3.1.3: Characteristic Polynomial
- Page ID
- 187448
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)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.
Let \(A\) be an \(n\times n\) matrix. The characteristic polynomial of \(A\) is the function \(f(\lambda)\) given by:
\[ f(\lambda) = \det(A-\lambda I) \nonumber \]
We will see below that the characteristic polynomial is in fact a polynomial. Finding the characteristic polynomial means computing the determinant of the matrix \(A-\lambda I\text{,}\) whose entries contain the unknown \(\lambda\).
Find the characteristic polynomial of the matrix
\[ A = \left[\begin{array}{cc}3&2\\2&0\end{array}\right] \nonumber \]
Solution
We have
\[\begin{aligned}f(\lambda)=\det(A-\lambda I)&=\det\left(\left[\begin{array}{cc}3&2\\2&0\end{array}\right]-\left[\begin{array}{cc}\lambda&0\\0&\lambda\end{array}\right]\right) \\ &=\left|\begin{array}{cc}3-\lambda&2\\2&0-\lambda\end{array}\right| \\ &=(3-\lambda)(-\lambda)-2\cdot 2=\lambda^{2}-3\lambda-4\end{aligned} \nonumber \]
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:
\[\begin{aligned}f(\lambda)=\det(A-\lambda I)&=\left|\begin{array}{ccc}-\lambda&6&8\\ \frac{1}{2}&-\lambda&0 \\ 0&\frac{1}{2}&-\lambda\end{array}\right|\\ &=8\left(\frac{1}{4}-0\cdot (-\lambda)\right)-\lambda\left(\lambda^{2}-6\cdot\frac{1}{2}\right) \\ &=-\lambda^{3}+3\lambda+2\end{aligned}\nonumber \]
The point of the characteristic polynomial is that we can use it to compute eigenvalues.
Let \(A\) be an \(n\times n\) matrix, and let \(f(\lambda) = \det(A-\lambda I)\) be its characteristic polynomial. Then a number \(\lambda_0\) is an eigenvalue of \(A\) if and only if \(f(\lambda_0) = 0\).
- Proof
-
The matrix equation \((A-\lambda_0 I)\vec{x}=\vec{0}\) has a nontrivial solution if and only if \(\det(A-\lambda_0 I) = 0\). Therefore,
\[ \begin{split} \lambda_0 \text{ is an eigenvalue of } A \amp\iff A\vec{x} = \lambda_0 \vec{x} \text{ has a nontrivial solution} \\ \amp\iff (A-\lambda_0 I)\vec{x} =\vec{0} \text{ has a nontrivial solution} \\ \amp\iff A - \lambda_0 I \text{ is not invertible} \\ \amp\iff \det(A - \lambda_0 I) = 0 \\ \amp\iff f(\lambda_0) = 0 \end{split} \nonumber \]
By the above, 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.
Find the eigenvalues and eigenvectors of the matrix
\[ A = \left[\begin{array}{cc}3&2\\2&0\end{array}\right] \nonumber \]
Solution
In the above Example \(\PageIndex{9}\) we computed the characteristic polynomial of \(A\) to be \(f(\lambda) = \lambda^2-3\lambda -4\). We can factor this to solve for its roots:
\[ (\lambda-4)(\lambda+1)=0\implies \lambda = -1,4 \nonumber \]
Therefore, the eigenvalues are \(-1\) and \(4\).
To compute the eigenvectors, we solve the homogeneous system of equations \((A-\lambda I)\vec{x}=\vec{0}\) for each eigenvalue \(\lambda\).
When \(\lambda=-1\text{,}\) we have
\[ A-(-1)I = \left[\begin{array}{cc}4&2\\2&1\end{array}\right] \xrightarrow{RREF} \left[\begin{array}{cc}1&\frac{1}{2}\\0&0\end{array}\right] \nonumber \]
The parametric form of the general solution is \(x=-\frac{1}{2}y\text{,}\) so the eigenspace \(E_{-1}\) is the line spanned by \( \left[\begin{array}{c}-\frac{1}{2} \\1\end{array}\right]\).
When \(\lambda=4\text{,}\) we have
\[ A-4I = \left[\begin{array}{cc}-1&2\\2&-4\end{array}\right] \xrightarrow{RREF} \left[\begin{array}{cc}1&-2\\0&0\end{array}\right] \nonumber \]
The parametric form of the general solution is \(x=2y\text{,}\) so the eigenspace \(E_{4}\) is the line spanned by \( \left[\begin{array}{c}2 \\1\end{array}\right]\).
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 \(\PageIndex{10}\) we computed the characteristic polynomial of \(A\) to be \(f(\lambda) = -\lambda^3+3\lambda+2\). We factor this completely:
\[ f(\lambda) = -(\lambda-2)(\lambda+1)^2 \nonumber \]
so the only eigenvalues are \(\lambda = 2,-1\).
We compute the \(2\)-eigenspace, \(E_2\) by solving the homogeneous system \((A-2I)\vec{x}=\vec{0}\). We have
\[A-2I=\left[\begin{array}{ccc}-2&6&8\\ \frac{1}{2}&-2&0\\0&\frac{1}{2}&-2\end{array}\right]\quad\xrightarrow{\text{RREF}}\quad\left[\begin{array}{ccc}1&0&-16\\0&1&-4\\0&0&0\end{array}\right]\nonumber\]
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 \((A+I)\vec{x}=\vec{0}\). We have
\[A+I=\left[\begin{array}{ccc}1&6&8\\ \frac{1}{2}&1&0\\0&\frac{1}{2}&1\end{array}\right]\quad\xrightarrow{\text{RREF}}\quad\left[\begin{array}{ccc}1&0&-4\\0&1&2\\0&0&0\end{array}\right]\nonumber\]
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
\[ \text{Span}\left\{\left[\begin{array}{c}4\\-2\\1\end{array}\right]\right\} \nonumber \]