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Eigenvalues and Eigenvectors Motivations• The static system problem of Ax = b has now been solved, e.g., by Gauss-Jordan method or Cramer’s rule.

### Transcript of Eigenvalues and Eigenvectors

Chapter6EigenvaluesandEigenvectorsPo-NingChen,ProfessorDepartmentofElectrical EngineeringNational ChiaoTungUniversityHsinChu,Taiwan30010,R.O.C.6.1 Introduction to eigenvalues 6-1MotivationsThestaticsystemproblemofAx=bhasnowbeensolved,e.g., byGauss-Jordan method or Cramers rule.However, a dynamic system problem such asAx = xcannot be solved by the static system method.Tosolvethedynamicsystemproblem, weneedtondthestaticfeatureofAthatisunchangedwiththemappingA. Inotherwords, Axmapstoitselfwithpossiblysomestretching(>1),shrinking(0 > [k[. Foranyvectorx=a1v1 + a2v2 ++ akvkthatisthelinearcombinationofall eigenvectors, thenormalizedmappingP=11A(namely, Pv1=v1)(when being applied repeatedly) converges to the eigenvector with the largestabsolute eigenvalue.I.e.,limkPkx =limk1k1Akx= limk1k1_a1Akv1 + a2Akv2 ++ akAkvk_= limk1k1_a1k1v1. .steadystate+a2k2v2 ++ kkAkvk. .transientstates_=a1v1.6.1 How to determine the eigenvalues? 6-4We wish to nd a non-zero vector vto satisfy Av= v; then(A I)v= 0 det(A I) = 0.So by solvingdet(A I) = 0, we can obtain all the eigenvalues of A.Example. Find the eigenvalues of A =_0.5 0.50.5 0.5_.Solution.det(A I) = det__0.5 0.50.5 0.5 __= (0.5 )20.52= 2 = ( 1) = 0 = 0 or 1.6.1 How to determine the eigenvalues? 6-5Proposition: Projection matrix (dened in Section 4.2) has eigenvalues either 1or 0.Proof:A projection matrix always satises P2= P. So P2v= Pv = v.By denition of eigenvalues and eigenvectors, we have P2v= 2v.Hence,v= 2vfor non-zero vector v, which immediately implies = 2.Accordingly, is either 1 or 0. 2Proposition: Permutationmatrixhaseigenvaluessatisfyingk=1forsomeinteger k.Proof:A purmutation matrix always satises Pk+1= Pfor some integer k.Example. P=__0 0 11 0 00 1 0__. Then, Pv =__v3v1v2__ and P3v= v. Hence,k = 3.Accordingly, k+1v= v, which gives k= 1 since the eigenvalue of Pcannotbe zero. 26.1 How to determine the eigenvalues? 6-6Proposition: MatrixmAm+ m1Am1++ 1A + 0Ihas the same eigenvectors as A, but its eigenvalues becomemm+ m1m1++ 1 + 0,where is the eigenvalue of A.Proof:Let viand ibe the eigenvector and eigenvalue of A. Then,_mAm+ m1Am1++ 0I_vi=_mm+ m1m1++ 0_viHence, vi and_mm+ m1m1++ 0_ are the eigenvector and eigen-value of the polynomial matrix. 26.1 How to determine the eigenvalues? 6-7Theorem(Cayley-Hamilton): (SupposeAhaslinearlyindependenteigen-vectors.)f(A) = An+ n1An1++ 0I= all-zero matrixiff() = det(A I) = n+ n1n1++ 0.Proof: Theeigenvaluesof f(A)areall zerosandtheeigenvectorsofAremainthe same as A. By denition of eigen-system, we havef(A)_v1v2 vn =_v1v2 vn__f(1) 000 f(2)0............0 0f(n)__.2Corollary(Cayley-Hamilton): (SupposeAhaslinearlyindependenteigen-vectors.)(1I A)(2I A)(nI A) = all-zero matrixProof: f() can be re-written as f() = (1)(2)(n). 26.1 How to determine the eigenvalues? 6-8(Problem 11, Section 6.1) Here is a strange fact about 2 by 2 matrices with eigen-values 1 ,= 2: The columns of A1Iare multiples of the eigenvector x2. Anyidea why this should be?Hint: (1I A)(2I A) =_0 00 0_ implies(1I A)w1 = 0 and (1I A)w2 = 0where (2I A) =_w1w2. Hence,the columns of (2I A) give the eigenvectors of 1if they are non-zero vectorsandthe columns of (1I A) give the eigenvectors of 2if they are non-zero vectors .So, the(non-zero)columnsof A 1I are(multiplesof )theeigen-vectorx2.6.1 Why Gauss-Jordan cannot solve Ax = x? 6-9The forward elimination may change the eigenvalues and eigenvectors?Example. Check eigenvalues and eigenvectors of A =_122 5_.Solution.The eigenvalues of A satisfy det(A I) = ( 3)2= 0.A = LU=_1 02 1_ _120 9_. The eigenvalues of Uapparently satisfydet(U I) = (1 )(9 ) = 0.Supposeu1andu2aretheeigenvectorsofU, respectivelycorrespondingto eigenvalues 1 and 9. Then, they cannot be the eigenvectors of A since ifthey were,___3u1 = Au1 = LUu1 = Lu1 =_1 02 1_u1=u1 = 03u2 = Au2 = LUu2 = 9Lu2 = 9_1 02 1_u2=u2 = 0.2Hence, the eigenvalues are nothing to do with pivots (except for a triangular A).6.1 How to determine the eigenvalues?(revisited) 6-10Solvedet(A I) = 0.det(A I) is a polynomial of order n.f() = det(A I) =det_________a1,1 a1,2a1,3 a1,na2,1a2,2 a2,3 a2,na3,1a3,2a3,3 a3,n...............an,1an,2an,3 an,n_________=(a1,1)(a2,2)(an,n) +(By Leibniz formula)=(1)(2)(n)ObservationsThe coecientof nis (1)n.The coecientof n1is

ni=1i =

ni=1ai,i = trace(A).. . .The coecientof 0is

ni=1i = f(0) = det(A).6.1 How to determine the eigenvalues?(revisited) 6-11These observations make easy the nding of the eigenvalues of 2 2 matrix.Example. Find the eigenvalues of A =_1 14 1_.Solution._1 + 2 = 1 + 1 = 212 = 1 4 = 3=(12)2= (1 + 2)2412 = 16=12 = 4= = 3, 1. 2Example. Find the eigenvalues of A =_1 12 2_.Solution._1 + 2 = 312 = 0= = 3, 0. 26.1 Imaginary eigenvalues 6-12In some cases, we have to allow imaginaryeigenvalues.In order to solve polynomial equations f() = 0, the mathematician was forcedto image that there exists a numberx satisfying x2= 1 .By this technique,a polynomial equations of order n have exactly n (possiblycomplex, not real) solutions.Example. Solve 2+ 1 = 0. = = i. 2Based on this, to solve the eigenvalues, we were forced to accept imaginaryeigenvalues.Example. Find the eigenvalues of A =_0 11 0_.Solution. det(A I) = 2+ 1 = 0 = = i. 26.1 Imaginary eigenvalues 6-13Proposition: TheeigenvaluesofasymmetricmatrixA(withreal entries)arereal, andtheeigenvaluesof askew-symmetric(orantisymmetric)matrixBarepure imaginary.Proof:Suppose Av= v. Then,Av = v(A real)= (Av)Tv= (v)Tv= (v)TATv= (v)Tv= (v)TAv= (v)Tv (symmetry means AT= A)= (v)Tv= (v)Tv (Av= v)= |v|2= |v|2(eigenvector must be non-zero, i.e., |v|2,= 0)= = = real6.1 Imaginary eigenvalues 6-14Suppose Bv= v. Then,Bv = v(Breal)= (Bv)Tv= (v)Tv= (v)TBTv= (v)Tv= (v)T(B)v = (v)Tv (skew-symmetry means BT= B)= (v)T()v = (v)Tv (Bv= v)= ()|v|2= |v|2(eigenvector must be non-zero, i.e., |v|2,= 0)= = = imaginary26.1 Eigenvalues/eigenvectors of inverse 6-15For invertible A,the relation betweeneigenvalues andeigenvectors of A and A1can be well-determined.Proposition: The eigenvalues and eigenvectors of A1are_ 11, v1_,_ 12, v2_, . . . ,_1n, vn_,where (i, vi)ni=1are the eigenvalues and eigenvectors of A.Proof:The eigenvalues of invertible A must be non-zerobecausedet(A) ,= 0.SupposeAv= v, where v ,= 0 and ,= 0. (I.e., and vare eigenvalue andeigenvector of A.)So, Av= v =A1(Av) = A1(v) =v= A1v =1v= A1v.2Note: The eigenvalues of Akand A1are kand 1with the same eigenvectorsas A, where is the eigenvalue of A.6.1 Eigenvalues/eigenvectors of inverse 6-16Proposition: The eigenvalues of ATare the same as the eigenvalues of A. (Butthey may have dierenteigenvectors.)Proof: det(A I) = det ((A I)T) = det (ATIT) = det (ATI). 2Corollary: TheeigenvaluesofaninvertiblematrixAsatisfyingA1=ATareon the unit circle of the complex plain.Proof: Suppose Av= v. Then,Av = v(A real)= (Av)Tv= (v)Tv= (v)TATv= (v)Tv= (v)TA1v= (v)Tv (AT= A1)= (v)T 1v= (v)Tv (A1v=1v)=1|v|2= |v|2(eigenvector must be non-zero, i.e., |v|2,= 0)= = [[2= 12Example. Find the eigenvalues of A =_0 11 0_ that satises AT= A1.Solution. det(A I) = 2+ 1 = 0 = = i. 26.1 Determination of eigenvectors 6-17After the identication of eigenvalues via det(AI) = 0, we can determinethe respective eigenvectors using the nullspacetechnique.Recall (from slide 3-41) how we completely solveBnnvn1= (AnnInn)vn1 = 0n1.Answer:R = rref(B) =_IrrFr(nr)0(nr)r0(nr)(nr)_(with no row exchange) Nn(nr)=_Fr(nr)I(nr)(nr)_Then, every solution vfor Bv= 0 is of the formv(null)n1= Nn(nr)w(nr)1for any w 1nr.Here, weusuallyndthe(n r) orthonormal basesforthenull spaceastherepresentativeeigenvectors, whichareexactlythe(n r)columnsofNn(nr)(with proper normalization).6.1 Determination of eigenvectors 6-18(Problem12, Section6.1)FindthreeeigenvectorsforthismatrixP(projectionmatrices have = 1 and 0):Projectionmatrix P=__0.2 0.4 00.4 0.8 00 0 1__.If two eigenvectors share the same , so do all their linear combinations. Findaneigenvector of Pwith no zero components.Solution.det(P I) = 0 gives (1 2) = 0; so = 0, 1. = 1: rref(P I) =__11200 0 00 0 0__;so F12 =_120, and N32 =__1201 00 1__, which implies eigenvectors =__1210__and__001__.6.1 Determination of eigenvectors 6-19 = 0: rref(P) =__1 2 00 0 00 0 1__ (with no row exchange);so F=__20__, and N31=__210__, whichimplies eigenvector =__210__.26.1 MATLAB revisited 6-20In MATLAB:We can nd the eigenvalues and eigenvectors of a matrix A by:[V D]= eig(A); % Find the eigenvalues/vectors of AThe columns of Vare the eigenvectors.The diagonals of Dare the eigenvalues.Proposition:The eigenvectors corresponding non-zero eigenvalues are in C(A).The eigenvectors corresponding zero eigenvalues are in N(A).Proof: The rst one can be seen from Av= vand the second one can be provedby Av= 0. 26.1 Some discussions on problems 6-21(Problem 25, Section 6.1) Suppose A and Bhave the same eigenvalues 1, . . . , nwiththesameindependenteigenvectorsx1, . . . , xn. ThenA=B. Reason:Any vector x is a combintion c1x1 ++ cnxn. What is Ax?What is Bx?ThinkingoverProblem25: SupposeAandBhavethesameeigenvaluesandeigenvectors (not necessarily independent). Can we claim that A = B.Answertothethinking: Not necessarily.Asacounterexample,bothA =_1111_and B=_2222_have eigenvalues0, 0and single eigenvector_11_ but they are not equal.Ifhowevertheeigenvectorsspanthen-dimensional space(suchastherearenofthem and they are linearly independent),then A = B.Hint for Problem 25: A_x1 xn =_x1 xn__1000 2 0............0 0n__.This important fact will be re-emphasized in Section 6.2.6.1 Some discussions on problems 6-22(Problem 26, Section 6.1) The block Bhas eigenvalues 1, 2 and Chas eigenvalues3, 4 and Dhas eigenvalues 5, 7. Findthe eigenvalues of the 4 by 4 matrix A:A =_B C0 D_ =__0 1 3 02 3 0 40 0 6 10 0 1 6__.ThinkingoverProblem26: The eigenvaluesof_B C0 D_are theeigenvalues ofBand Dbecause we can showdet__B C0 D__ = det(B)det(D).(See Problems23and 25in Section 5.2.)6.1 Some discussions on problems 6-23(Problem 23, Section 5.2) With 2 by 2 blocks in 4 by 4 matrices, you cannot alwaysuse block dete