Systems of Differential Equations
6.3: Review: Eigenvalues and Eigenvectors
Understanding eigenvalues and eigenvectors is essential for solving systems of differential equations, particularly in finding solutions to homogeneous systems. This section aims to review these concepts and demonstrate how to find them.
A. Definition
Consider a square matrix [asciimath]A[/asciimath] of size [asciimath]nxxn[/asciimath] and a vector [asciimath]bb(v)[/asciimath] with [asciimath]n[/asciimath] elements. Multiplying matrix [asciimath]A[/asciimath] by the vector [asciimath]bb(v)[/asciimath] yields a new vector [asciimath]bb(u)[/asciimath] with [asciimath]n[/asciimath] elements. Geometrically, this operation can be viewed as transforming the vector [asciimath]bb(v)[/asciimath] by matrix [asciimath]A[/asciimath], which may involve rotation, scaling, reflection, or a combination of these, depending on the properties of [asciimath]A[/asciimath]. The resulting vector [asciimath]bb(u)[/asciimath] might differ in direction and magnitude from the original vector [asciimath]bb(v)[/asciimath].
In many applications, we seek a special scalar [asciimath]\lambda[/asciimath] and a corresponding nonzero vector [asciimath]bb(v)[/asciimath] such that when matrix [asciimath]A[/asciimath] multiplies [asciimath]bb(v)[/asciimath], the result is a scalar multiple of [asciimath]bb(v)[/asciimath], not yielding a new vector. This relationship is expressed as
In that case, scalar[asciimath]lambda[/asciimath] is called the eigenvalue, and vector [asciimath]bb{v}[/asciimath] is the eigenvector of matrix [asciimath]A[/asciimath]. An eigenvalue, thus, represents the factor by which an eigenvector is scaled when undergoing the linear transformation represented by [asciimath]A[/asciimath].
To find the eigenvalues of matrix [asciimath]A[/asciimath], we need to solve Equation 6.3.1 for a nonzero [asciimath]lambda[/asciimath]. Rewriting the equation, we obtain
[asciimath]A bb{v} - lambda bb{v}=bb(0)[/asciimath]
[asciimath]A bb{v} - lambda I_nbb{v}=bb(0)[/asciimath]
[asciimath](A- lambda I_n)bb{v} =bb(0)[/asciimath]
Here [asciimath]I_n[/asciimath] is the identity matrix of the same size as [asciimath]A[/asciimath]. The determinant of [asciimath]A- lambda I_n[/asciimath] must be zero for this system to have non-trivial solutions. We define [asciimath]c_A(lambda)=det(A-lambda I)[/asciimath] as the characteristic polynomial of matrix [asciimath]A[/asciimath]. The roots of the characteristic polynomial are the eigenvalues, which can be expressed as
[asciimath]det(A - \lambda I) = 0.[/asciimath] (6.3.2)
Once the eigenvalues are determined, the corresponding eigenvectors are obtained by solving the system [asciimath](A- lambda I_n)bb{v} =bb(0)[/asciimath] for each eigenvalue [asciimath]lambda[/asciimath]. These vectors are not unique, as any scalar multiple of an eigenvector is also a valid eigenvector.
B. Properties of Eigenvalues and Eigenvectors
- Algebraic Multiplicity: Refers to the number of times an eigenvalue appears as a root in the characteristic polynomial of a matrix. It provides a count of how many times an eigenvalue is repeated.
- Geometric Multiplicity: Indicates the number of linearly independent eigenvectors associated with an eigenvalue. It is always less than or equal to the algebraic multiplicity.
- Eigenvectors Linear Independence: Eigenvectors corresponding to different eigenvalues of a matrix are linearly independent. This is a key property that helps in forming a basis in the vector space spanned by these eigenvectors. If the algebraic and geometric multiplicities of an eigenvalue are equal, then there exists a full set of linearly independent eigenvectors for that eigenvalue.
- Complex Conjugate Eigenvalues and Eigenvectors: In systems that have complex eigenvalues, these eigenvalues and their corresponding eigenvectors occur in conjugate pairs. This means if [asciimath]lambda[/asciimath] is a complex eigenvalue with an associated eigenvector [asciimath]bb(v)[/asciimath], then [asciimath]bar{lambda}[/asciimath] (the complex conjugate of [asciimath]lambda[/asciimath]) is also an eigenvalue, with the corresponding eigenvector being [asciimath]bar{bb{v}}[/asciimath] (the complex conjugate of [asciimath]bb(v)[/asciimath]).
- Diagonalization: A matrix is diagonalizable if and only if, for each eigenvalue, the algebraic multiplicity equals the geometric multiplicity. This means there are enough linearly independent eigenvectors to form a basis for the space. If a matrix is not diagonalizable, it is called a defective matrix.
For the given matrix, a) find the characteristic polynomial of the matrix and b) all the eigenvalues and their associated eigenvectors.
[asciimath]A=[(3,5),(1,-1)][/asciimath]
Show/Hide Solution
[asciimath]A-lambda I[/asciimath] [asciimath]=[(3,5),(1,-1)] -lambda[(1,0),(0,1)][/asciimath] [asciimath]=[(3-lambda,5),(1,-1-lambda) ][/asciimath]
Thus, the characteristic polynomial of [asciimath]A[/asciimath] is
[asciimath]c_A(lambda)=det(A-lambda I)[/asciimath]
[asciimath]=|(3-lambda,5),(1,-1-lambda) |[/asciimath]
[asciimath]=(lambda-3)(lambda+1)-5[/asciimath]
[asciimath]=lambda^2-2lambda-8[/asciimath]
[asciimath]=(lambda-4)(lambda+2)[/asciimath]
b) The roots of [asciimath]c_A(lambda)[/asciimath] , which are [asciimath]lambda_1=4[/asciimath] and [asciimath]lambda_2=-2[/asciimath] , are the eigenvalues of [asciimath]A[/asciimath]. To find the corresponding eigenvectors, we need to find the solution to the system [asciimath](A-lambda I)bb(u)=bb(0)[/asciimath] for each eigenvalue.
For [asciimath]lambda_1=4[/asciimath] , we have
[asciimath](A-lambda I)bb(u)=bb(0)[/asciimath]
[asciimath][(3-lambda_1,5),(1,-1-lambda_1)][/asciimath] [asciimath][(u_1),(u_2)]=[(0),(0)][/asciimath]
[asciimath][(3-4,5),(1,-1-4) ][/asciimath] [asciimath][(u_1),(u_2)]=[(0),(0)][/asciimath]
[asciimath][(-1,5),(1,-5) ][/asciimath] [asciimath][(u_1),(u_2)]=[(0),(0)][/asciimath]
To solve the system, we form the augmented matrix and bring it to RREF using row operations.
[asciimath][(-1,5,|,0),(1,-5,|,0) ][/asciimath] [asciimath]stackrel(R2harrR1)(->)[/asciimath] [asciimath][(1,-5,|,0),(-1,5,|,0) ][/asciimath] [asciimath]stackrel(R2+R1)(->)[/asciimath] [asciimath][(1,-5,|,0),(0,0,|,0) ][/asciimath]
The second column lacks a leading 1, and therefore [asciimath]u_2[/asciimath] is a free variable. It is customary to let the free variable be represented by a parameter, say [asciimath]t[/asciimath]. We then write [asciimath]u_1[/asciimath] and [asciimath]u_2[/asciimath] in terms of the parameter [asciimath]t[/asciimath].
[asciimath]u_1-5u_2=0[/asciimath] [asciimath]-> u_1=5t[/asciimath]
[asciimath]u_2=t[/asciimath]
Thus the general solution is [asciimath]bb(u_1)=[(u_1),(u_2)]=t[(5),(1)][/asciimath] where [asciimath]t[/asciimath] is a nonzero arbitrary real number. We usually look for a basic (without a parameter) eigenvector. We can choose a value for [asciimath]t[/asciimath] to find a basic eigenvector. Using [asciimath]t=1[/asciimath], the eigenvectors corresponding to [asciimath]lambda_1=4[/asciimath] is [asciimath]bb(u_1)[/asciimath] [asciimath]=[(5),(1)][/asciimath].
For [asciimath]lambda_2=-2[/asciimath], we have
[asciimath][(3-lambda_2,5),(1,-1-lambda_2)][/asciimath] [asciimath][(u_1),(u_2)]=[(0),(0)][/asciimath]
[asciimath][(3+2,5),(1,-1+2) ][/asciimath] [asciimath][(u_1),(u_2)]=[(0),(0)][/asciimath]
[asciimath][(5,5),(1,1) ][/asciimath] [asciimath][(u_1),(u_2)]=[(0),(0)][/asciimath]
Similarly, to solve the system, we form the augmented matrix and bring it to RREF using row operations.
The general solution is
[asciimath]bb(u_2)=[(u_1),(u_2)]=t[(-1),(1)][/asciimath]
where [asciimath]t[/asciimath] is a nonzero arbitrary real number. Using [asciimath]t=1[/asciimath], the basic eigenvectors corresponding to [asciimath]lambda_2=-2[/asciimath] is [asciimath]bb(u_2)[/asciimath] [asciimath]=[(-1),(1)][/asciimath].
Both eigenvalues are simple eigenvalues with the algebraic multiplicity of one and therefore their eigenvectors are linearly independent.
Try an Example
For the given matrix, a) find the characteristic polynomial of the matrix and b) all the eigenvalues and their associated eigenvectors.
[asciimath]A=[(-7,-1),(5,-9)][/asciimath]
Show/Hide Solution
[asciimath]A-lambda I[/asciimath] [asciimath]=[(-7,-1),(5,-9)] -lambda[(1,0),(0,1)][/asciimath] [asciimath]=[(-7-lambda,-1),(5,-9-lambda)][/asciimath]
Thus the characteristic polynomial of [asciimath]A[/asciimath] is
[asciimath]c_A(lambda)=det(A-lambda I)[/asciimath]
[asciimath]=|(-7-lambda,-1),(5,-9-lambda) |[/asciimath]
[asciimath]=(7+lambda)(9+lambda)+5[/asciimath]
[asciimath]=lambda^2+16lambda+68[/asciimath]
Completing the square, we get
[asciimath]=(lambda+8)^2+4[/asciimath]
b) The roots of [asciimath]c_A(lambda)[/asciimath], are the eigenvalues of [asciimath]A[/asciimath].
[asciimath](lambda+8)^2+4=0[/asciimath]
[asciimath](lambda+8)^2=-4[/asciimath]
[asciimath]lambda+8=+-2i[/asciimath]
[asciimath]lambda=-8+-2i[/asciimath]
Next, to find the corresponding eigenvectors, we follow the same steps as we did for the previous example, solving system [asciimath](A-lambda I)bb(u)=bb(0)[/asciimath]. However, since the eigenvalues are complex conjugates, their corresponding eigenvectors will also be conjugates. Therefore, we only need to find the eigenvector associated with one of the eigenvalues.
We find the eigenvector associated with [asciimath]lambda_1=-8-2i[/asciimath].
[asciimath](A-lambda I)bb(u)=bb(0)[/asciimath]
[asciimath][(-7-(-8-2i),-1),(5,-9-(-8-2i)) ][/asciimath] [asciimath][(u_1),(u_2)]=[(0),(0)][/asciimath]
[asciimath][(1+2i,-1),(5,-1+2i) ][/asciimath] [asciimath][(u_1),(u_2)]=[(0),(0)][/asciimath]
To solve the system, we form the augmented matrix and bring it to RREF using row operations.
[asciimath][(1+2i,-1 ,|,0),(5,-1+2i ,|,0) ][/asciimath] [asciimath]stackrel((1-2i)R1)(->)[/asciimath] [asciimath][(5,-1+2i ,|,0),(5,-1+2i ,|,0) ][/asciimath]
[asciimath]stackrel(R2-R1)(->)[/asciimath] [asciimath][(5,-1+2i ,|,0) ,(0,0,|,0) ][/asciimath] [asciimath]stackrel(1/5R1)(->)[/asciimath] [asciimath][(1,-1/5+2/5i ,|,0) ,(0,0,|,0) ][/asciimath]
The second column lacks a leading 1, and therefore [asciimath]u_2[/asciimath] is a free variable. We let the free variable be represented by a parameter [asciimath]t[/asciimath]. We then write [asciimath]u_1[/asciimath] and [asciimath]u_2[/asciimath] in terms of the parameter [asciimath]t[/asciimath].
[asciimath]u_1+(-1/5+2/5i )u_2=0[/asciimath] [asciimath]stackrel(u_2=t)(->)[/asciimath] [asciimath]u_1=(1/5-2/5i ) t[/asciimath]
[asciimath]u_2=t[/asciimath]
Therefore, the eigenvectors corresponding to eigenvalue [asciimath]lambda_1=-8-2i[/asciimath] are [asciimath]bb(u_1)=t[(1/5-2/5i ),(1)][/asciimath] for [asciimath]t!=0[/asciimath]. Letting [asciimath]t=5[/asciimath], we have
[asciimath]bb(u_1)=[(1-2i),(5)][/asciimath] [asciimath]=[(1),(5)]-i[(2),(0)][/asciimath]
The eigenvector corresponding to the conjugate eigenvalue is the conjugate of eigenvector [asciimath]bb(u_1)[/asciimath]. Thus, the eigenvector associated with eigenvalue [asciimath]lambda_2=-8+2i[/asciimath] is
[asciimath]bb(u_2)[/asciimath] [asciimath]=[(1+2i),(5)] =[(1),(5)]+i[(2),(0)][/asciimath]
Both eigenvalues are simple eigenvalues with the algebraic multiplicity of one and therefore their eigenvectors are linearly independent.
Try an Example
For the given matrix, a) find the characteristic polynomial of the matrix and b) all the eigenvalues and their associated eigenvectors.
[asciimath]A=[(0,6,-2),(0,-2,0),(1,3,-3)][/asciimath]
Show/Hide Solution
[asciimath]A-lambda I[/asciimath] [asciimath]=[(0,6,-2),(0,-2,0),(1,3,-3)] -lambda[(1,0,0),(0,1,0),(0,0,1)][/asciimath] [asciimath]=[(-lambda,6,-2),(0,-2-lambda,0) ,(1,3,-3-lambda)][/asciimath]
Thus the characteristic polynomial of [asciimath]A[/asciimath] is
[asciimath]c_A(lambda)=det(A-lambda I)[/asciimath]
[asciimath]=|(-lambda,6,-2),(0,-2-lambda,0) ,(1,3,-3-lambda) |[/asciimath]
[asciimath]=-lambda|(-2-lambda,0) ,(3,-3-lambda) |+|(6,-2),(-2-lambda,0) |[/asciimath]
[asciimath]=-lambda(-2-lambda)(-3-lambda)-(-2-lambda)(-2)[/asciimath]
[asciimath]=(-2-lambda)(lambda^2+3lambda+2)[/asciimath]
[asciimath]=-(lambda+2)^2(lambda+1)[/asciimath]
The roots of [asciimath]c_A(lambda)[/asciimath], [asciimath]lambda_(1,2)=-2[/asciimath], with multiplicity 2, and [asciimath]lambda_3=-1[/asciimath], with multiplicity 1, are the eigenvalues of [asciimath]A[/asciimath].
To find the corresponding eigenvectors, we need to find the solution to the system [asciimath](A-lambda I)bb(u)=bb(0)[/asciimath] for each eigenvalue as we did in previous examples.
For [asciimath]lambda_(1,2)=-2[/asciimath] , we have
[asciimath]|(-lambda,6,-2),(0,-2-lambda,0) ,(1,3,-3-lambda) |[/asciimath] [asciimath][(u_1),(u_2),(u_3)]=[(0),(0),(0)][/asciimath]
[asciimath]|(2,6,-2),(0,-2+2,0) ,(1,3,-3+2) |[/asciimath] [asciimath][(u_1),(u_2),(u_3)]=[(0),(0),(0)][/asciimath]
[asciimath]|(2,6,-2),(0,0,0) ,(1,3,-1) |[/asciimath] [asciimath][(u_1),(u_2),(u_3)]=[(0),(0),(0)][/asciimath]
To solve the system, we form the augmented matrix and bring it to RREF using row operations.
[asciimath]|(2,6,-2),(0,0,0) ,(1,3,-1) |[/asciimath] [asciimath]stackrel(R2harrR3)(->)[/asciimath] [asciimath]|(2,6,-2),(1,3,-1),(0,0,0) |[/asciimath]
[asciimath]stackrel(R1harrR2)(->)[/asciimath] [asciimath]|(1,3,-1),(2,6,-2) ,(0,0,0) |[/asciimath] [asciimath]stackrel(R2-2R1)(->)[/asciimath] [asciimath]|(1,3,-1),(0,0,0) ,(0,0,0) |[/asciimath]
The second and third columns lack a leading 1, and therefore [asciimath]u_2[/asciimath] and [asciimath]u_3[/asciimath] are free variables. We let [asciimath]u_2[/asciimath] and [asciimath]u_3[/asciimath] be represented by parameters [asciimath]s[/asciimath] and [asciimath]t[/asciimath], respectively. We then write [asciimath]u_1[/asciimath] in terms of the parameters.
[asciimath]u_1+3s-t=0[/asciimath] [asciimath]-> u_1=-3s+t[/asciimath]
[asciimath]u_2=s[/asciimath]
[asciimath]u_3=t[/asciimath]
Then, the eigenvector [asciimath]bb(u_(1,2))[/asciimath] can be expressed as
[asciimath]bb(u_(1,2))=[(u_1),(u_2),(u_3)]=[(-3s+t),(s),(t)][/asciimath] [asciimath]=s[(-3),(1),(0)]+t[(1),(0),(1)],[/asciimath] [asciimath]s,t!=0[/asciimath] at the same time
where [asciimath]s[/asciimath] and [asciimath]t[/asciimath] can’t be equal to zero at the same time because that would result in a zero vector and eigenvectors never equal zero. The eigenspace is spanned by two vectors [asciimath]{[(-3),(1),(0)], \ [(1),(0),(1)]}[/asciimath].
Therefore, the basic eigenvectors associated with eigenvalue [asciimath]lambda_(1,2)[/asciimath] are [asciimath]bb(u_1)=[(-3),(1),(0)][/asciimath] and [asciimath]bb(u_2)= [(1),(0),(1)][/asciimath].
For [asciimath]lambda_3=-1[/asciimath], we have
[asciimath]|(1,6,-2),(0,-1,0) ,(1,3,-2) |[/asciimath] [asciimath][(u_1),(u_2),(u_3)]=[(0),(0),(0)][/asciimath]
Similarly, we form the augmented matrix and bring it to RREF using row operations.
[asciimath]|(1,6,-2),(0,-1,0) ,(1,3,-2) |[/asciimath] [asciimath]stackrel(R3-R1)(->)[/asciimath] [asciimath]|(1,6,-2),(0,-1,0) ,(0,-3,0) |[/asciimath]
[asciimath]stackrel(-R2)(->)[/asciimath] [asciimath]|(1,6,-2),(0,1,0) ,(0,-3,0) |[/asciimath] [asciimath]stackrel(R3+3R2)(->)[/asciimath] [asciimath]|(1,6,-2),(0,1,0) ,(0,0,0) |[/asciimath] [asciimath]stackrel(R1-6R2)(->)[/asciimath] [asciimath]|(1,0,-2),(0,1,0) ,(0,0,0) |[/asciimath]
The third column lacks a leading 1, and therefore [asciimath]u_3[/asciimath] is a free variable. We let the free variable be represented by a parameter [asciimath]t[/asciimath]. We then write [asciimath]u_1[/asciimath] and [asciimath]u_2[/asciimath] in terms of the parameter [asciimath]t[/asciimath].
[asciimath]u_1-2t=0[/asciimath] [asciimath]-> u_1=2t[/asciimath]
[asciimath]u_2=0[/asciimath]
[asciimath]u_3=t[/asciimath]
Then, the eigenvector [asciimath]bb(u_3)[/asciimath] can be expressed as
[asciimath]bb(u_3)=[(u_1),(u_2),(u_3)]=[(2t),(0),(t)][/asciimath] [asciimath]=t[(2),(0),(1)],[/asciimath] [asciimath]t!=0[/asciimath]
For both eigenvalues, the algebraic multiplicity equals the geometric multiplicity and thus their eigenvectors are linearly independent.
Try an Example
Section 6.3 Exercises
- Find the eigenvalues of the matrix
[asciimath][(0,-5),(4,6)][/asciimath].
Show/Hide Answer
[asciimath]lambda_(1,2)=[/asciimath] [asciimath]3+-isqrt(11)[/asciimath]
- Find the eigenvalues and eigenvectors of the matrix [asciimath][(5,2,8),(-4,-1,-16),(0,0,3)][/asciimath].
Show/Hide Answer
[asciimath]lambda_1=1, \ bb(v_1)=[(-1),(2),(0)][/asciimath] or any scalar multiple.
[asciimath]lambda_(2,3)=3, \ bb(v_2)=[(-1),(1),(0)], \ bb(v_3)=[(-4),(0),(1)][/asciimath] or any scalar multiple.
- Find the eigenvalues and eigenvectors of the matrix [asciimath][(-5,-4),(14,10)].[/asciimath]
Show/Hide Answer
[asciimath]lambda_1=3, \ bb(v_1)=[(1),(-2)][/asciimath] or any scalar multiple.
[asciimath]lambda_2=2, \ bb(v_2)=[(4),(-7)][/asciimath] or any scalar multiple.
