Applications of Eigenvalues and Eigenvectors

Applications ofEigenvalues andEigenvectors✓✏22.2✑✒IntroductionMany applications of matrices in both engineering and science utilize eigenvalues and, sometimes,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 transforminga 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 ofboth first order and second order.★✧PrerequisitesBefore starting this Section you should ...Learning OutcomesAfter completing this Section you should beable to ...1 have a knowledge of determinants andmatrices2 have a knowledge of linear first orderdifferential equations✓ diagonalize a matrix with distinct eigenvaluesusing the modal matrix✓ solve systems of linear differential equationsby the ‘decoupling’ method✥✦

1. Applications of Eigenvalues and EigenvectorsDiagonalization 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, eigenvectorswhich are linearly independent. If we form a matrix P whose columns are these eigenvectors,it can then be shown thatdet(P ) ≠0so that P −1 exists.The product P −1 AP is then a diagonal matrix D whose diagonal elements are the eigenvaluesof A. Thus if λ 1 ,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⎡⎤λ 1 0 ... 0P −1 0 λ 2 ... 0AP = D = ⎢⎥⎣ .⎦0 ... ... λ nWe 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, asadiagonal matrix, has eigenvalues λ 1 ,λ 2 ,...,λ n which are the same asthose of A then the matrices D and A are said to be similar. The transformation ofA into D usingP −1 AP = Dis said to be a similarity transformation[ ] 2 3Example Let A = . Obtain the modal matrix P and calculate the product3 2P −1 AP . (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 Eigenvectors2

SolutionThe matrix [ A]has two distinct [ ] eigenvalues λ 1 = −1, λ 2 =5with corresponding eigenvectorsxxX 1 = and X−x2 = . We can therefore form the modal matrix from the simplestxeigenvectors of these forms:[ ]1 1P =−1 1[ ]2 3(Other eigenvectors would be acceptable e.g. we could use P =but there is no−2 3reason 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 =[1det(P ) adj(P )=1 21 1−1 1] T= 1 2[ 1 −11 1]We can now construct the product P −1 AP :∴ P −1 AP = 1 [ 1 −12 1 1[ 1 −1][ ][ 2 33 2][ ] −1 51 1−1 1]= 1 2 1 1= 1 [ −2 02 0 10[ ] −1 0=0 5]1 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 find1 −1[ ] 5 0P −1 AP =(i.e. the resulting diagonal elements would also be interchanged).0 −1[ ] −1 4The matrix A =has eigenvalues −1 and 3 and associated eigenvectorsand respectively.[ ] [0]31 10 1[ ] [ ] [ ]1 12 21 1If P 1 = , P0 12 = , P0 23 =1 0write down the productsP −11 AP 1 , P −12 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[ −1 0]P1 −1 AP 1 =0 3= D 1[ −1 0]P2 −1 AP 2 =0 3= D 2P3 −1 AP 3 =[ 3 00 −1]= D 3Note that D 1 = D 2 , demonstrating that any eigenvectors of A can be used to form P . Note alsothat since the columns of P 1 have been interchanged in forming P 3 then so have the eigenvaluesin D 3 as compared with D 1 .Matrix PowersIf 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 obtainPP −1 AP P −1 = PDP −1ButPP −1 = P −1 P = I∴ IAI = PDP −1 and so A = PDP −1We can use this result to obtain the powers of a square matrix, a process which is sometimesuseful in control theory. Note thatA 2 = A.A A 3 = 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.Obtain D 2 and D 3 if D =[ 3 00 −2].Write down D 10 .Your solutionHELM (VERSION 1: March 18, 2004): Workbook Level 222.2: Eigenvalues and Eigenvectors4

D 2 =D 3 =Continuing in this way: D 10 =[ 3 00 −2][ 3 00 −2]=[ ][ 320 3 00 (−2) 2 0 (−2)[ ]31000 (−2) 10[ ]3200 (−2) 2] [ ]330=0 (−2) 3We now use the relationA = PDP −1to obtain a formula for powers of A in terms of the easily calculated powers of the diagonalmatrix DA 2 = A.A =(PDP −1 )(PDP −1 )=PD(P −1 P )DP −1 = PDIDP −1 = PD 2 P −1Similarly:A 3 = A 2 .A =(PD 2 P −1 )(PDP −1 )=PD 2 (P −1 P )DP −1 = PD 3 P −1or, in general,A k = PD k P −1[ ] 2 3Example If A = find A3 223 .(Use the results of the worked example).Solution[We know from the previous worked example that if P =[ ] −1 0P −1 AP == D0 5where P −1 = 1 [ ] 1 −12 1 1∴ A = PDP −11 1−1 1∴ A 23 = PD 23 P −1 using the general result shown above[ ][ ][ ]1 1 −1 0 1 −1i.e. A =−1 1 0 5 23 1 1which is easily determined.]5 HELM (VERSION 1: March 18, 2004): Workbook Level 222.2: Eigenvalues and Eigenvectors

Exercises1. Find a diagonalising matrix P if⎡ ⎤[ ]1 0 04 2(a) A =(b) A = ⎣1 2 0⎦−1 12 −2 3Verify, in each case, that P −1 AP is diagonal, with the eigenvalues of A as its diagonalelements.Answers1. (a) P =[ −1 −21 1](b) P =⎡⎣1 0 0−1 1 0−2 2 1⎤⎦Systems of First Order Differential EquationsSystems of first 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 first step will be to recast the system of ordinarydifferential equations in the matrix form Ẋ = AX where A is an n × n coefficient matrix ofconstants, X is the n × 1 column vector of unknown functions and Ẋ is the n × 1 column vectorcontaining the derivatives of the unknowns.. The main step will be to use the modal matrix ofA to diagonalise the system of differential equations. This process will transform Ẋ = AX intothe form Ẏ = DY where D is a diagonal matrix. We shall find 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 first order differential equations}ẋ = −2x(∗)ẏ = −5ygiven the initial conditionsx(0) =3 i.e. x =3 at t =0y(0) =2 i.e. y =2 at t =0(The notation is that ẋ = dxdt , ẏ = dydt )Recall, from your course on differential equations, that the general solution ofthe differential equation dydt = Ky is y = y 0e Kt .HELM (VERSION 1: March 18, 2004): Workbook Level 222.2: Eigenvalues and Eigenvectors6

Your solutionUsing the hint: x = x 0 e −2t y = y 0 e −5t where x 0 = x(0) and y 0 = y(0).From the given initial condition x 0 =3 y 0 =2 so finally x =3e −2t y =2e −5t .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.[ ] [ ] [ ]x ẋ−2 0Thus if X = Ẋ = A =yẏ0 −5the 2 equations (*) can be written as[ ] [ ][ ]ẋ −2 0 x=ẏ 0 −5 yi.e.Ẋ = AX.Write the pair of coupled differential equations}ẋ =4x +2yẏ = −x + yin matrix form.(∗∗)Your solution]][ xyẊ = AX4 2−1 1The essential difference between the two pairs of differential equations just considered is thatthe first pair (∗) were really separate equations, the first equation of (∗) involving only theunknown x, the second involving only y. In matrix terms this corresponded to a diagonalmatrix A in the system Ẋ = AX. The second system (∗∗) ofequations were coupled in that[=][ ẋẏ7 HELM (VERSION 1: March 18, 2004): Workbook Level 222.2: Eigenvalues and Eigenvectors

oth equations involved both x and y. This corresponded to the non-diagonal matrix A inthe system Ẋ = AX.Clearly the second system here is more difficult to deal with than the first and there is wherewe can use our knowledge of diagonalisation.ExampleFind the solution of the coupled differential equationsẋ = 4x +2yẏ = −x + ywith initial conditions x(0) = 1 y(0) =0Here ẋ ≡ dxdtand ẏ ≡dydt .SolutionDefining as above[ ][ ]x(t)ẋX =and Ẋ = .y(t)ẏthe original system of differential equations can be written, as we have seen,[ ]4 2Ẋ = AX where A =in the present example.−1 1[ ] r(t)We now introduce a new column vector of unknowns Y = through the relations(t)X = PYwhere P is the modal matrix of A. Then, since P is a matrix of constants:Ẋ = P Ẏso Ẋ = AX becomes P Ẏ = AX = A(PY)Then, multiplying by P −1 on the left,Ẏ = (P −1 AP )YBut, because of the properties of the modal matrix, we know that P −1 AP is a diagonalmatrix. Thusifλ 1 ,λ 2 are distinct eigenvalues of A then:[ ]P −1 λ1 0AP =0 λ 2Hence Ẏ =(P −1 AP )Y becomes[ ] [ ][ ]ṙ λ1 0 r=.ṡ 0 λ 2 sHELM (VERSION 1: March 18, 2004): Workbook Level 222.2: Eigenvalues and Eigenvectors8

Solution (contd.)That is, when written out we haveṙ = λ 1 rṡ = λ 2 s.These equations are de-coupled. The first equation only involves the unknown function r(t)and has solution r(t) =Ce λ 1t . The second equation only involves the unknown function s(t)and has solution s(t) =Ke λ 2t where C, K are arbitrary constants.Once r, s are known the original unknowns x, y can be found from the relation X = PY.Note that the theory outlined above is applicable to any system of differential equations of theformẊ = AXwhere A is an n × n matrix with distinct eigenvalues λ 1 ,λ 2 ,...,λ n .Consider the present example in which[ ]4 2A = .−1 1It is easily [ checked ] that [ A has ] distinct eigenvalues λ 1 =3λ 2 =2and corresponding eigenvectors−2 1X 1 = ,X1 2 = . Therefore, if−1[ ][ ]−2 13 0P =then P1 −1−1 AP =0 2and (from above),Sor(t) =Ce 3t s(t) =Ke 2t .[ xy]≡ X = PY ===[ ][ ]−2 1 r1 −1 s[ ][ ]−2 1 Ce3t1 −1 Ke 2t[ ]−2Ce 3t + Ke 2tCe 3t − Ke 2t .9 HELM (VERSION 1: March 18, 2004): Workbook Level 222.2: Eigenvalues and Eigenvectors

Solution (contd.)Thereforex = −2Ce 3t + Ke 2ty = Ce 3t − Ke 2t .We can now impose the initial conditions x(0) = 1 and y(0) = 0 to give1 = −2C + K0 = C − K.Thus C = K = −1 and the solution to the original system of differential equations isx(t) = 2e 3t − e 2ty(t) = −e 3t + e 2t .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 equationsThe only restriction, as we have said, is that the matrix A in the system Ẋ = 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.Atypical example of such a system with two unknowns has the formẍ = ax + byÿ = cx + dyor, in matrix form,Ẍ = AX[ ] xwhere X =y[ a bA =c d], ẍ = d2 xdt 2 ,ÿ = d2 ydt 2[ ] r(t)Make the substitution X = PY where Y = and P is the modal matrixs(t)of A, A being assumed here to have distinct eigenvalues λ 1 and λ 2 . Solve theresulting pair of ‘decoupled’ equations for the case, which arises in practice,where λ 1 and λ 2 are both negative.HELM (VERSION 1: March 18, 2004): Workbook Level 222.2: Eigenvalues and Eigenvectors10

Your solutionExactly as with a first order system putting X = PY into the second order system Ẍ = AX[ ]Ÿ = P −1 AP Y that is Ÿ = DY D =giveswhere λ1 00 λ 2In full[ ¨r¨s]=[ ][λ1 0 r0 λ 2 s]That is, ¨r = λ 1 r = −ω 2 1r and ¨s = λ 2 s = −ω 2 2s(for the case where λ 1 and λ 2 are both negative.) The two uncoupled equations are of the formof the differential equation governing simple harmonic motion. Hence the general solutions arer = A cos ω 1 t + B sin ω 1 ts = C cos ω 2 t + E sin ω 2 tThe solutions for x and y are then obtained by use ofX = PY.Note that in this second order case four initial conditions, two each for both x and y, arerequired because four constants A, B, C, E arise in the solution.11 HELM (VERSION 1: March 18, 2004): Workbook Level 222.2: Eigenvalues and Eigenvectors

Exercises1. Solve by decoupling each of the following systems:(a)[ ][dX3 41dt = AX where A = , X(0) =4 −33](b) ẋ 1 = x 2ẋ 2 = x 1 +3x 3ẋ 3 = x 2with x 1 (0) =2, x 2 (0) =0, x 3 (0) =2(c)⎡ ⎤⎡ ⎤2 2 11dXdt = ⎣1 3 1⎦ X with X(0) = ⎣0⎦1 2 20(d) ẋ 1 = x 1ẋ 2 = −2x 2 + x 3ẋ 3 =4x 2 + x 3with x 1 (0) = x 2 (0) = x 3 (0) =12. 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 bym 1 ÿ 1 = −k 1 y 1 + k 2 (y 2 − y 1 )m 2 ÿ 2 = −k 2 (y 2 − y 1 )where dots denote derivatives with respect to time. Write this system as a matrix equationŸ = AY and use the decoupling method to find Y if(i) m 1 = m 2 =1,k 1 =3,k 2 =2and the initial conditions are y 1 (0) =1,y 2 (0) = 2, ẏ(0) = −2 √ 6, ẏ 2 (0) = √ 6(ii) m 1 = m 2 =1, k 1 =6,k 2 =4and the initial conditions are y 1 (0) = y 2 (0) =0, ẏ 1 (0) = √ 2, ẏ 2 (0) =2 √ 2Verify your solutions by substitution in each case.HELM (VERSION 1: March 18, 2004): Workbook Level 222.2: Eigenvalues and Eigenvectors12

Answers1. (a)X =(c)X = 1 42. (i) Y =[ 2e5t− e −5te 5t + 2e −5t ]⎡⎣(b) X =⎤e 5t + 3e te 5t − e t ⎦ (d) X = 1e 5t − e t 5[ √cos t − 2 sin 6t2 cos t + sin √ 6t](ii) Y =⎡⎣2 cosh 2t4 sinh 2t2 cosh 2t⎡⎣[ √sin 2t2 sin √ 2t⎤⎦⎤5e t2e 2t + 3e −3t ⎦8e 2t − 3e −3t]13 HELM (VERSION 1: March 18, 2004): Workbook Level 222.2: Eigenvalues and Eigenvectors