hw8 - Week 8 1 6.1 1st order systems of DE(briefly 2 5.4 Eigenvalues and Eigenvectors 3 5.5 Eigenspaces and Diagonalization 2 A vector v = 0 in Rn(or in

# hw8 - Week 8 1 6.1 1st order systems of DE(briefly 2 5.4...

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Week 8:1.6.1 1st order systems of DE (briefly!)2.5.4 Eigenvalues and Eigenvectors3.5.5 Eigenspaces and Diagonalization
2A vectorv= 0 inRn(or inCn)is aneigenvector with eigenvalueλofann-by-nmatrixAifAv=λv.We re-write the vector equation as (A-λIn)v= 0,which is a homogeneous system with coefmatrix (A-λI),and we wantλso that the system has a non-trivial solution.We see that the eigenvalues are the rootsof thecharacteristic polynomialP(λ) = 0,whereP(λ) = det(A-λI).To find the eigenvectors we find the distinctrootsλ=λi,and for eachisolve (A-λiI)v= 0.
Problem 1.Find the eigenvalues and eigenvectorsofA=3-1-5-1Solution.P(λ) =det(A-λI2) =flflflfl3-λ-1-5-1-λflflflfl= (-1-λ)(3-λ)-5 =λ2-2λ-3-5=λ2-2λ-8 = (λ-4)(λ+ 2),soλ1= 4, λ2=-2.Forλ1= 4,we reduce the coef matrix of thesystem (A-4I)x= 0,A-4I=3-4-1-5-1-4=-1-1-5-51100,so forx2=awe getx1=-a,and theeigenvectors forλ1= 4 are the vectors(x1, x2) =a(-1,1) witha= 0,fromthe space with basis (-1,1).
4Forλ2=-2,we start over with coefA+ 2I=3 + 2-1-5-1 + 25-100.To find a spanning set without fractions, weanticipate that findingx1will use divisionby 5, and takex2= 5b.Then 5x1-x2= 0 gives5x1= 5b,sox1=b,and (x1, x2) =b(1,5),spanned by (1,5) [or, if you must, (15,1)].Problem 2.Find the eigenvalues and eigenvectorsofA=10-128020-812-6.Solution.SinceAis 3-by-3,the characteristicpolynomialP(λ) is cubic, and we can makethis problem difficult by neglecting to think
strategically. How shouldP(λ) be given;for example, do we need the coef ofλ2?We’re looking for the roots, so we need thefactors (λ-λi).For that objective there is only one goodmethod of computing the determinant; expansionon the 2nd row. Look why the 2nd row exp is best:we getP(λ) = (2-λ)flflflfl10-λ8-8-6-λflflflfl= (2-λ)[(10-λ)(-6-λ) + 64]= (2-λ)[λ2-4λ-60 + 64] =-(λ-2)3.
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