Week 8:1.6.1 1st order systems of DE (briefly!)2.5.4 Eigenvalues and Eigenvectors3.5.5 Eigenspaces and Diagonalization—————A 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=3-1-5-1¶Solution.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 the system (A-4I)x= 0,
2A-4I=3-4-1-5-1-4¶=-1-1-5-5¶→1100¶,so forx2=awe getx1=-a,and the eigenvectors forλ1= 4 are thevectors (x1, x2) =a(-1,1) witha= 0,from the space with basis (-1,1).—————Forλ2=-2,we start over with coefA+ 2I=3 + 2-1-5-1 + 2¶→5-100¶.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 characteristic polynomialP(λ) is cubic,and we can make this problem difficult by neglecting to thinkstrategically. How shouldP(λ) be given; for example,do we need the coef ofλ2? We’re looking for the roots, sowe need the factors (λ-λi). For that objectivethere is only one good method of computing the determinant;expansion on 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.
3Can we agree thatP(λ) =-λ3+ 12λ2-24λ+ 8,[which we would have gotten from the other methods]is not as useful as having a factor [since the other factors onlyneed a quadratic; not solving a cubic]?