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5.2: The Characteristic Polynomial

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    70207
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    Objectives
    1. Learn that the eigenvalues of a triangular matrix are the diagonal entries.
    2. Find all eigenvalues of a matrix using the characteristic polynomial.
    3. Learn some strategies for finding the zeros of a polynomial.
    4. Recipe: the characteristic polynomial of a \(2\times 2\) matrix.
    5. Vocabulary words: characteristic polynomial, trace.

    In Section 5.1 we discussed how to decide whether a given number \(\lambda\) 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 eigenvalues of a matrix. This does not reduce to solving a system of linear equations: indeed, it requires solving a nonlinear equation in one variable, namely, finding the roots of the characteristic polynomial.

    Definition \(\PageIndex{1}\): Characteristic Polynomial

    Let \(A\) be an \(n\times n\) matrix. The characteristic polynomial of \(A\) is the function \(f(\lambda)\) given by

    \[ f(\lambda) = \det(A-\lambda I_n). \nonumber \]

    We will see below, Theorem \(\PageIndex{2}\), that the characteristic polynomial is in fact a polynomial. Finding the characterestic polynomial means computing the determinant of the matrix \(A-\lambda I_n\text{,}\) whose entries contain the unknown \(\lambda\).

    Example \(\PageIndex{1}\)

    Find the characteristic polynomial of the matrix

    \[ A = \left(\begin{array}{cc}5&2\\2&1\end{array}\right). \nonumber \]

    Solution

    We have

    \[\begin{aligned}f(\lambda)=\det(A-\lambda I_{2})&=\det\left(\left(\begin{array}{cc}5&2\\2&1\end{array}\right)-\left(\begin{array}{cc}\lambda&0\\0&\lambda\end{array}\right)\right) \\ &=\det\left(\begin{array}{cc}5-\lambda&2\\2&1-\lambda\end{array}\right) \\ &=(5-\lambda)(1-\lambda)-2\cdot 2=\lambda^{2}-6\lambda+1.\end{aligned}\]

    Example \(\PageIndex{2}\)

    Find the characteristic polynomial of the matrix

    \[ A = \left(\begin{array}{ccc}0&6&8\\ \frac{1}{2}&0&0\\0&\frac{1}{2}&0\end{array}\right). \nonumber \]

    Solution

    We compute the determinant by expanding cofactors along the third column:

    \[\begin{aligned}f(\lambda)=\det(A-\lambda I_{3})&=\det\left(\begin{array}{ccc}-\lambda&6&8\\ \frac{1}{2}&-\lambda&0 \\ 0&\frac{1}{2}&-\lambda\end{array}\right) \\ &=8\left(\frac{1}{4}-0\cdot -\lambda\right)-\lambda\left(\lambda^{2}-6\cdot\frac{1}{2}\right) \\ &=-\lambda^{3}+3\lambda+2.\end{aligned}\]

    The point of the characteristic polynomial is that we can use it to compute eigenvalues.

    Theorem \(\PageIndex{1}\): Eigenvalues are Roots of the Characteristic Polynomial

    Let \(A\) be an \(n\times n\) matrix, and let \(f(\lambda) = \det(A-\lambda I_n)\) be its characteristic polynomial. Then a number \(\lambda_0\) is an eigenvalue of \(A\) if and only if \(f(\lambda_0) = 0\).

    Proof

    By the Theorem 5.1.1 in Section 5.1, the matrix equation \((A-\lambda_0 I_n)x=0\) has a nontrivial solution if and only if \(\det(A-\lambda_0 I_n) = 0\). Therefore,

    \[ \begin{split} \lambda_0 \text{ is an eigenvalue of } A \amp\iff Ax = \lambda_0 x \text{ has a nontrivial solution} \\ \amp\iff (A-\lambda_0 I_n)x = 0 \text{ has a nontrivial solution} \\ \amp\iff A - \lambda_0 I_n \text{ is not invertible} \\ \amp\iff \det(A - \lambda_0 I_n) = 0 \\ \amp\iff f(\lambda_0) = 0. \end{split} \nonumber \]

    Example \(\PageIndex{3}\): Finding eigenvalues

    Find the eigenvalues and eigenvectors of the matrix

    \[ A = \left(\begin{array}{cc}5&2\\2&1\end{array}\right). \nonumber \]

    Solution

    In the above Example \(\PageIndex{1}\) we computed the characteristic polynomial of \(A\) to be \(f(\lambda) = \lambda^2-6\lambda + 1\). We can solve the equation \(\lambda^2-6\lambda+1=0\) using the quadratic formula:

    \[ \lambda = \frac{6\pm\sqrt{36-4}}2 = 3\pm 2\sqrt 2. \nonumber \]

    Therefore, the eigenvalues are \(3+2\sqrt 2\) and \(3-2\sqrt 2\).

    To compute the eigenvectors, we solve the homogeneous system of equations \((A-\lambda I_2)x=0\) for each eigenvalue \(\lambda\). When \(\lambda=3+2\sqrt 2\text{,}\) we have

    \[ \begin{split} A-(3+\sqrt 2)I_2 = \amp\left(\begin{array}{cc}2-2\sqrt{2}&2\\2&-2-2\sqrt{2}\end{array}\right) \\ \xrightarrow{R_1 = R_1\times(2+2\sqrt2)}\; \amp\left(\begin{array}{cc}-4&4+4\sqrt{2}\\2&-2-2\sqrt{2}\end{array}\right) \\ \xrightarrow{R_2 = R_2+R_1/2}\; \amp\left(\begin{array}{cc}-4&4+4\sqrt{2}\\0&0\end{array}\right) \\ \xrightarrow{R_1 = R_1 \div -4}\; \amp\left(\begin{array}{cc}1&-1-\sqrt{2}\\0&0\end{array}\right). \end{split} \nonumber \]

    The parametric form of the general solution is \(x=(1+\sqrt 2)y\text{,}\) so the \((3+2\sqrt 2)\)-eigenspace is the line spanned by \({1+\sqrt 2\choose 1}\). We compute in the same way that the \((3-2\sqrt 2)\)-eigenspace is the line spanned by \({1-\sqrt 2\choose 1}\).

    clipboard_edbb8c405921d48286828b31c52a94f46.png

    Figure \(\PageIndex{1}\): The green line is the \((3-2\sqrt 2)\)-eigenspace, and the violet line is the \((3+2\sqrt 2)\)-eigenspace.
    Example \(\PageIndex{4}\): Finding eigenvalues

    Find the eigenvalues and eigenvectors of the matrix

    \[ A = \left(\begin{array}{ccc}0&6&8\\ \frac{1}{2}&0&0\\0&\frac{1}{2}&0\end{array}\right). \nonumber \]

    Solution

    In the above Example \(\PageIndex{2}\) we computed the characteristic polynomial of \(A\) to be \(f(\lambda) = -\lambda^3+3\lambda+2\). We eyeball that \(f(2) = -8 + 3\cdot 2 + 2 = 0\). Thus \(\lambda-2\) divides \(f(\lambda)\text{;}\) to find the other roots, we perform polynomial long division:

    \[ \frac{-\lambda^3+3\lambda+2}{\lambda-2} = -\lambda^2-2\lambda-1 = -(\lambda+1)^2. \nonumber \]

    Therefore,

    \[ f(\lambda) = -(\lambda-2)(\lambda+1)^2, \nonumber \]

    so the only eigenvalues are \(\lambda = 2,-1\).

    We compute the \(2\)-eigenspace by solving the homogeneous system \((A-2I_3)x=0\). We have

    \[A-2I_{3}=\left(\begin{array}{ccc}-2&6&8\\ \frac{1}{2}&-2&0\\0&\frac{1}{2}&-2\end{array}\right)\quad\xrightarrow{\text{RREF}}\quad\left(\begin{array}{ccc}1&0&-16\\0&1&-4\\0&0&0\end{array}\right).\nonumber\]

    The parametric form and parametric vector form of the solutions are:

    \[\left\{\begin{array}{rrr}x&=&16z \\ y&=&4z \\ z&=&z\end{array}\right.\quad\longrightarrow\quad\left(\begin{array}{c}x\\y\\z\end{array}\right)=z\left(\begin{array}{c}16\\4\\1\end{array}\right).\nonumber\]

    Therefore, the \(2\)-eigenspace is the line

    \[ \text{Span}\left\{\left(\begin{array}{c}16\\4\\1\end{array}\right)\right\}. \nonumber \]

    We compute the \(-1\)-eigenspace by solving the homogeneous system \((A+I_3)x=0\). We have

    \[A+I_{3}=\left(\begin{array}{ccc}1&6&8\\ \frac{1}{2}&1&0\\0&\frac{1}{2}&1\end{array}\right)\quad\xrightarrow{\text{RREF}}\quad\left(\begin{array}{ccc}1&0&-4\\0&1&2\\0&0&0\end{array}\right).\nonumber\]

    The parametric form and parametric vector form of the solutions are:

    \[\left\{\begin{array}{rrr}z&=&4z\\ y&=&-2z\\ z&=&z\end{array}\right.\quad\longrightarrow\quad\left(\begin{array}{c}x\\y\\z\end{array}\right)=z\left(\begin{array}{c}4\\-2\\1\end{array}\right).\nonumber\]

    Therefore, the \(-1\)-eigenspace is the line

    \[ \text{Span}\left\{\left(\begin{array}{c}4\\-2\\1\end{array}\right)\right\}. \nonumber \]

    clipboard_eef96504ffd881236db143cac5799e459.png

    Figure \(\PageIndex{2}\): The green line is the \(-1\)-eigenspace, and the violet line is the \(2\)-eigenspace.

    Form of the Characteristic Polynomial

    It is time that we justified the use of the term “polynomial.” First we need a vocabulary word.

    Definition \(\PageIndex{2}\): Trace

    The trace of a square matrix \(A\) is the number \(\text{Tr}(A)\) obtained by summing the diagonal entries of \(A\text{:}\)

    \[\text{Tr}\left(\begin{array}{ccccc}\color{red}{a_{11}}&\color{black}{a_{12}}&\cdots&a_{1,n-1}&a_{1n} \\ a_{21}&\color{red}{a_{22}}&\color{black}{\cdots}&a_{2,n-1}&a_{2n} \\ \vdots &\vdots &\ddots &\vdots &\vdots \\ a_{n-1,1}&a_{n-1,2}&\cdots&\color{red}{a_{n-1,n-1}}&\color{black}{a_{n-1,n}} \\ a_{n1}&a_{n2}&\cdots &a_{n,n-1}&\color{red}{a_{nn}}\end{array}\right)=\color{red}{a_{11}+a_{22}+\cdots +a_{nn}}\color{black}{.}\nonumber\]

    Theorem \(\PageIndex{2}\)

    Let \(A\) be an \(n\times n\) matrix, and let \(f(\lambda)=\det(A-\lambda I_n)\) be its characteristic polynomial. Then \(f(\lambda)\) is a polynomial of degree \(n\). Moreover, \(f(\lambda)\) has the form

    \[ f(\lambda) = (-1)^n\lambda^n + (-1)^{n-1}\text{Tr}(A)\lambda^{n-1} + \cdots + \det(A). \nonumber \]

    In other words, the coefficient of \(\lambda^{n-1}\) is \(\pm\text{Tr}(A)\text{,}\) and the constant term is \(\det(A)\) (the other coefficients are just numbers without names).

    Proof

    First we notice that

    \[ f(0) = \det(A - 0I_n) = \det(A), \nonumber \]

    so that the constant term is always \(\det(A)\).

    We will prove the rest of the theorem only for \(2\times 2\) matrices; the reader is encouraged to complete the proof in general using cofactor expansions. We can write a \(2\times 2\) matrix as \(A = \left(\begin{array}{cc}a&b\\c&d\end{array}\right)\text{;}\) then

    \[ \begin{split} f(\lambda) \amp= \det(A-\lambda I_2) = \det\left(\begin{array}{cc}a-\lambda&b\\c&d-\lambda\end{array}\right) = (a-\lambda)(d-\lambda)-bc \\ \amp=\lambda^2 - (a+d)\lambda + (ad-bc) = \lambda^2 - \text{Tr}(A)\lambda + \det(A). \end{split} \nonumber \]

    Recipe: The Characteristic Polynomial of a \(2\times 2\) Matrix

    When \(n=2\text{,}\) the previous Theorem \(\PageIndex{2}\) tells us all of the coefficients of the characteristic polynomial:

    \[ f(\lambda) = \lambda^2 - \text{Tr}(A)\lambda + \det(A). \nonumber \]

    This is generally the fastest way to compute the characteristic polynomial of a \(2\times 2\) matrix.

    Example \(\PageIndex{5}\)

    Find the characteristic polynomial of the matrix

    \[ A = \left(\begin{array}{cc}5&2\\2&1\end{array}\right). \nonumber \]

    Solution

    We have

    \[ f(\lambda) = \lambda^2 - \text{Tr}(A)\lambda + \det(A) = \lambda^2 - (5+1)\lambda + (5\cdot 1-2\cdot 2) = \lambda^2 - 6\lambda + 1, \nonumber \]

    as in the above Example \(\PageIndex{1}\).

    Remark

    By the above Theorem \(\PageIndex{2}\), the characteristic polynomial of an \(n\times n\) matrix is a polynomial of degree \(n\). Since a polynomial of degree \(n\) has at most \(n\) roots, this gives another proof of the fact that an \(n\times n\) matrix has at most \(n\) eigenvalues. See Note 5.1.3 in Section 5.1.

    Eigenvalues of a Triangular Matrix

    It is easy to compute the determinant of an upper- or lower-triangular matrix; this makes it easy to find its eigenvalues as well.

    Corollary \(\PageIndex{1}\)

    If \(A\) is an upper- or lower-triangular matrix, then the eigenvalues of \(A\) are its diagonal entries.

    Proof

    Suppose for simplicity that \(A\) is a \(3\times 3\) upper-triangular matrix:

    \[ A = \left(\begin{array}{ccc}a_{11}&a_{12}&a_{13}\\0&a_{22}&a_{23}\\0&0&a_{33}\end{array}\right). \nonumber \]

    Its characteristic polynomial is

    \[ f(\lambda) = \det(A-\lambda I_3) = \det\left(\begin{array}{ccc}a_{11}-\lambda&a_{12}&a_{13}\\0&a_{22}-\lambda&a_{23}\\0&0&a_{33}-\lambda\end{array}\right). \nonumber \]

    This is also an upper-triangular matrix, so the determinant is the product of the diagonal entries:

    \[ f(\lambda) = (a_{11}-\lambda)(a_{22}-\lambda)(a_{33}-\lambda). \nonumber \]

    The zeros of this polynomial are exactly \(a_{11},\,a_{22},\,a_{33}.\)

    Example \(\PageIndex{6}\)

    Find the eigenvalues of the matrix

    \[ A = \left(\begin{array}{cccc}1&7&2&4\\0&1&3&11\\0&0&\pi&101\\0&0&0&0\end{array}\right). \nonumber \]

    Solution

    The eigenvalues are the diagonal entries \(1,\pi,0\). (The eigenvalue \(1\) occurs twice, but it counts as one eigenvalue; in Section 5.4 we will define the notion of algebraic multiplicity of an eigenvalue.)

    Factoring the Characteristic Polynomial

    If \(A\) is an \(n\times n\) matrix, then the characteristic polynomial \(f(\lambda)\) has degree \(n\) by the above Theorem \(\PageIndex{2}\). When \(n=2\text{,}\) one can use the quadratic formula to find the roots of \(f(\lambda)\). There exist algebraic formulas for the roots of cubic and quartic polynomials, but these are generally too cumbersome to apply by hand. Even worse, it is known that there is no algebraic formula for the roots of a general polynomial of degree at least \(5\).

    In practice, the roots of the characteristic polynomial are found numerically by computer. That said, there do exist methods for finding roots by hand. For instance, we have the following consequence of the rational root theorem (which we also call the rational root theorem):

    Theorem \(\PageIndex{3}\): Rational Root Theorem

    Suppose that \(A\) is an \(n\times n\) matrix whose characteristic polynomial \(f(\lambda)\) has integer (whole-number) entries. Then all rational roots of its characteristic polynomial are integer divisors of \(\det(A)\).

    For example, if \(A\) has integer entries, then its characteristic polynomial has integer coefficients. This gives us one way to find a root by hand, if \(A\) has an eigenvalue that is a rational number. Once we have found one root, then we can reduce the degree by polynomial long division.

    Example \(\PageIndex{7}\)

    Find the eigenvalues of the matrix

    \[ A = \left(\begin{array}{ccc}7&0&3\\-3&2&-3\\-3&0&-1\end{array}\right). \nonumber \]

    Hint: one eigenvalue is an integer.

    Solution

    We compute the characteristic polynomial by expanding cofactors along the first row:

    \[ \begin{split} f(\lambda) = \det(A - \lambda I_3) \amp= \det\left(\begin{array}{ccc}7-\lambda&0&3\\-3&2-\lambda&-3\\-3&0&-1-\lambda\end{array}\right) \\ \amp= (7-\lambda)(2-\lambda)(-1-\lambda) + 3\cdot 3(2-\lambda) \\ \amp= -\lambda^3 + 8\lambda^2 - 14\lambda + 4. \end{split} \nonumber \]

    The determinant of \(A\) is the constant term \(f(0) = 4\text{;}\) its integer divisors are \(\pm 1,\pm 2,\pm 4\). We check which are roots:

    \[ f(1) = -3 \quad f(-1) = 27 \quad f(2) = 0 \quad f(-2) = 72 \quad f(4) = 12 \quad f(-4) = 252. \nonumber \]

    The only rational root of \(f(\lambda)\) is \(\lambda=2\). We divide by \(\lambda-2\) using polynomial long division:

    \[ \frac{-\lambda^3 + 8\lambda^2 - 14\lambda + 4}{\lambda-2} = -\lambda^2 + 6\lambda - 2. \nonumber \]

    We can use the quadratic formula to find the roots of the quotient:

    \[ \lambda = \frac{-6 \pm\sqrt{36 - 4\cdot 2}}{-2} = 3\pm\sqrt 7. \nonumber \]

    We have factored \(f\) completely:

    \[ f(\lambda) = -(\lambda-2)\bigl(\lambda-(3+\sqrt7)\bigr)\bigl(\lambda-(3-\sqrt7)\bigr). \nonumber \]

    Therefore, the eigenvalues of \(A\) are \(2,\,3+\sqrt7,\,3-\sqrt7.\)

    In the above example, we could have expanded cofactors along the second column to obtain

    \[ f(\lambda) = (2-\lambda)\det\left(\begin{array}{cc}7-\lambda&3\\-3&-1-\lambda\end{array}\right). \nonumber \]

    Since \(2-\lambda\) was the only nonzero entry in its column, this expression already has the \(2-\lambda\) term factored out: the rational root theorem was not needed. The determinant in the above expression is the characteristic polynomial of the matrix \(\left(\begin{array}{cc}7&3\\-3&-1\end{array}\right)\text{,}\) so we can compute it using the trace and determinant:

    \[ f(\lambda) = (2-\lambda)\bigl(\lambda^2-(7-1)\lambda+(-7+9)\bigr) = (2-\lambda)(\lambda^2-6\lambda+2). \nonumber \]

    Example \(\PageIndex{8}\)

    Find the eigenvalues of the matrix

    \[ A = \left(\begin{array}{ccc}7&0&3\\-3&2&-3\\4&2&0\end{array}\right). \nonumber \]

    Solution

    We compute the characteristic polynomial by expanding cofactors along the first row:

    \[ \begin{split} f(\lambda) = \det(A - \lambda I_3) \amp= \det\left(\begin{array}{ccc}7-\lambda&0&3\\-3&2-\lambda&-3\\4&2&-\lambda\end{array}\right) \\ \amp= (7-\lambda)\bigl(-\lambda(2-\lambda)+6\bigr) + 3\bigl(-6-4(2-\lambda)\bigr) \\ \amp= -\lambda^3 + 9\lambda^2 - 8\lambda. \end{split} \nonumber \]

    The constant term is zero, so \(A\) has determinant zero. We factor out \(\lambda\text{,}\) then eyeball the roots of the quadratic factor:

    \[ f(\lambda) = -\lambda(\lambda^2-9\lambda+8) = -\lambda(\lambda-1)(\lambda-8). \nonumber \]

    Therefore, the eigenvalues of \(A\) are \(0,1,\) and \(8\).

    Note \(\PageIndex{1}\): Finding Eigenvalues of a Matrix Larger than \(2\times 2\)

    Let \(A\) be an \(n\times n\) matrix. Here are some strategies for factoring its characteristic polynomial \(f(\lambda)\). First, you must find one eigenvalue:

    1. 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 Example \(\PageIndex{7}\).
    2. If there is no constant term, you can factor out \(\lambda\text{,}\) as in this Example \(\PageIndex{8}\).
    3. If the matrix is triangular, the roots are the diagonal entries.
    4. Guess one eigenvalue using the Theorem \(\PageIndex{3}\): if \(\det(A)\) is an integer, substitute all (positive and negative) divisors of \(\det(A)\) into \(f(\lambda)\).
    5. Find an eigenvalue using the geometry of the matrix. For instance, a reflection, Example 5.1.4 in Section 5.1, has eigenvalues \(\pm 1\).

    After obtaining an eigenvalue \(\lambda_1\text{,}\) use polynomial long division to compute \(f(\lambda)/(\lambda-\lambda_1)\). This polynomial has lower degree. If \(n=3\) then this is a quadratic polynomial, to which you can apply the quadratic formula to find the remaining roots.


    This page titled 5.2: The Characteristic Polynomial is shared under a GNU Free Documentation License 1.3 license and was authored, remixed, and/or curated by Dan Margalit & Joseph Rabinoff via source content that was edited to the style and standards of the LibreTexts platform.