Objectives: Topics
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The characteristic polynomial of a matrix
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Algebraic and geometric multiplicity of eigenvalues
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Section 5.2 The Characteristic Equation
Handout 5.2 The Characteristic Equation
Objectives: Goals
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Construct the characteristic polynomial of a matrix and use it to identify eigenvalues and their multiplicities.
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Characterize the long-term behaviour of dynamical systems using eigenvalue decompositions.
Note 5.14.
Similar matrices are introduced in Section 5.2, but are not used until Section 5.3. We will wait to introduce them in Section 5.3.
Recall:
\(\lambda\) is an eigenvalue of \(A\) if and only if \((A-\lambda I)\) is not invertible.
Therefore, to calculate the eigenvalues of \(A\text{,}\) we can need to find all \(\lambda\) so that
\begin{equation*}
\det(A-\lambda I)=0.
\end{equation*}
Definition 5.15 Charteristic Polynomial and Characteristic Equation.
The quantity \(\det(A-\lambda I)\) is the characteristic polynomial of \(A\text{.}\)
The equation \(\det(A - \lambda I) = 0\) is the characteristic equation of \(A\text{.}\)
The roots of the characteristic polynomial are the eigenvalues of \(A\text{.}\)
Example 5.16.
Find the characteristic polynomial of
\begin{equation*}
A=\begin{bmatrix}
5 \amp 2\\
2 \amp 1
\end{bmatrix}.
\end{equation*}
Then determine the eigenvalues of \(A\text{.}\)
Solution.
Note that
\(\begin{eqnarray*} \det(A - \lambda I) \amp = \amp \begin{vmatrix} 5-\lambda \amp 2 \\ 2 \amp 1 - \lambda \end{vmatrix} \\ \amp = \amp (5-\lambda)(1-\lambda) - 2(2) \\ \amp = \amp \lambda^2 - 6\lambda + 5 - 4 \\ \amp = \amp \lambda^2 - 6\lambda + 1. \end{eqnarray*}\)
Therefore, the characteristic polynomial of \(A\) is \(\lambda^2 - 6\lambda + 1\text{.}\) We know from Definition 5.15 that the roots of this polynomial are the eigenvalues of \(A\text{.}\) Since the polynomial doesn’t factor nicely, we use the quadratic formula to find the eigenvalues:
\begin{equation*}
\lambda = \dfrac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(1)}}{2(1)} = \dfrac{6 \pm \sqrt{32}}{2} = 3 \pm 2\sqrt{2}\text{.}
\end{equation*}
Therefore, the eigenvalues of \(A\) are \(\lambda_1 = 3 + 2\sqrt{2}\) and \(\lambda_2 = 3 - 2\sqrt{2}\text{.}\)
Example 5.17.
Let
\begin{equation*}
M=\begin{bmatrix}
a \amp b\\
c \amp d
\end{bmatrix}.
\end{equation*}
Express the characteristic equation of \(M\) in terms of \(\det(M)\text{.}\) What is the equation when \(M\) is singular?
Solution.
Note that
\(\begin{eqnarray*} \det(M - \lambda I) \amp = \amp \begin{vmatrix} a-\lambda \amp b \\ c \amp d - \lambda \end{vmatrix} \\ \amp = \amp (a-\lambda)(d-\lambda) - bc \\ \amp = \amp \lambda^2 - a\lambda - d\lambda + ad - bc \\ \amp = \amp \lambda^2 - (a+d)\lambda + (ad - bc). \end{eqnarray*}\)
We see that the constant term of the characteristic polynomial is \(ad - bc = \det(M)\text{.}\) Furthermore, the linear coefficient is \(-(a + d) = -\operatorname{tr}(M)\text{,}\) where \(\operatorname{tr}(M)\) is the trace of \(M\) and is the sum of its diagonal elements. When \(M\) is singular, we have \(\det(M) = 0\text{,}\) so the characteristic polynomial is \(\lambda^2 - (a+d)\lambda\text{,}\) which has a factor of \(\lambda\text{.}\) Therefore, we see that \(\lambda_1 = 0\) is an eigenvalue of \(M\text{.}\)
Suppose \(\lambda_1\) and \(\lambda_2\) are the eigenvalues of \(M\text{.}\) Then, we have that
\begin{equation*}
\lambda^2 - \operatorname{tr}(M)\lambda + \det(M) = (\lambda - \lambda_1)(\lambda - \lambda_2) = \lambda^2 - (\lambda_1 + \lambda_2)\lambda + \lambda_1\lambda_2\text{.}
\end{equation*}
It follows that \(\operatorname{tr}(M)\) is the sum of the eigenvalues of \(M\) and \(\det(M)\) is the product of the eigenvalues of \(M\text{.}\) This is no coincidence and is true for all \(n\times n\) matrices.
Theorem 5.18 Determinant, Trace, and Eigenvalues.
Let \(A\) be an \(n \times n\) matrix with eigenvalues \(\lambda_1, \dots, \lambda_n\text{.}\) Then, \(\det(A) = \lambda_1\cdots\lambda_n\) and \(\operatorname{tr}(A) = \lambda_1 + \cdots + \lambda_n\text{.}\)
Definition 5.19 Algebraic Multiplicity.
The algebraic multiplicity of an eigenvalue is its multiplicity as a root of the characteristic polynomial.
Example 5.20.
Compute the algebraic multiplicities of the eigenvalues for
\begin{equation*}
A = \begin{bmatrix}
1 \amp 0 \amp 0 \amp 0\\
0 \amp 2 \amp 0 \amp 0\\
0 \amp 0 \amp -1 \amp 1\\
0 \amp 0 \amp 0 \amp -1
\end{bmatrix}.
\end{equation*}
Solution.
Note that
\(\begin{eqnarray*}\det(A - \lambda I) \amp = \amp \begin{vmatrix} 1 - \lambda \amp 0 \amp 0 \amp 0 \\ 0 \amp 2-\lambda \amp 0 \amp 0 \\ 0 \amp 0 \amp -1 - \lambda \amp 1 \\ 0 \amp 0 \amp 0 \amp -1-\lambda \end{vmatrix}\\ \amp = \amp (1-\lambda)(2-\lambda)(-1-\lambda)(-1-\lambda) \\ \amp = \amp (1-\lambda)(2-\lambda)(-1-\lambda)^2.\end{eqnarray*}\)
It follows that \(\lambda_1 = 1\) is an eigenvalue of \(A\) with algebraic multiplicity equal to 1 and \(\lambda_2 = 2\) is an eigenvalue of \(A\) with algebraic multiplicity equal to 1. Since the characteristic polynomials has a factor of \((-1-\lambda)^2\text{,}\) we see that \(\lambda_3 = -1\) is an eigenvalue of \(A\) with algebraic multiplicity equal to 2.
Definition 5.21 Geometric Multiplicity.
The geometric multiplicity of an eigenvalue \(\lambda\) is the dimension of \(\operatorname{Nul}(A-\lambda I)\text{.}\)
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The geometric multiplicity of an eigenvalue \(\lambda\) is always between 1 and the algebraic multiplicity of \(\lambda\text{.}\) It can never be 0 nor be greater than the algebraic mutliplicy.
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Basic example:\begin{equation*} \begin{bmatrix} 1 \amp 1\\ 0 \amp 1 \end{bmatrix} \end{equation*}has only eigenvalue \(\lambda=1\text{.}\) Its algebraic multiplicity is 2, but its geometric multiplicity is 1.
Example 5.22.
Compute the geometric multiplicities of the eigenvalues for
\begin{equation*}
A = \begin{bmatrix}
1 \amp 0 \amp 0 \amp 0\\
0 \amp 2 \amp 0 \amp 0\\
0 \amp 0 \amp -1 \amp 1\\
0 \amp 0 \amp 0 \amp -1
\end{bmatrix}.
\end{equation*}
Solution.
As we saw in Example 5.20, the eigenvalues of \(A\) are \(\lambda_1 = 1\) (with algebraic multiplicity = 1), \(\lambda_2 = 2\) (with algebraic multiplicity = 1), and \(\lambda_3 = -1\) (with algebraic multiplicity = 2). For \(\lambda_1 = 1\text{,}\) we see that
\begin{equation*}
A - 1I = \begin{bmatrix} 0 \amp 0 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp -2 \amp 1 \\ 0 \amp 0 \amp 0 \amp -2 \end{bmatrix}
\end{equation*}
which has a single free variable. Thus, the geometric multiplicity of \(\lambda_1 = 1\) is 1 (which is actually forced since the geometric multiplicity has to be between 1 and the algebraic multiplicity which was also 1 for this eigenvalue). By similar reasoning, we see that the geometric multiplicity of \(\lambda_2 = 2\) is 1.
Now consider \(\lambda_3 = -1\text{.}\) The algebraic multiplicity of \(\lambda_3 = -1\) is 2. Therefore, the geometric multiplicity could be 1 or 2. Note that
\begin{equation*}
A - (-1)I = \begin{bmatrix} 2 \amp 0 \amp 0 \amp 0 \\ 0 \amp 3 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0 \amp 1 \\ 0 \amp 0 \amp 0 \amp 0 \end{bmatrix}
\end{equation*}
which has a single free variable. Therefore, the geometric multiplicity for \(\lambda_3 = -1\) is 1, not 2.
Example 5.23.
For each part below, give an example of a \(5\times 5\) matrix withe the stated properties, if possible.
(a)
The only eigenvalue is \(\lambda=0\text{,}\) but the geometric multiplicity of \(\lambda=0\) is one.
Solution.
For simplicity, we will attempt to form an upper triangular matrix with the properties. This makes things easier since we know that the eigenvalues of a triangular matrix are exactly the entries on the diagonal. Therefore, each diagonal element of our upper triangular matrix needs to be 0. This leads us to
\begin{equation*}
A = \begin{bmatrix} 0 \amp \amp \amp \amp \\ 0 \amp 0 \amp \amp \amp \\ 0 \amp 0 \amp 0 \amp \amp \\ 0 \amp 0 \amp 0 \amp 0 \amp \\ 0 \amp 0 \amp 0 \amp 0 \amp 0\end{bmatrix}
\end{equation*}
where the blank entries are undecided. So far, we have guaranteed that every eigenvalue of \(A\) is \(\lambda = 0\text{.}\) Now, we need to force \(\lambda = 0\) to have geometric multiplicity equal to 1. This means that the matrix \(A - 0I = A\) needs to have exactly one free variable. Note that, as is, Column 1 of \(A\) already corresponds to a free variable. Thus, we need to make sure that Columns 2, 3, 4, and 5 are pivot columns. To do that, we need to put nonzero entries directly above the diagonal entry. This give the matrix
\begin{equation*}
A = \begin{bmatrix} 0 \amp 1 \amp \amp \amp \\ 0 \amp 0 \amp 2 \amp \amp \\ 0 \amp 0 \amp 0 \amp 3 \amp \\ 0 \amp 0 \amp 0 \amp 0 \amp 4 \\ 0 \amp 0 \amp 0 \amp 0 \amp 0\end{bmatrix}\text{.}
\end{equation*}
We can fill in the remaining entries with whatever we want and \(A\) will still have the required properties.
(b)
The only eigenvalues are \(\lambda_1=0\) and \(\lambda_2 = 1\text{.}\) The algebraic multiplicity of \(\lambda_1=0\) is three and the geometric multiplicity of \(\lambda_2 = 1\) is three.
Solution.
We can proceed in a similar manner to Task 5.23.a. Since \(\lambda_1 = 0\) has algebraic multiplicity 3, we must list it three times on the diagonal. We can fill in the remaining two entries on the diagonal with the other eigenvalue \(\lambda_2 = 1\text{.}\) This gives us
\begin{equation*}
B = \begin{bmatrix} 0 \amp \amp \amp \amp \\ 0 \amp 0 \amp \amp \amp \\ 0 \amp 0 \amp 0 \amp \amp \\ 0 \amp 0 \amp 0 \amp 1 \amp \\ 0 \amp 0 \amp 0 \amp 0 \amp 1\end{bmatrix}\text{.}
\end{equation*}
Note that \(B\) is supposed to have the geometric multiplicity of \(\lambda_2 = 1\) to be 3. Unfortunately, the algebraic multiplicity of \(\lambda_2 = 1\) is 2. So it is impossible to have the geometric multiplicity of \(\lambda_2 = 1\) to be 3 since the geometric multiplicity of an eigenvalue can never exceed its algebraic multiplicity. It follows that creating a matrix with the desired properties is impossible. It would require at least a \(6 \times 6\) matrix.
(c)
Exactly three eigenvalues, all with geometric multiplicity equal to 1.
Solution.
Note that the two matrices below both satisfy the requirement and have eigenvalues \(\lambda = 1,2,3\text{.}\)
\begin{equation*}
C = \begin{bmatrix} 1 \amp 0 \amp 0 \amp 0 \amp 0 \\ 0 \amp 2 \amp 0 \amp 0 \amp 0 \\ 0 \amp 0 \amp 3 \amp 1 \amp 0 \\ 0 \amp 0 \amp 0 \amp 3 \amp 1 \\ 0 \amp 0 \amp 0 \amp 0 \amp 3 \end{bmatrix} \quad \quad D = \begin{bmatrix} 1 \amp 1 \amp 0 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \amp 0 \amp 0 \\ 0 \amp 0 \amp 2 \amp 1 \amp 0 \\ 0 \amp 0 \amp 0 \amp 2 \amp 0 \\ 0 \amp 0 \amp 0 \amp 0 \amp 3 \end{bmatrix}\text{.}
\end{equation*}
Example 5.24.
Consider the Markov Chain given by
\begin{equation*}
\vec x_0=
\begin{bmatrix}
1\\
0
\end{bmatrix},
\quad \vec x_{k+1}=P\vec x_k
=
\begin{bmatrix}
0.6 \amp 0.4\\
0.4 \amp 0.6
\end{bmatrix}
\vec x_k.
\end{equation*}
Nodes A and B with transitions 0.6 for staying and 0.4 for switching; matrix \(P=\begin{bmatrix}0.6 \amp 0.4\\ 0.4 \amp 0.6\end{bmatrix}\text{.}\)
In the parts below, we will use eigenvalues to describe the long-term behavior of the system.
(a)
Find the eigenvalues of \(P\) and corresponding eigenvectors.
Solution.
Note that \(\det(P) = 0.6^2 - 0.4^2 = 0.2\) and \(\operatorname{tr}(P) = 0.6 + 0.6 = 1.2\text{.}\) It follows by the result in Example 5.17 that the characteristic polynomial of \(P\) is
\begin{equation*}
\det(P-\lambda I) = \lambda^2 - \operatorname{tr}(P)\lambda + \det(P) = \lambda^2 -1.2\lambda + 0.2\text{.}
\end{equation*}
Setting the characteristic polynomial equal to 0, multiplying both sides by 5, and factoring yields:
\begin{equation*}
5\lambda^2 - 6\lambda + 1 = 0 \quad \implies \quad (\lambda - 1)(5\lambda - 1) = 0\text{.}
\end{equation*}
Therefore, the eigenvalues of \(P\) are \(\lambda_1 = 1\) and \(\lambda_2 = \frac{1}{5}\text{.}\) (We knew that \(\lambda = 1\) had to be an eigenvalue of \(P\) since \(P\) is a stochastic matrix.)
For \(\lambda_1 = 1\text{,}\) note that
\begin{equation*}
P - I = \begin{bmatrix} -0.4 \amp 0.4 \\ 0.4 \amp -0.4 \end{bmatrix} \sim \begin{bmatrix} 1\amp -1 \\ 0 \amp 0 \end{bmatrix}\text{.}
\end{equation*}
It follows that a nonzero vector in \(\Nul(P - I)\) is \(\vec v_1 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}\text{.}\) Furthermore, since \(P - I\) had a single free variable, we have that
\begin{equation*}
1\text{-eigenspace}(P) = \Nul(P - I) = \Span\left\{\begin{bmatrix} 1 \\ 1 \end{bmatrix}\right\}\text{.}
\end{equation*}
For \(\lambda_2 = \frac{1}{5}\text{,}\) note that
\begin{equation*}
P - \frac{1}{5}I = P - 0.2I = \begin{bmatrix} 0.4 \amp 0.4 \\ 0.4 \amp 0.4 \end{bmatrix} \sim \begin{bmatrix} 1 \amp 1 \\ 0 \amp 0 \end{bmatrix}\text{.}
\end{equation*}
It follows that a nonzero vector in \(\Nul(P - \frac{1}{5}I)\) is \(\vec v_2 = \begin{bmatrix} 1 \\ -1 \end{bmatrix}\text{.}\) Furthermore, since \(P - \frac{1}{5}I\) had a single free variable, we have that
\begin{equation*}
\frac{1}{5}\text{-eigenspace}(P) = \Nul(P - \frac{1}{5}I) = \Span\left\{\begin{bmatrix} 1 \\ -1 \end{bmatrix}\right\}\text{.}
\end{equation*}
(b)
Use the eigenvalues and eigenvectors to determine what \(\vec x_k\) tends to as \(k\to\infty\text{.}\)
Solution.
Note that
\begin{equation*}
\vec x_0 = \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \frac{1}{2}\begin{bmatrix} 1 \\ 1 \end{bmatrix} + \frac{1}{2}\begin{bmatrix} 1 \\ -1 \end{bmatrix} = \frac{1}{2}\vec v_1 + \frac{1}{2}\vec v_2\text{,}
\end{equation*}
where \(\vec v_1\) and \(\vec v_2\) are eigenvectors corresponding to \(\lambda_1 = 1\) and \(\lambda_2 = \frac{1}{5}\text{,}\) respectively. Note that
\begin{equation*}
\vec x_1 = P\vec x_0 = P\left(\frac{1}{2}\vec v_1 + \frac{1}{2}\vec v_2\right) = \frac{1}{2}P\vec v_1 + \frac{1}{2}P\vec v_2 = \frac{1}{2}\left(1\right)\vec v_1 + \frac{1}{2}\left(\frac{1}{5}\right)\vec v_2\text{,}
\end{equation*}
where we replaced \(P\vec v_1\) with \(1\vec v_1\) and replaced \(P\vec v_2\) with \(\frac{1}{5}\vec v_2\) since \(\vec v_1\) and \(\vec v_2\) are eigenvectors of \(P\text{.}\) Similarly, we have that
\begin{equation*}
\vec x_2 = P\vec x_1 = \frac{1}{2}\left(1\right)P\vec v_1 + \frac{1}{2}\left(\frac{1}{5}\right)P\vec v_2 = \frac{1}{2}\left(1\right)^2\vec v_1 + \frac{1}{2}\left(\frac{1}{5}\right)^2\vec v_2\text{,}
\end{equation*}
and in general, \(\frac{1}{5}\vec v_2\) since \(\vec v_1\) and \(\vec v_2\) are eigenvectors of \(P\text{.}\) Similarly, we have that
\begin{equation*}
\vec x_k = P\vec x_{k-1} = \frac{1}{2}\left(1\right)^{k-1}P\vec v_1 + \frac{1}{2}\left(\frac{1}{5}\right)^{k-1}P\vec v_2 = \frac{1}{2}\left(1\right)^k\vec v_1 + \frac{1}{2}\left(\frac{1}{5}\right)^k\vec v_2\text{.}
\end{equation*}
Therefore, we have that
\begin{equation*}
\vec \ell = \lim_{k \rightarrow \infty} \vec x_k = \lim_{k \rightarrow \infty} \left(\frac{1}{2}\left(1\right)^k\vec v_1 + \frac{1}{2}\left(\frac{1}{5}\right)^k\vec v_2\right) = \frac{1}{2}\vec v_1 = \begin{bmatrix} 0.5 \\ 0.5 \end{bmatrix}\text{.}
\end{equation*}
Example 5.25.
True or False: A row replacement operation on a matrix does not change its eigenvalues.Solution.
False. Note that the matrix \(A = \begin{bmatrix}1 \amp 2 \\ 2 \amp 1 \end{bmatrix}\) has eigenvalues \(\lambda = -1, 3\) (check this), but doing the row replacement \(R_2 - 2R_1 \rightarrow R_2\) yields the matrix \(B = \begin{bmatrix} 1 \amp 2 \\ 0 \amp -3 \end{bmatrix}\text{,}\) which has eigenvalues \(\lambda = 1,-3\text{.}\)
Example 5.26.
For what values of \(k\) does the matrix have one real eigenvalue with algebraic multiplicity 2?
\begin{equation*}
A = \begin{bmatrix}
-3 \amp k\\
2 \amp -6
\end{bmatrix}
\end{equation*}
Solution.
Note that the characteristic polynomial of \(A\) is
\(\begin{eqnarray*}\det(A - \lambda I) \amp = \amp \begin{vmatrix} -3 - \lambda \amp k \\ 2 \amp -6-\lambda \end{vmatrix}\\ \amp = \amp (-3-\lambda)(-6-\lambda) - 2k \\ \amp = \amp \lambda^2 + 9\lambda + 18 - 2k.\end{eqnarray*}\)
Since we want \(A\) to have a single eigenvalue with algebraic multiplicity 2, we want the characteristic polynomial of \(A\) to be of the form
\begin{equation*}
(\lambda - \lambda_1)^2 = \lambda^2 - 2\lambda_1 + \lambda_1^2\text{.}
\end{equation*}
Therefore, setting the linear coefficients of the two polynomials equal, we find
\begin{equation*}
9 = -2\lambda_1 \quad \implies \quad \lambda_1 = -\frac{9}{2}\text{.}
\end{equation*}
Setting the constants of the two polynomials equal, we see that
Therefore, if \(k = -\frac{9}{8}\text{,}\) then \(A\) will have a single real eigenvalue with algebraic multiplicity 2.
