Applications of Eigenvalues and Eigenvectors









22.2

Introduction Many applications of matrices in both engineering and science utilize eigenvalues and, sometimes, eigenvectors. Control theory, vibration analysis, electric circuits, advanced dynamics and quantum mechanics are just a few of the application areas. Many of the applications involve the use of eigenvalues and eigenvectors in the process of transforming a given matrix into a diagonal matrix and we discuss this process in this Section. We then go on to show how this process is invaluable in solving coupled differential equations of both first order and second order.

#

Prerequisites Before starting this Section you should . . . "

① have a knowledge of determinants and matrices ② have a knowledge of linear first order differential equations

Learning Outcomes

✓ diagonalize a matrix with distinct eigenvalues using the modal matrix

After completing this Section you should be able to . . .

✓ solve systems of linear differential equations by the ‘decoupling’ method

!

1. Applications of Eigenvalues and Eigenvectors

Diagonalization of a Matrix with distinct eigenvalues Diagonalization means transforming a non-diagonal matrix into an equivalent matrix which is diagonal and hence is simpler to deal with. A matrix A with distinct eigenvalues has, as we mentioned in property 3 in Section 22.1, eigenvectors which are linearly independent. If we form a matrix P whose columns are these eigenvectors, it can then be shown that det(P )
= 0 so that P −1 exists. The product P −1 AP is then a diagonal matrix D whose diagonal elements are the eigenvalues of A. Thus if λ1 , λ2 , . . . λn are the distinct eigenvalues of A with associated eigenvectors X (1) , X (2) , . . . , X (n) respectively: 
.. (1) .. (2) .. (n) . X . ··· . X P = X will produce a product  λ1 0 . . . 0  0 λ2 . . . 0  P −1 AP = D =  ..  . 0 . . . . . . λn

    

We see that the order of the eigenvalues in D matches the order in which P is formed from the eigenvectors. N.B. (a) The matrix P is called the modal matrix of A (b) Since D, as a diagonal matrix, has eigenvalues λ1 , λ2 , . . . , λn which are the same as those of A then the matrices D and A are said to be similar. The transformation of A into D using P −1 AP = D is said to be a similarity transformation



2 3 Example Let A = . Obtain the modal matrix P and calculate the product 3 2 P −1 AP . (The eigenvalues and eigenvectors of this particular matrix A were obtained earlier in this workbook).

HELM (VERSION 1: March 18, 2004): Workbook Level 2 22.2: Eigenvalues and Eigenvectors

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Solution The matrix A

has two distinct eigenvalues λ1 = −1, λ2 = 5 with corresponding eigenvectors x x and X2 = . We can therefore form the modal matrix from the simplest X1 = −x x eigenvectors of these forms:

1 1 P = −1 1

2 3 (Other eigenvectors would be acceptable e.g. we could use P = but there is no −2 3 reason to over complicate the calculation). It is easy to obtain the inverse of this 2 × 2 matrix P and the reader should confirm that: P

−1

1 1 = adj(P ) = det(P ) 2



1 1 −1 1

T

1 = 2



1 −1 1 1



We can now construct the product P −1 AP :





1 1 −1 2 3 1 1 −1 ∴ P AP = 1 3 2 −1 1 2 1



1 1 −1 −1 5 = 1 1 5 2 1

1 −2 0 = 0 10 2

−1 0 = 0 5 as expected. Show(by repeating the method outlined above) that had we defined P =

1 1 (i.e. interchanged the order in which the eigenvectors were taken) we would find 1 −1

5 0 −1 (i.e. the resulting diagonal elements would also be interchanged). P AP = 0 −1



−1 4 The matrix A = has eigenvalues −1 and 3 and associated eigen

0 3 1 1 vectors and respectively. 0 1





1 1 2 2 1 1 If P1 = , P2 = , P3 = 0 1 0 2 1 0 write down the products

P1−1 AP1 ,

P2−1 AP2 ,

P3−1 AP3

(You may not need to do detailed calculations).

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HELM (VERSION 1: March 18, 2004): Workbook Level 2 22.2: Eigenvalues and Eigenvectors

Your solution

Note that D1 = D2 , demonstrating that any eigenvectors of A can be used to form P . Note also that since the columns of P1 have been interchanged in forming P3 then so have the eigenvalues in D3 as compared with D1 . P3−1 AP3 = P2−1 AP2 = P1−1 AP1 =

3 0 0 −1 −1 0 0 3 −1 0 0 3

= D3

= D2

= D1

Matrix Powers

If P −1 AP = D then we can obtain A as the subject of this matrix equation as follows: multiply on the left by P and on the right by P −1 to obtain P P −1 AP P −1 = P DP −1 But P P −1 = P −1 P = I ∴

IAI = P DP −1

and so

A = P DP −1

We can use this result to obtain the powers of a square matrix, a process which is sometimes useful in control theory. Note that A2 = A.A

A3 = A.A.A. etc.

as we would expect: clearly obtaining high powers of A directly would involve many multiplications. The process is quite straightforward, however, for a diagonal matrix D.

2

3

Obtain D and D if D =

3 0 . Write down D10 . 0 −2

Your solution

HELM (VERSION 1: March 18, 2004): Workbook Level 2 22.2: Eigenvalues and Eigenvectors

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Continuing in this way: D10 = D3 = D2 =

0 310 0 (−2)10

32 0 0 (−2)2 3 0 0 −2



3 0 0 (−2)

3 0 0 −2



=

=



33 0 0 (−2)3

0 32 0 (−2)2



We now use the relation A = P DP −1

to obtain a formula for powers of A in terms of the easily calculated powers of the diagonal matrix D A2 = A.A = (P DP −1 )(P DP −1 ) = P D(P −1 P )DP −1 = P DIDP −1 = P D2 P −1 Similarly: A3 = A2 .A = (P D2 P −1 )(P DP −1 ) = P D2 (P −1 P )DP −1 = P D3 P −1 or, in general, Ak = P Dk P −1

Example If A =

2 3 3 2

find A23 .

(Use the results of the worked example).

Solution



We know from the previous worked example that if P = P −1 AP = where P

−1

1 = 2



−1 0 0 5

∴ ∴ i.e.

1 −1 1 1



1 1 −1 1



=D

A = P DP −1 A23 = P D23 P −1 using the general result shown above





1 −1 1 1 −1 0 A = 1 1 −1 1 0 523

which is easily determined.

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HELM (VERSION 1: March 18, 2004): Workbook Level 2 22.2: Eigenvalues and Eigenvectors

Exercises 1. Find a diagonalising matrix P if

4 2 (a) A = −1 1

  1 0 0 2 0 (b) A = 1 2 −2 3

Verify, in each case, that P −1 AP is diagonal, with the eigenvalues of A as its diagonal elements.

−1 −2 1. (a) P = 1 1 Answers

  1 0 0 (b) P = −1 1 0 −2 2 1

Systems of First Order Differential Equations Systems of first order ordinary differential equations arise in many areas of mathematics and engineering, for example in control theory and in the analysis of electrical circuits. In each case the basic unknowns are each a function of the time variable t. A number of techniques have been developed to solve such systems of equations; for example the Laplace transform or the use of the exponential matrix (outside the scope of this discussion). Here we shall use eigenvalues and eigenvectors to obtain the solution. Our first step will be to recast the system of ordinary differential equations in the matrix form X˙ = AX where A is an n × n coefficient matrix of constants, X is the n × 1 column vector of unknown functions and X˙ is the n × 1 column vector containing the derivatives of the unknowns.. The main step will be to use the modal matrix of A to diagonalise the system of differential equations. This process will transform X˙ = AX into the form Y˙ = DY where D is a diagonal matrix. We shall find that this new diagonal system of differential equations can be easily solved. This special solution will allow us to obtain the solution of the original system.

Obtain the solutions of the pair of first order differential equations  x˙ = −2x y˙ = −5y

(∗)

given the initial conditions x(0) = 3 y(0) = 2

i.e. x = 3 at t = 0 i.e. y = 2 at t = 0

dx dy , y˙ = ) dt dt Recall, from your course on differential equations, that the general solution of dy the differential equation = Ky is y = y0 eKt . dt

(The notation is that x˙ =

HELM (VERSION 1: March 18, 2004): Workbook Level 2 22.2: Eigenvalues and Eigenvectors

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Your solution

From the given initial condition Using the hint:

x = x0 e−2t

x0 = 3 y = y0 e−5t

y0 = 2

so finally x = 3e−2t

y = 2e−5t .

where x0 = x(0) and y0 = y(0).

In the above example although we had two differential equations to solve they were really quite separate. We needed no knowledge of matrix theory to solve them. However, we should note that the two differential equations here can be written in matrix form.





x x˙ −2 0 ˙ Thus if X = X= A= y y˙ 0 −5 the 2 equations (*) can be written as



x˙ −2 0 x = y˙ 0 −5 y i.e. X˙ = AX.

Write the pair of coupled differential equations  x˙ = 4x + 2y y˙ = −x + y

(∗∗)

in matrix form.

Your solution

X˙

x˙ y˙

=

AX

=



4 2 −1 1



x y



The essential difference between the two pairs of differential equations just considered is that the first pair (∗) were really separate equations, the first equation of (∗) involving only the unknown x, the second involving only y. In matrix terms this corresponded to a diagonal matrix A in the system X˙ = AX. The second system (∗∗) of equations were coupled in that 7

HELM (VERSION 1: March 18, 2004): Workbook Level 2 22.2: Eigenvalues and Eigenvectors

both equations involved both x and y. This corresponded to the non-diagonal matrix A in the system X˙ = AX. Clearly the second system here is more difficult to deal with than the first and there is where we can use our knowledge of diagonalisation.

Example

Find the solution of the coupled differential equations

with initial conditions Here x˙ ≡

x˙ = 4x + 2y y˙ = −x + y x(0) = 1 y(0) = 0

dx dy and y˙ ≡ . dt dt

Solution Defining as above

x(t) X= y(t)

and

X˙ =

x˙ y˙

.

the original system of differential equations can be written, as we have seen,

4 2 X˙ = AX where A= in the present example. −1 1

r(t) We now introduce a new column vector of unknowns Y = through the relation s(t) X = PY where P is the modal matrix of A. Then, since P is a matrix of constants: X˙ = P Y˙ so X˙ = AX becomes P Y˙ = AX = A(P Y ) Then, multiplying by P −1 on the left, Y˙ = (P −1 AP )Y But, because of the properties of the modal matrix, we know that P −1 AP is a diagonal matrix. Thus if λ1 , λ2 are distinct eigenvalues of A then:

λ1 0 −1 P AP = 0 λ2 Hence Y˙ = (P −1 AP )Y becomes



r˙ λ1 0 r = . 0 λ2 s˙ s

HELM (VERSION 1: March 18, 2004): Workbook Level 2 22.2: Eigenvalues and Eigenvectors

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Solution (contd.) That is, when written out we have r˙ = λ1 r s˙ = λ2 s. These equations are de-coupled. The first equation only involves the unknown function r(t) and has solution r(t) = Ceλ1 t . The second equation only involves the unknown function s(t) and has solution s(t) = Keλ2 t where C, K are arbitrary constants. Once r, s are known the original unknowns x, y can be found from the relation X = P Y . Note that the theory outlined above is applicable to any system of differential equations of the form X˙ = AX where A is an n × n matrix with distinct eigenvalues λ1 , λ2 , . . . , λn . Consider the present example in which

4 2 A= . −1 1 It is easily that checked

A has

distinct eigenvalues λ1 = 3 λ2 = 2 and corresponding eigenvectors −2 1 , X2 = . Therefore, if X1 = 1 −1



−2 1 3 0 −1 P = then P AP = 1 −1 0 2 and (from above), r(t) = Ce3t So

s(t) = Ke2t .

x y



≡ X = PY

=

−2 1 1 −1 −2 1 1 −1





r s



Ce3t = Ke2t

−2Ce3t + Ke2t . = Ce3t − Ke2t

9



HELM (VERSION 1: March 18, 2004): Workbook Level 2 22.2: Eigenvalues and Eigenvectors

Solution (contd.) Therefore x = −2Ce3t + Ke2t y = Ce3t − Ke2t . We can now impose the initial conditions x(0) = 1 and y(0) = 0 to give 1 = −2C + K 0 = C − K. Thus C = K = −1 and the solution to the original system of differential equations is x(t) = 2e3t − e2t y(t) = −e3t + e2t .

The approach we have demonstrated in this example can be extended to (a) Systems of first order differential equations containing more than 2 unknowns (b) systems of second order differential equations The only restriction, as we have said, is that the matrix A in the system X˙ = AX has distinct eigenvalues.

Systems of second order differential equations The ‘decoupling method’ discussed above can be readily extended to this situation which could arise, for example, in a mechanical system consisting of coupled springs. A typical example of such a system with two unknowns has the form x¨ = ax + by y¨ = cx + dy or, in matrix form, ¨ = AX X

x where X = y

A=

a b c d

,

x¨ =

d2 x , dt2

y¨ =

d2 y dt2



r(t) Make the substitution X = P Y where Y = and P is the modal matrix s(t) of A, A being assumed here to have distinct eigenvalues λ1 and λ2 . Solve the resulting pair of ‘decoupled’ equations for the case, which arises in practice, where λ1 and λ2 are both negative.

HELM (VERSION 1: March 18, 2004): Workbook Level 2 22.2: Eigenvalues and Eigenvectors

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HELM (VERSION 1: March 18, 2004): Workbook Level 2 22.2: Eigenvalues and Eigenvectors

Exactly as with a first order system putting X = P Y into the second order system gives

Y¨ = P −1 AP Y In full

r¨ s¨



=

that is Y¨ = DY

λ1 0 0 λ2



r s

That is, r¨ = λ1 r = −ω12 r

where

D=

11 ¨ = AX X

λ1 0 0 λ2



and s¨ = λ2 s = −ω22 s

(for the case where λ1 and λ2 are both negative.) The two uncoupled equations are of the form of the differential equation governing simple harmonic motion. Hence the general solutions are r = A cos ω1 t + B sin ω1 t s = C cos ω2 t + E sin ω2 t The solutions for x and y are then obtained by use of X = P Y. Note that in this second order case four initial conditions, two each for both x and y, are required because four constants A, B, C, E arise in the solution.

Your solution

Exercises 1. Solve by decoupling each of the following systems:



dX 3 4 1 = AX where A = , X(0) = (a) 4 −3 3 dt (b) x˙ 1 = x2 x˙ 2 = x1 + 3x3 x˙ 3 = x2 with x1 (0) = 2,  2 dX  (c) = 1 dt 1 (d) x˙ 1 = x1

x2 (0) = 0, x3 (0) = 2    1 2 1 3 1 X with X(0) = 0 0 2 2

x˙ 2 = −2x2 + x3 x˙ 3 = 4x2 + x3 with x1 (0) = x2 (0) = x3 (0) = 1 2. Matrix methods can also be used as we have discussed to solve systems of second-order differential equations such as might arise with coupled electrical or mechanical systems. For example the motion of two masses m1 and m2 vibrating on coupled springs, neglecting damping and spring masses, is governed by m1 y¨1 = −k1 y1 + k2 (y2 − y1 ) m2 y¨2 = −k2 (y2 − y1 ) where dots denote derivatives with respect to time. Write this system as a matrix equation Y¨ = AY and use the decoupling method to find Y if (i) m1 = m2 = 1, k1 = 3, k2 = 2 √ √ and the initial conditions are y1 (0) = 1, y2 (0) = 2, y(0) ˙ = −2 6, y˙ 2 (0) = 6 (ii) m1 = m2 = 1, k1 = 6, k2 = 4 and the initial conditions are y1 (0) = y2 (0) = 0, y˙ 1 (0) =

√

√ 2, y˙ 2 (0) = 2 2

Verify your solutions by substitution in each case.

HELM (VERSION 1: March 18, 2004): Workbook Level 2 22.2: Eigenvalues and Eigenvectors

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HELM (VERSION 1: March 18, 2004): Workbook Level 2 22.2: Eigenvalues and Eigenvectors

Answers

5t 2e − e−5t 1. (a)X = e5t + 2e−5t

13

  2 cosh 2t (b) X =  4 sinh 2t  2 cosh 2t

  t   e5t + 3et 5e 1 1 (c)X = e5t − et  (d) X = 2e2t + 3e−3t  4 5 e5t − et 8e2t − 3e−3t √

√

cos t − 2 sin 6t sin √2t √ (ii) Y = 2 cos t + sin 6t 2 sin 2t

2. (i) Y =