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Chapter IVEigenvalues and Eigenvectors105
IV Eigenvalues and EigenvectorsIV.1 Eigenvalues and EigenvectorsIV.1.1 DefinitionLetAbe ann×nmatrix. A numberλand non-zero vectorvare an eigenvalue eigenvectorpair forAifAv=λvAlthoughvis required to be nonzero,λ= 0 is possible. Ifvis an eigenvector, so issvforany numbers6= 0.Rewrite the eigenvalue equation as(λI-A)v=0Then we see thatvis a non-zero vector in the nullspaceN(λI-A).Such a vector onlyexists ifλI-Ais a singular matrix, or equivalently ifdet(λI-A) = 0IV.1.2 Standard procedureThis leads to the standard textbook method of finding eigenvalues. The function ofλdefinedbyp(λ) = det(λI-A) is a polynomial of degreen, called the characteristic polynomial, whosezeros are the eigenvalues. So the standard procedure is:Compute the characteristic polynomialp(λ)Find all the zeros (roots) ofp(λ). This is equivalent to completely factoringp(λ) asp(λ) = (λ-λ1)(λ-λ2)· · ·(λ-λn)Such a factorization always exists if we allow the possibility that the zerosλ1, λ2, . . .are complex numbers.But it may be hard to find.In this factorization there maybe repetitions in theλi’s. The number of times aλiis repeated is called itsalgebraicmultiplicity.For each distinctλifindN(λI-A), that is, all the solutions to(λiI-A)v=0The non-zero solutions are the eigenvectors forλi.106
IV.1 Eigenvalues and EigenvectorsIV.1.3 Example 1This is the typical case where all the eigenvalues are distinct. LetA=3-6-7185-1-21Then, expanding the determinant, we finddet(λI-A) =λ3-12λ2+ 44λ-48This can be factored asλ3-12λ2+ 44λ-48 = (λ-2)(λ-4)(λ-6)So the eigenvalues are 2, 4 and 6.These steps can be done with MATLAB/Octave usingpolyandroot.IfAis a squarematrix, the commandpoly(A)computes the characteristic polynomial, or rather, its coeffi-cients.> A=[3 -6 -7; 1 8 5; -1 -2 1];> p=poly(A)p =1.0000-12.000044.0000-48.0000Recall that the coefficient of the highest power comes first.The functionrootstakes asinput a vector representing the coefficients of a polynomial and returns the roots.>roots(p)ans =6.00004.00002.0000To find the eigenvector(s) forλ1= 2 we must solve the homogeneous equation (2I-A)v=0.Recall thateye(n)is then×nidentity matrixI107
IV Eigenvalues and Eigenvectors>rref(2*eye(3) - A)ans =10-1011000From this we can read off the solutionv1=1-11Similarly we find forλ2= 4 andλ3= 6 that the corresponding eigenvectors arev2=-1-11v3=-210The three eigenvectorsv1,v2andv3are linearly independent and form a basis forR3.The MATLAB/Octave command for finding eigenvalues and eigenvectors iseig.Thecommandeig(A)lists the eigenvalues>eig(A)ans =4.00002.00006.0000while the variant[V,D] = eig(A)returns a matrixVwhose columns are eigenvectors and adiagonal matrixDwhose diagonal entries are the eigenvalues.>[V,D] = eig(A)V =5.7735e-015.7735e-01-8.9443e-015.7735e-01-5.7735e-014.4721e-01-5.7735e-015.7735e-012.2043e-16108
IV.1 Eigenvalues and EigenvectorsD =4.000000.000000.000000.000002.000000.000000.000000.000006.00000Notice that the eigenvalues have been normalized to have length one.Also, since they

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