Section 8.2 : Homogeneous Linear Systems Review: Eigenvalues and Eigenvectors Let a Be an N × N Matrix with Constant Real Compo

Section 8.2 : Homogeneous Linear Systems

Review: Eigenvalues and Eigenvectors

Let A be an n n matrix with constant real components aij. × An eigenvector of A is a nonzero n 1 column vector v such that Av = λv × for some scalar λ.

A scalar λ is called an eigenvalue of A if there is a nontrivial solution v of Av = λv. Any such v , 0 is called an eigenvector of A corresponding to λ.

Eigenvalue Problem: Find values of the scalar λ and nonzero n 1 column vectors v such × that the matrix equation Av = λv (1) is satisﬁed.

Equation (1) can be written equivalently as the homogeneous matrix equation: (A λI) v = 0 (2) − where I is the n n identity matrix. ×

1 (A λI) v = 0 (2) −

Fact: The homogeneous matrix system (2) has nontrivial solutions if and only if the determinant of the coeﬃcient matrix A λI vanishes, that is, if − and only if

a11 λ a12 a1n − ··· a21 a22 λ a2n det(A λI) = . .− ··· . = 0. (3) − ......

an an ann λ 1 2 ··· −

det(A λI) is a polynomial of degree n in λ, called the characteristic − polynomial of A. det(A λI) = 0 is called the characteristic equation of A. − The eigenvalues of A are the roots λ of the characteristic equation.

Given the eigenvalues of A, the eigenvectors can be determined by ﬁnding all nontrivial solutions to Eq. (2).

2 Homogeneous Linear System of ODEs with Constant Coeﬃcients

We are interested in solving:

X0 = AX (4) where A is an n n matrix with real constant entries, and X = X(t) is an × n 1 column vector. ×

For n = 1: Equation (4) becomes the scalar equation: x10 (t) = a11 x1(t), and the general solution is

a11t x1(t) = c1 e .

For n 2: Look for a solution vector X = X(t) of the form ≥ X = V eλt (5) where V , 0 is a constant n 1 column vector, and λ is a constant. × Substituting Eq. (5) into system (4) yields:

V λeλt = AV eλt

AV eλt V λeλt = 0 − (AV λV ) eλt = 0 − so that AV = λV since eλt > 0.

Therefore, in order to ﬁnd solutions of system (4), we need to ﬁnd eigenvalues and eigenvectors of matrix A.

3 8.2.1: Distinct Real Eigenvalues

Theorem: Let A be a real, constant, n n matrix. × If A has k n distinct real eigenvalues λ , λ , ..., λk with corresponding ≤ 1 2 eigenvectors V1, V2, ..., Vk, then the functions

λ1t λ2t λkt X1(t) = V1 e , X2(t) = V2 e , ..., Xk(t) = Vk e

are linearly independent on ( , ). −∞ ∞

Here we consider the case of k = n.

Theorem 8.2.1: General Solution of Homogeneous Systems Consider the homogeneous system of diﬀerential equations

X0 = AX (4) where the coeﬃcient matrix A is a real, constant, n n matrix. × Let λ1, λ2, ..., λn be n distinct real eigenvalues of A with corresponding

eigenvectors V1, V2, ..., Vn. Then

λit Xi(t) = Vi e , i = 1, ..., n,

form a fundamental set of solutions of system (4) on ( , ), and the −∞ ∞ general solution on ( , ) is: −∞ ∞

λ1t λ2t λnt X = c V e + c V e + + cn Vn e (6) 1 1 2 2 ···

where ci, i = 1, ..., n, are arbitrary constants.

4 Example: Find the general solution of dx = 9x 3y dt − dy = 16x 7y dt −

5 Example (cont):

6 8.2.2: Repeated Eigenvalues

Deﬁnition:

We say that λ1 is an eigenvalue of (algebraic) multiplicity k, where k is a positive integer, if (λ λ )k is a factor of the characteristic − 1 equation det(A λI) = 0, but (λ λ )k+1 is not. − − 1

Fact: An eigenvalue of multiplicity k can have anywhere between 1 and k (inclusive) linearly independent eigenvectors corresponding to it.

Let λ be an eigenvalue of A of multiplicity k n. Diﬀerent cases can 1 ≤ occur for the eigenvectors corresponding to λ1:

1. There are k linearly independent eigenvectors V1, V2, ..., Vk correspond-

ing to λ1.

2. There is only one linearly independent eigenvector corresponding to λ1.

3. There are m linearly independent eigenvectors corresponding to λ1, where 1 < m < k.

Note: We only consider cases 1 and 2.

7 Case 1: Eigenvalue of multiplicity k with k linearly independent eigenvector

Theorem: Consider the homogeneous system of diﬀerential equations

X0 = AX (4) where the coeﬃcient matrix A is a real, constant, n n matrix. × If V1, V2, ..., Vk are k linearly independent eigenvectors corresponding to an eigenvalue λ of multiplicity k n, then ≤ λt λt λt X1(t) = V1 e , X2(t) = V2 e , ..., Xk(t) = Vk e

are linearly independent solutions of system (4) on ( , ). −∞ ∞ In this case, the general solution to system (4) contains the linear combination

λt λt λt c V e + c V e + + ck Vk e 1 1 2 2 ···

where ci, i = 1, ..., k, are arbitrary constants.

Note: If k = n, that is, if λ is the only eigenvalue of A, then the functions

Xi(t), i = 1, ..., n, given above form a fundamental set of solutions to system (4) on ( , ), and the linear combination above represents the −∞ ∞ entire general solution on ( , ). −∞ ∞

8 Example: Find the general solution of X0 = AX, where    13 12 9     −  A =  15 14 9   −   0 0 2 

9 Example (cont):

10 Case 2: Eigenvalue of multiplicity k with 1 linearly independent eigenvector

Let λ be an eigenvalue of multiplicity k with 1 linearly independent eigenvector.

(A) Suppose k = 2: We are looking for 2 linearly independent solutions to system (4) from this eigenvalue λ.

λt One solution: X1 = V1 e , where V1 is an eigenvector corresponding to λ.

To ﬁnd a second solution: Try solution of the form

λt λt X2 = W1 t e + W2 e (7)

Substituting this X2 into system (4), X0 = AX, simplifying, and rearrang- ing terms yields:

(AW λW ) t eλt + (AW λW W ) eλt = 0 1 − 1 2 − 2 − 1 This equation holds for all t if and only if each term in parenthesis is 0. This gives: (A λI) W = 0 (8) − 1 (A λI) W = W (9) − 2 1

Therefore, solving Eq. (8) for W1 and then solving Eq. (9) for W2 yields the second solution X2 in Eq. (7).

11 Note: From Eq. (8), W1 is an eigenvector of A corresponding to eigenvalue

λ. Therefore, set W1 = V1. For consistency, also set W2 = V2.

We can now write the second solution X2 as:

λt λt X2 = V1 t e + V2 e (10) where

(A λI) V = 0 (11) − 1 (A λI) V = V (12) − 2 1

That is, V1 is an eigenvector of A corresponding to eigenvalue λ, and V2, which solves Eq. (12), is called a generalized eigenvector.

(B) Suppose k = 3: We are looking for 3 linearly independent solutions to system (4) from this eigenvalue λ.

We know two solutions:

λt X1 = V1 e

λt λt X2 = V1 t e + V2 e where V1 and V2 solve Eqs. (11) and (12).

12 To ﬁnd a third solution: Try solution of the form

t2 X = W eλt + W t eλt + W eλt (13) 3 1 2 2 3

Substituting this X3 into system (4), X0 = AX, simplifying, and rearrang- ing terms, we ﬁnd that the vectors W1, W2 and W3 must satisfy: (A λI) W = 0 (14) − 1 (A λI) W = W (15) − 2 1 (A λI) W = W (16) − 3 2 Note: Equations (14) and (15) are the same as Eqs. (8) and (9), respec- tively. Therefore, set W1 = V1 and W2 = V2. For consistency, also set

W3 = V3.

We can now write the third solution X3 as:

t2 X = V eλt + V t eλt + V eλt (17) 3 1 2 2 3 where

(A λI) V = 0 (18) − 1 (A λI) V = V (19) − 2 1 (A λI) V = V (20) − 3 2

That is, V1 is an eigenvector of A corresponding to eigenvalue λ, and V2 and V3, which solve Eqs. (19) and (20), are generalized eigenvectors.

13 (C) In general, for multiplicity k: We are looking for k linearly independent solutions to system (4) from this eigenvalue λ. They are:

λt X1 = V1 e

λt λt X2 = V1 t e + V2 e t2 X = V eλt + V t eλt + V eλt 3 1 2 2 3 . k 1 k 2 t − λt t − λt λt Xk = V e + V e + + Vk e 1 (k 1)! 2 (k 2)! ··· − − where V1, V2, ..., Vk is a chain of generalized eigenvectors satisfying:

(A λI) V = 0 − 1 (A λI) V = V − 2 1 (A λI) V = V − 3 2 .

(A λI) Vk = Vk 1 − −

14 Example: Find the general solution of X0 = AX, where " # 4 1 A = − 1 2

15 Example (cont):

16 Example: Suppose a 3 3 matrix A has eigenvalue λ = 2 of multiplicity × − 3 with only one linearly independent eigenvector V1. Let        1   2   15           −    V1 =  2  , V2 =  1  , V3 =  2         7   9   7  be a chain of generalized eigenvectors.

Find the general solution to X0 = AX.

17 8.2.3: Complex Eigenvalues

Notes:

1. Given a complex number z = α + i β: The conjugate of z is z = α i β. • − The real part of z is Re(z) = α. • The imaginary part of z is Im(z) = β. •

2. Given a vector V with complex entries vk = αk + i βk: The conjugate of V is denoted by V and has kth entry • vk = αk i βk. − The real part of V is denoted by Re(V ) and has kth entry • Re(vk) = αk.

The real part of V is denoted by Im(V ) and has kth entry • Im(vk) = βk.

3. Given a square matrix A whose characteristic equation has real coeﬃ- cients: Complex eigenvalues always appear in conjugate pairs. That is, if • λ = α + i β is an eigenvalue of A, then so is λ = α i β. − If V is an eigenvector corresponding to a complex eigenevalue λ, • then V is an eigenvector corresponding to λ.

18 Theorem 8.2.2: Solutions Corresponding to a Complex Eigenvalue

Given the homogeneous system X0 = AX (Eq. (4)), where A is an n n matrix with real entries. Suppose A has the complex eigenvalue × λ1 = α + i β (α and β are real) with corresponding eigenvector V1. Then: λ1t λ1t V1 e and V1 e (21) are solutions to system (4).

We can use Euler’s formula to derive real solutions to system (4) from the complex solutions in Eq. (21).

Theorem 8.2.3: Real Solutions Corresponding to a Complex Eigen- value

Let B1 = Re(V1) and B2 = Im(V1). Then: X = B cos(βt) B sin(βt) eαt (22) 1 1 − 2 αt X2 = B2 cos(βt) + B1 sin(βt) e (23)

are linearly independent solution to system (4) on ( , ). −∞ ∞

19 Example: Find the general solution of dx = 4x + 5y dt dy = 2x + 6y dt −

20 Example (cont):