Application of Eigenvectors and Eigenvalues

8/11/2019 Application of Eigenvectors and Eigenvalues

http://slidepdf.com/reader/full/application-of-eigenvectors-and-eigenvalues 1/13

Applications of

Eigenvalues andEigenvectors

22.2

IntroductionMany applications of matrices in both engineering and science utilize eigenvalues and, some-times, eigenvectors. Control theory, vibration analysis, electric circuits, advanced dynamics andquantum mechanics are just a few of the application areas.

Many of the applications involve the use of eigenvalues and eigenvectors in the process of trans-forming a given matrix into a diagonal matrix and we discuss this process in this Section. Wethen go on to show how this process is invaluable in solving coupled differential equations of both rst order and second order.

Prerequisites

Before starting this Section you should . . .

① have a knowledge of determinants andmatrices

② have a knowledge of linear rst orderdifferential equations

Learning OutcomesAfter completing this Section you should beable to . . .

✓ diagonalize a matrix with distinct eigen-values using the modal matrix

✓ solve systems of linear differential equa-tions by the ‘decoupling’ method



1. Applications of Eigenvalues and Eigenvectors

Diagonalization of a Matrix with distinct eigenvaluesDiagonalization means transforming a non-diagonal matrix into an equivalent matrix which isdiagonal and hence is simpler to deal with.

A matrix A with distinct eigenvalues has, as we mentioned in property 3 in Section 22.1, eigenvec-tors 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 − 1AP is then a diagonal matrix D whose diagonal elements are the eigenval-ues of A. Thus if λ1, λ 2, . . . λ n are the distinct eigenvalues of A with associated eigenvectorsX (1) , X (2) , . . . , X (n ) respectively:

P = X (1) ... X (2) ... · · · ... X (n )

will produce a product

P − 1AP = D =

λ 1 0 . . . 00 λ2 . . . 0

...0 . . . . . . λ n

We see that the order of the eigenvalues in D matches the order in which P is formed from theeigenvectors.N.B.

(a) The matrix P is called the modal matrix of A

(b) Since D , as a diagonal matrix, has eigenvalues λ1, λ 2, . . . , λ n which are the same asthose of A then the matrices D and A are said to be similar . The transformation of A into D using

P − 1AP = D

is said to be a similarity transformation

Example Let A = 2 33 2 . Obtain the modal matrix P and calculate the product

P − 1AP . (The eigenvalues and eigenvectors of this particular matrix A wereobtained earlier in this workbook).

HELM (VERSION 1: March 18, 2004): Workbook Level 222.2: Eigenvalues and Eigenvectors

2



Solution

The matrix A has two distinct eigenvalues λ1 = −1, λ2 = 5 with corresponding eigenvectors

X 1 = x

−x and X 2 =

xx . We can therefore form the modal matrix from the simplest

eigenvectors of these forms:

P = 1 1

−1 1

(Other eigenvectors would be acceptable e.g. we could use P = 2 3

−2 3 but there is no

reason to over complicate the calculation).

It is easy to obtain the inverse of this 2 ×2 matrix P and the reader should conrm that:

P − 1

= 1

det( P ) adj(P ) = 12

1 1

−1 1

T

= 1

2 1

−1

1 1

We can now construct the product P − 1AP :

∴ P − 1AP = 1

2 1 −11 1

2 33 2

1 1

−1 1

= 1

2 1 −11 1 −1 5

1 5

= 1

2 −2 0

0 10= −1 0

0 5

as expected. Show(by repeating the method outlined above) that had we dened P = 1 11 −1 (i.e. interchanged the order in which the eigenvectors were taken) we would nd

P − 1AP = 5 00 −1 (i.e. the resulting diagonal elements would also be interchanged).

The matrix A = −1 40 3 has eigenvalues −1 and 3 and associated eigen-

vectors 10 and 1

1 respectively.

If P 1 = 1 10 1 , P 2 = 2 2

0 2 , P 3 = 1 11 0

write down the products P − 11 AP 1, P − 1

2 AP 2, P − 13 AP 3

(You may not need to do detailed calculations).

3 HELM (VERSION 1: March 18, 2004): Workbook Level 222.2: Eigenvalues and Eigenvectors



Your solution

P − 1

1 A P 1 =

−

1 0

0 3

= D 1

P − 1

2 A P 2 = − 1 0 0 3 = D 2

P − 1

3 A P 3 = 3 0 0 − 1 = D 3

N o t e t h a t D 1 = D 2 , d e m o n s t r a t i n g t h a t a n y e i g e n v e c t o r s o f A c a n b e u s e d t o f o r m P . N o t e a l s o t h a t s i n c e t h e c o l u m n s o f P 1 h a v e b e e n i n t e r c h a n g e d i n f o r m i n g P 3 t h e n s o h a v e t h e e i g e n v a l u e s i n D

3

a s c o m p a r e d w i t h D 1

.

Matrix Powers

If P − 1AP = 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 − 1AP P − 1 = P DP − 1

But P P − 1 = P − 1P = 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 sometimesuseful in control theory. Note that

A2 = A.A A 3 = A.A.A. etc.

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

Obtain D 2 and D 3 if D = 3 00 −2 . Write down D 10 .

Your solution


4



D 2 = 3 0 0 − 2 3 0

0 − 2 = 3 2 0 0 ( − 2 ) 2

D 3 =

3 2 0

0 ( − 2 ) 2

3 0

0 ( − 2 )

=

3 3 0

0 ( − 2 ) 3

C o n t i n u i n g i n t h i s w a y : D 1 0 = 3 1 0 0 0 ( − 2 ) 1 0

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 diagonalmatrix D

A2 = A.A = ( P DP − 1)(P DP − 1) = P D (P − 1P )DP − 1 = P DIDP − 1 = P D 2P − 1

Similarly:

A3 = A 2.A = ( P D 2P − 1)(P DP − 1) = P D 2(P − 1P )DP − 1 = P D 3P − 1

or, in general,

Ak = P D k P − 1

Example If A = 2 33 2 nd A23 .

(Use the results of the worked example).

Solution

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

−1 1

P − 1AP = −1 00 5 = D

where P − 1 = 12

1 −11 1

∴ A = P DP − 1

∴ A23 = P D 23P − 1 using the general result shown above

i.e. A = 1 1

−1 1 −1 00 523

1 −11 1

which is easily determined.




Exercises

1. Find a diagonalising matrix P if

(a) A = 4 2

−1 1 (b) A =

1 0 0

1 2 02 −2 3

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

A n s w e r s

1 . ( a ) P = − 1 − 2 1 1 ( b ) P =

1 0 0

− 1 1 0

− 2 2 1

Systems of First Order Differential EquationsSystems of rst order ordinary differential equations arise in many areas of mathematics andengineering, for example in control theory and in the analysis of electrical circuits. In each casethe basic unknowns are each a function of the time variable t. A number of techniques havebeen developed to solve such systems of equations; for example the Laplace transform or the useof the exponential matrix (outside the scope of this discussion). Here we shall use eigenvaluesand eigenvectors to obtain the solution. Our rst 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 vectorcontaining 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 intothe form Y = DY where D is a diagonal matrix. We shall nd that this new diagonal systemof differential equations can be easily solved. This special solution will allow us to obtain thesolution of the original system.

Obtain the solutions of the pair of rst order differential equationsx = −2xy = −5y (

∗)

given the initial conditions

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

(The notation is that ˙x = dxdt

, y = dydt

)

Recall, from your course on differential equations, that the general solution of the differential equation

dydt

= K y is y = y0eKt .


6



Your solution

U s i n g t h e h i n t : x = x 0 e − 2 t y = y 0 e −

5 t w h e r e x 0 = x ( 0 ) a n d y 0 = y ( 0 ) .

F r o m t h e g i v e n i n i t i a l c o n d i t i o n x 0 = 3 y 0 = 2 s o n a l l y x = 3 e − 2 t y = 2 e −

5 t .

In the above example although we had two differential equations to solve they were really quiteseparate. 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.

Thus if X = xy X = xy A = −2 00 −5

the 2 equations (*) can be written as

xy = −2 0

0 −5 xy

i.e. X = AX .

Write the pair of coupled differential equations

x = 4x + 2yy = −x + y (

∗∗)

in matrix form.

Your solution

˙ x ˙ y = 4 2

− 1 1 x y

˙ X = A X

The essential difference between the two pairs of differential equations just considered is thatthe rst pair (

∗

) were really separate equations, the rst equation of (

∗

) involving only theunknown x, the second involving only y. In matrix terms this corresponded to a diagonalmatrix A in the system X = AX . The second system (

∗∗) of equations were coupled in that




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

Example Find the solution of the coupled differential equations

x = 4x + 2yy = −x + y

with initial conditions x(0) = 1 y(0) = 0

Here x ≡ dxdt

and y ≡ dydt

.

Solution

Dening as above

X = x(t)y(t) and X = x

y .

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

X = AX where A = 4 2

−1 1 in the present example.

We now introduce a new column vector of unknowns Y = r (t)s(t) through the relation

X = P Y

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 − 1AP )Y

But, because of the properties of the modal matrix, we know that P − 1AP is a diagonalmatrix . Thus if λ1, λ 2 are distinct eigenvalues of A then:

P − 1AP = λ1 0

0 λ2

Hence Y = ( P − 1AP )Y becomes

rs =

λ1 00 λ2

r

s .


8



Solution (contd.)

That is, when written out we have

r = λ1rs = λ2s.

These equations are de-coupled . The rst equation only involves the unknown function r (t )and has solution r (t) = Ceλ 1 t . The second equation only involves the unknown function s(t)and has solution s(t) = K e λ 2 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 theform

X = AX

where A is an n ×n matrix with distinct eigenvalues λ1, λ 2, . . . , λ n .

Consider the present example in which

A = 4 2

−1 1 .

It is easily checked that A has distinct eigenvalues λ 1 = 3 λ 2 = 2 and corresponding eigenvectors

X 1 = −21 , X 2 = 1

−1 . Therefore, if

P = −2 11 −1 then P

− 1

AP = 3 00 2

and (from above),

r (t) = C e3t s (t) = K e 2t .

So

xy ≡X = P Y = −2 1

1 −1 rs

= −2 11

−1

Ce3t

Ke 2t

= −2Ce 3t + Ke 2t

Ce 3t −Ke 2t .




Solution (contd.)

Therefore

x =

−2Ce 3t + Ke 2t

y = Ce3t −Ke 2t .

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) = 2 e3t −e2t

y(t) = −e3t

+ e2t

.

The approach we have demonstrated in this example can be extended to

(a) Systems of rst 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 distincteigenvalues.

Systems of second order differential equationsThe ‘decoupling method’ discussed above can be readily extended to this situation which couldarise, 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,

X = AX

where X = xy

A = a bc d

, x = d2 xd t 2 , y = d2 y

d t 2

Make the substitution X = P Y where Y = r (t )s (t) and P is the modal matrix

of A, A being assumed here to have distinct eigenvalues λ1 and λ2. Solve theresulting pair of ‘decoupled’ equations for the case, which arises in practice,where λ1 and λ2 are both negative.


10



Your solution

E x a c t l y a s w i t h a r s t o r d e r s y s t e m p u t t i n g X = P Y i n t o t h e s e c o n d o r d e r s y s t e m ¨ X = A X g i v e s

¨ Y = P − 1 A P Y t h a t i s ¨ Y = D Y w h e r e D = λ 1 0

0 λ 2 I n f u l l ¨ r

¨ s

= λ 1 0 0 λ 2

r s

T h a t i s , ¨ r = λ 1 r = − ω 2 1 r a n d ¨ s = λ 2 s = − ω 2

2 s

( f o r t h e c a s e w h e r e λ 1 a n d λ 2 a r e b o t h n e g a t i v e . ) T h e t w o u n c o u p l e d e q u a t i o n s a r e o f t h e f o r m o f t h e d i ff e r e n t i a l e q u a t i o n g o v e r n i n g s i m p l e h a r m o n i c m o t i o n . H e n c e t h e g e n e r a l s o l u t i o n s a r e

r = A c o s ω 1 t + B s i n ω 1 t s = C c o s ω 2 t + E s i n ω 2 t

T h e s o l u t i o n s f o r x a n d y a r e t h e n o b t a i n e d b y u s e o f

X = P Y .

N o t e t h a t i n t h i s s e c o n d o r d e r c a s e f o u r i n i t i a l c o n d i t i o n s , t w o e a c h f o r b o t h x a n d y , a r e r e q u i r e d b e c a u s e f o u r c o n s t a n t s A , B , C , E a r i s e i n t h e s o l u t i o n .




Exercises

1. Solve by decoupling each of the following systems:

(a) dX

dt = AX where A = 3 4

4 −3, X (0) = 1

3(b) x1 = x2

x2 = x1 + 3 x3

x3 = x2

with x1(0) = 2 , x2(0) = 0 , x3(0) = 2

(c) dX

dt =

2 2 11 3 11 2 2

X with X (0) =100

(d) x1 = x1

x2 = −2x2 + x3

x3 = 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 -orderdifferential equations such as might arise with coupled electrical or mechanical systems.For example the motion of two masses m 1 and m 2 vibrating on coupled springs, neglectingdamping and spring masses, is governed by

m 1y1 = −k1y1 + k2(y2 −y1)m 2y2 = −k2(y2 −y1)

where dots denote derivatives with respect to time. Write this system as a matrix equationY = AY and use the decoupling method to nd Y if

(i) m1 = m 2 = 1 , k1 = 3 , k2 = 2

and the initial conditions are y1(0) = 1 , y2(0) = 2 , y(0) = −2√ 6, y2(0) = √ 6(ii) m1 = m 2 = 1 , k1 = 6 , k2 = 4

and the initial conditions are y1(0) = y2(0) = 0 , y1(0) = √ 2, y2(0) = 2√ 2Verify your solutions by substitution in each case.


12



A n s w e r s

1 . ( a ) X = 2 e 5 t − e −

5 t

e 5 t + 2 e − 5 t ( b ) X =

2 c o s h 2 t 4 s i n h 2 t 2 c o s h 2 t

( c ) X = 1 4 e 5 t + 3 e t

e 5 t − e t

e 5 t − e t

( d ) X = 1 5 5 e t

2 e 2 t + 3 e − 3 t

8 e 2 t − 3 e −

3 t

2 . ( i ) Y = c o s t − 2 s i n √ 6 t 2 c o s t + s i n √ 6 t ( i i ) Y = s i n √ 2 t

2 s i n √ 2 t


Date post:	03-Jun-2018
Category:	Documents
Upload:	anonymous-zfmlsb2gja
View:	230 times
Download:	0 times

Application of Eigenvectors and Eigenvalues

Documents