x = 0 \text{ has a nontrivial solution} \\
            \amp\iff A - \lambda_0 I_n \text{ is not invertible} \\
            \amp\iff \det(A - \lambda_0 I_n) = 0 \\
            \amp\iff f(\lambda_0) = 0.
            \end{split}
          </me>
        </p>
      </proof>
    </theorem>
  
    <example>
      <title>Finding eigenvalues</title>
      <statement>
        <p>
          Find the eigenvalues and eigenvectors of the matrix
          <me>A = \mat{5 2; 2 1}.</me>
        </p>
      </statement>
      <solution>
        <p>
          In the above <xref ref="charpoly-eg-22"/> we computed the characteristic polynomial of <m>A</m> to be <m>f(\lambda) = \lambda^2-6\lambda + 1</m>.  We can solve the equation <m>\lambda^2-6\lambda+1=0</m> using the quadratic formula:
          <me>
            \lambda = \frac{6\pm\sqrt{36-4}}2 = 3\pm 2\sqrt 2.
          </me>
          Therefore, the eigenvalues are <m>3+2\sqrt 2</m> and <m>3-2\sqrt 2</m>.
        </p>
        <p>
          To compute the eigenvectors, we solve the homogeneous system of equations <m>(A-\lambda I_2)x=0</m> for each eigenvalue <m>\lambda</m>.  When <m>\lambda=3+2\sqrt 2</m>, we have
          <me>
            \begin{split}
            A-(3+\sqrt 2)I_2 = \amp\mat{2-2\sqrt2 2; 2 -2-2\sqrt2} \\
            \xrightarrow{R_1 = R_1\times(2+2\sqrt2)}\;
            \amp\mat{-4 4+4\sqrt2; 2 -2-2\sqrt2} \\
            \xrightarrow{R_2 = R_2+R_1/2}\;
            \amp\mat{-4 4+4\sqrt2; 0 0} \\
            \xrightarrow{R_1 = R_1 \div -4}\;
            \amp\mat{1 -1-\sqrt2; 0 0}.
            \end{split}
          </me>
          The parametric form of the general solution is <m>x=(1+\sqrt 2)y</m>, so the <m>(3+2\sqrt 2)</m>-eigenspace is the line spanned by <m>{1+\sqrt 2\choose 1}</m>.  We compute in the same way that the <m>(3-2\sqrt 2)</m>-eigenspace is the line spanned by <m>{1-\sqrt 2\choose 1}</m>.
        </p>
        <figure>
          <caption>The green line is the <m>(3-2\sqrt 2)</m>-eigenspace, and the violet line is the <m>(3+2\sqrt 2)</m>-eigenspace.</caption>
          <mathbox source="demos/eigenspace.html?mat=5,2:2,1&amp;nomult" height="500px"/>
        </figure>
      </solution>
    </example>
  
    <example>
      <title>Finding eigenvalues</title>
      <statement>
        <p>
          Find the eigenvalues and eigenvectors of the matrix
          <me>
            A = \mat{0 6 8; \frac 12 0 0; 0 \frac 12 0}.
          </me>
        </p>
      </statement>
      <solution>
        <p>
          In the above <xref ref="charpoly-eg-33"/> we computed the characteristic polynomial of <m>A</m> to be <m>f(\lambda) = -\lambda^3+3\lambda+2</m>.  We eyeball that <m>f(2) = -8 + 3\cdot 2 + 2 = 0</m>.  Thus <m>\lambda-2</m> divides <m>f(\lambda)</m>; to find the other roots, we perform <url href="https://en.wikipedia.org/wiki/Polynomial_long_division">polynomial long division</url>:
          <me>
            \frac{-\lambda^3+3\lambda+2}{\lambda-2}
            = -\lambda^2-2\lambda-1 = -(\lambda+1)^2.
          </me>
          Therefore,
          <me>
            f(\lambda) = -(\lambda-2)(\lambda+1)^2,
          </me>
          so the only eigenvalues are <m>\lambda = 2,-1</m>.
        </p>
        <p>
          We compute the <m>2</m>-eigenspace by solving the homogeneous system <m>(A-2I_3)x=0</m>.  We have
          <me>
            A-2I_3 = \mat{-2 6 8; \frac 12 -2 0; 0 \frac 12 -2}
            \quad\xrightarrow{\text{RREF}}\quad
            \mat{1 0 -16; 0 1 -4; 0 0 0}.
          </me>
          The parametric form and parametric vector form of the solutions are:
          <me>
            \syseq{x = 16z; y = 4z; z = z}
            \quad\To\quad
            \vec{x y z} = z\vec{16 4 1}.
          </me>
          Therefore, the <m>2</m>-eigenspace is the line
          <me>
            \Span\left\{\vec{16 4 1}\right\}.
          </me>
        </p>
        <p>
          We compute the <m>-1</m>-eigenspace by solving the homogeneous system <m>(A+I_3)x=0</m>.  We have
          <me>
            A+I_3 = \mat{1 6 8; \frac 12 1 0; 0 \frac 12 1}
            \quad\xrightarrow{\text{RREF}}\quad
            \mat{1 0 -4; 0 1 2; 0 0 0}.
          </me>
          The parametric form and parametric vector form of the solutions are:
          <me>
            \syseq{x = 4z; y = -2z; z = z}
            \quad\To\quad
            \vec{x y z} = z\vec{4 -2 1}.
          </me>
          Therefore, the <m>-1</m>-eigenspace is the line
          <me>
            \Span\left\{\vec{4 -2 1}\right\}.
          </me>
        </p>
        <figure>
          <caption>The green line is the <m>-1</m>-eigenspace, and the violet line is the <m>2</m>-eigenspace.</caption>
          <mathbox source="demos/eigenspace.html?mat=0,6,8:1/2,0,0:0,1/2,0&amp;nomult" height="500px"/>
        </figure>
      </solution>
    </example>
  
    <paragraphs>
      <title>Form of the characteristic polynomial</title>
  
    <p>
      It is time that we justified the use of the term <q>polynomial.</q> First we need a vocabulary word.
    </p>
  
    <definition>
      <idx><h>Trace</h><see>Matrix</see></idx>
      <idx><h>Matrix</h><h>trace of</h></idx>
      <notation><usage>\Tr(A)</usage><description>Trace of a matrix</description></notation>
      <statement>
        <p>
          The <term>trace</term> of a square matrix <m>A</m> is the number <m>\Tr(A)</m> obtained by summing the diagonal entries of <m>A</m>:
          <me>
            \def\r{\color{red}}
            \Tr\mat{\r a_{11} a_{12} \cdots, a_{1,n-1} a_{1n} ;
            a_{21} \r a_{22} \cdots, a_{2,n-1} a_{2n} ;
            \vdots, \vdots, \ddots, \vdots, \vdots ;
            a_{n-1,1} a_{n-1,2} \cdots, \r a_{n-1,n-1} a_{n-1,n} ;
            a_{n1} a_{n2} \cdots, a_{n,n-1} \r a_{nn}}
            =
            \r a_{11} + a_{22} + \cdots + a_{nn}\color{black}.
          </me>
        </p>
      </statement>
    </definition>
  
    <theorem xml:id="charpoly-is-poly">
      <idx><h>Characteristic polynomial</h><h>form of</h></idx>
      <statement>
        <p>
          Let <m>A</m> be an <m>n\times n</m> matrix, and let <m>f(\lambda)=\det(A-\lambda I_n)</m> be its characteristic polynomial.  Then <m>f(\lambda)</m> is a polynomial of degree <m>n</m>.  Moreover, <m>f(\lambda)</m> has the form
          <me>
            f(\lambda) = (-1)^n\lambda^n + (-1)^{n-1}\Tr(A)\lambda^{n-1} + \cdots + \det(A).
          </me>
          In other words, the coefficient of <m>\lambda^{n-1}</m> is <m>\pm\Tr(A)</m>, and the constant term is <m>\det(A)</m> (the other coefficients are just numbers without names).
        </p>
      </statement>
      <proof visible="true">
        <p>
          First we notice that
          <me>
            f(0) = \det(A - 0I_n) = \det(A),
          </me>
          so that the constant term is always <m>\det(A)</m>.
        </p>
        <p>
          We will prove the rest of the theorem only for <m>2\times 2</m> matrices; the reader is encouraged to complete the proof in general using cofactor expansions.  We can write a <m>2\times 2</m> matrix as <m>A = \smallmat  abcd</m>; then
          <me>
            \begin{split}
            f(\lambda) \amp= \det(A-\lambda I_2)
            = \det\mat{a-\lambda, b; c d-\lambda}
            = (a-\lambda)(d-\lambda)-bc \\
            \amp=\lambda^2 - (a+d)\lambda + (ad-bc)
            = \lambda^2 - \Tr(A)\lambda + \det(A).
            \end{split}
          </me>
        </p>
      </proof>
    </theorem>
  
    <bluebox>
      <title>Recipe: The characteristic polynomial of a <m>2\times 2</m> matrix</title>
      <idx><h>Characteristic polynomial</h><h>of a <m>2\times 2</m> matrix</h></idx>
      <p>
        When <m>n=2</m>, the previous <xref ref="charpoly-is-poly"/> tells us all of the coefficients of the characteristic polynomial:
        <me>
          f(\lambda) = \lambda^2 - \Tr(A)\lambda + \det(A).
        </me>
        This is generally the fastest way to compute the characteristic polynomial of a <m>2\times 2</m> matrix.
      </p>
    </bluebox>
  
    <example>
      <statement>
        <p>
          Find the characteristic polynomial of the matrix
          <me>
            A = \mat{5 2; 2 1}.
          </me>
        </p>
      </statement>
      <solution>
        <p>
          We have
          <me>
            f(\lambda) = \lambda^2 - \Tr(A)\lambda + \det(A)
            = \lambda^2 - (5+1)\lambda + (5\cdot 1-2\cdot 2)
            = \lambda^2 - 6\lambda + 1,
          </me>
          as in the above <xref ref="charpoly-eg-22"/>.
        </p>
      </solution>
    </example>
  
    <remark>
      <p>
        By the above <xref ref="charpoly-is-poly"/>, the characteristic polynomial of an <m>n\times n</m> matrix is a polynomial of degree <m>n</m>.  Since a polynomial of degree <m>n</m> has at most <m>n</m> roots, this gives another proof of the fact that an <m>n\times n</m> matrix has at most <m>n</m> eigenvalues.  See this <xref ref="evecs-max-nr"/>.
      </p>
    </remark>
  
    </paragraphs>
  
    <paragraphs>
      <title>Eigenvalues of a triangular matrix</title>
  
      <p>
        It is easy to compute the determinant of an upper- or lower-triangular matrix; this makes it easy to find its eigenvalues as well.
      </p>
  
      <corollary>
        <idx><h>Eigenvalue</h><h>of a triangular matrix</h></idx>
        <idx><h>Matrix</h><h>upper-triangular</h><h>eigenvalues of</h></idx>
        <idx><h>Matrix</h><h>lower-triangular</h><h>eigenvalues of</h></idx>
        <statement>
          <p>
            If <m>A</m> is an upper- or lower-triangular matrix, then the eigenvalues of <m>A</m> are its diagonal entries.
          </p>
        </statement>
        <proof visible="true">
          <p>
            Suppose for simplicity that <m>A</m> is a <m>3\times 3</m> upper-triangular matrix:
            <me>
              A = \mat{a_{11} a_{12} a_{13}; 0 a_{22} a_{23}; 0 0 a_{33}}.
            </me>
            Its characteristic polynomial is
            <me>
              f(\lambda) = \det(A-\lambda I_3)
              = \det\mat{a_{11}-\lambda, a_{12} a_{13};
              0 a_{22}-\lambda, a_{23};
              0 0 a_{33}-\lambda}.
            </me>
            This is also an upper-triangular matrix, so the determinant is the product of the diagonal entries:
            <me>
              f(\lambda) = (a_{11}-\lambda)(a_{22}-\lambda)(a_{33}-\lambda).
            </me>
            The zeros of this polynomial are exactly <m>a_{11},\,a_{22},\,a_{33}.</m>
          </p>
        </proof>
      </corollary>
  
      <example>
        <statement>
          <p>
            Find the eigenvalues of the matrix
            <me>
              A = \mat{1 7 2 4; 0 1 3 11; 0 0 \pi, 101; 0 0 0 0}.
            </me>
          </p>
        </statement>
        <solution>
          <p>
            The eigenvalues are the diagonal entries <m>1,\pi,0</m>.  (The eigenvalue <m>1</m> occurs twice, but it counts as one eigenvalue; in <xref ref="diagonalization"/> we will define the notion of <em>algebraic multiplicity</em> of an eigenvalue.)
          </p>
        </solution>
      </example>
  
    </paragraphs>
  
  
    <paragraphs>
      <title>Factoring the characteristic polynomial</title>
      <idx><h>Characteristic polynomial</h><h>factoring by hand</h></idx>
      <idx><h>Polynomial</h><h>factoring by hand</h></idx>
  
      <p>
        If <m>A</m> is an <m>n\times n</m> matrix, then the characteristic polynomial <m>f(\lambda)</m> has degree <m>n</m> by the above <xref ref="charpoly-is-poly"/>.  When <m>n=2</m>, one can use the quadratic formula to find the roots of <m>f(\lambda)</m>.  There exist algebraic formulas for the roots of <url href="https://en.wikipedia.org/wiki/Cubic_function#General_solution_to_the_cubic_equation_with_real_coefficients">cubic</url> and <url href="https://en.wikipedia.org/wiki/Quartic_function#Solving_a_quartic_equation">quartic</url> polynomials, but these are generally too cumbersome to apply by hand.  Even worse, it is known that there is <url href="https://en.wikipedia.org/wiki/Abel%E2%80%93Ruffini_theorem">no algebraic formula</url> for the roots of a general polynomial of degree at least <m>5</m>.
      </p>
  
      <p>
        In practice, the roots of the characteristic polynomial are found <url href="https://en.wikipedia.org/wiki/Newton%27s_method">numerically by computer</url>.  That said, there do exist methods for finding roots by hand.  For instance, we have the following consequence of the <url href="https://en.wikipedia.org/wiki/Rational_root_theorem">rational root theorem</url> (which we also call the rational root theorem):
      </p>
  
      <fact type-name="Rational Root Theorem" xml:id="rational-root-thm">
        <idx><h>Rational Root Theorem</h></idx>
        <idx><h>Polynomial</h><h>rational roots</h></idx>
        <p>
          Suppose that <m>A</m> is an <m>n\times n</m> matrix whose characteristic polynomial <m>f(\lambda)</m> has integer (whole-number) entries.  Then all rational roots of its characteristic polynomial are integer divisors of <m>\det(A)</m>.
        </p>
      </fact>
  
      <p>
        For example, if <m>A</m> has integer entries, then its characteristic polynomial has integer coefficients.  This gives us one way to find a root by hand, if <m>A</m> has an eigenvalue that is a rational number.  Once we have found one root, then we can reduce the degree by <url href="https://en.wikipedia.org/wiki/Polynomial_long_division">polynomial long division</url>.
      </p>
  
      <example xml:id="charpoly-eg-rat-root">
        <statement>
          <p>
            Find the eigenvalues of the matrix
            <me>A = \mat{7 0 3; -3 2 -3; -3 0 -1}.</me>
            Hint: one eigenvalue is an integer.
          </p>
        </statement>
        <solution>
          <p>
            We compute the characteristic polynomial by expanding cofactors along the first row:
            <me>
              \begin{split}
              f(\lambda) = \det(A - \lambda I_3)
              \amp= \det\mat{7-\lambda, 0 3; -3 2-\lambda, -3; -3 0 -1-\lambda} \\
              \amp= (7-\lambda)(2-\lambda)(-1-\lambda) + 3\cdot 3(2-\lambda) \\
              \amp= -\lambda^3 + 8\lambda^2 - 14\lambda + 4.
              \end{split}
            </me>
            The determinant of <m>A</m> is the constant term <m>f(0) = 4</m>; its integer divisors are <m>\pm 1,\pm 2,\pm 4</m>.  We check which are roots:
            <me>
              f(1) = -3 \quad f(-1) = 27 \quad f(2) = 0 \quad f(-2) = 72 \quad f(4) = 12 \quad f(-4) = 252.
            </me>
            The only rational root of <m>f(\lambda)</m> is <m>\lambda=2</m>.  We divide by <m>\lambda-2</m> using polynomial long division:
            <me>
              \frac{-\lambda^3 + 8\lambda^2 - 14\lambda + 4}{\lambda-2} = -\lambda^2 + 6\lambda - 2.
            </me>
            We can use the quadratic formula to find the roots of the quotient:
            <me>
              \lambda = \frac{-6 \pm\sqrt{36 - 4\cdot 2}}{-2}
              = 3\pm\sqrt 7.
            </me>
            We have factored <m>f</m> completely:
            <me>
              f(\lambda) = -(\lambda-2)\bigl(\lambda-(3+\sqrt7)\bigr)\bigl(\lambda-(3-\sqrt7)\bigr).
            </me>
            Therefore, the eigenvalues of <m>A</m> are <m>2,\,3+\sqrt7,\,3-\sqrt7.</m>
          </p>
        </solution>
      </example>
  
      <p>
        In the above example, we could have expanded cofactors along the second column to obtain
        <me>
          f(\lambda) = (2-\lambda)\det\mat{7-\lambda, 3; -3 -1-\lambda}.
        </me>
        Since <m>2-\lambda</m> was the only nonzero entry in its column, this expression already has the <m>2-\lambda</m> term factored out: the rational root theorem was not needed.  The determinant in the above expression is the characteristic polynomial of the matrix <m>\smallmat 73{-3}{-1}</m>, so we can compute it using the trace and determinant:
        <me>
          f(\lambda) = (2-\lambda)\bigl(\lambda^2-(7-1)\lambda+(-7+9)\bigr)
          = (2-\lambda)(\lambda^2-6\lambda+2).
        </me>
      </p>
  
      <example xml:id="charpoly-eg-0-eval">
        <statement>
          <p>
            Find the eigenvalues of the matrix
            <me>A = \mat{7 0 3; -3 2 -3; 4 2 0}.</me>
          </p>
        </statement>
        <solution>
          <p>
            We compute the characteristic polynomial by expanding cofactors along the first row:
            <me>
              \begin{split}
              f(\lambda) = \det(A - \lambda I_3)
              \amp= \det\mat{7-\lambda, 0 3; -3 2-\lambda, -3; 4 2 -\lambda} \\
              \amp= (7-\lambda)\bigl(-\lambda(2-\lambda)+6\bigr) + 3\bigl(-6-4(2-\lambda)\bigr) \\
              \amp= -\lambda^3 + 9\lambda^2 - 8\lambda.
              \end{split}
            </me>
            The constant term is zero, so <m>A</m> has determinant zero.  We factor out <m>\lambda</m>, then eyeball the roots of the quadratic factor:
            <me>
              f(\lambda) = -\lambda(\lambda^2-9\lambda+8) = -\lambda(\lambda-1)(\lambda-8).
            </me>
            Therefore, the eigenvalues of <m>A</m> are <m>0,1,</m> and <m>8</m>.
          </p>
        </solution>
      </example>
  
      <bluebox>
        <title>Finding Eigenvalues of a Matrix Larger than <m>2\times 2</m></title>
        <p>
          Let <m>A</m> be an <m>n\times n</m> matrix.  Here are some strategies for factoring its characteristic polynomial <m>f(\lambda)</m>.  First, you must find one eigenvalue:
          <ol>
            <li>
              Do not multiply out the characteristic polynomial if it is already partially factored!  This happens if you expand cofactors along the second column in this <xref ref="charpoly-eg-rat-root"/>.
            </li>
            <li>
              If there is no constant term, you can factor out <m>\lambda</m>, as in this <xref ref="charpoly-eg-0-eval"/>.
            </li>
            <li>
              If the matrix is triangular, the roots are the diagonal entries.
            </li>
            <li>
              Guess one eigenvalue using the <xref ref="rational-root-thm"/>: if <m>\det(A)</m> is an integer, substitute all (positive and negative) divisors of <m>\det(A)</m> into <m>f(\lambda)</m>.
            </li>
            <li>
              Find an eigenvalue using the geometry of the matrix.  For instance, a <xref ref="evec-eg-reflection" text="title">reflection</xref> has eigenvalues <m>\pm 1</m>.
            </li>
          </ol>
          After obtaining an eigenvalue <m>\lambda_1</m>, use <url href="https://en.wikipedia.org/wiki/Polynomial_long_division">polynomial long division</url> to compute <m>f(\lambda)/(\lambda-\lambda_1)</m>.  This polynomial has lower degree.  If <m>n=3</m> then this is a quadratic polynomial, to which you can apply the quadratic formula to find the remaining roots.
        </p>
      </bluebox>
  
    </paragraphs>
  
  
  </section>
  