Now a can be diagonalized if we can find an

  • University of Texas
  • M 340L
  • Homework Help
  • Lieutenant62
  • 12
  • 100% (8) 8 out of 8 people found this document helpful

This preview shows page 7 - 10 out of 12 pages.

NowAcan be diagonalized if we can findan eigenbasis ofR2of eigenvectorsv1,v2ofAcorresponding to eigenvaluesλ1,λ2, for then:A=Pλ100λ2P1,P= [v1v2].Butdet[AλI] =3λ423λ= 8(3λ)(3 +λ) =λ21 = 0,i.e.,λ1= 1 andλ2=1.Correspondingeigenvectors arev1=21,v2=11,soP=2111,P1=1112.Thusf(A) =2111f(1)00f(1)1112.Nowf(1) = 4x3+x2+ 2x3x=1= 4,whilef(1) = 4x3+x2+ 2x3x=1=8.Consequently,f(A) =211140081112=16241220.01410.0 points
jones (dzj62) – HW10 – gilbert – (53525)8Using the fact thatex= 1 +x+12!x2+. . .+1n!xn+. . . ,computeetAas a matrix-valued function oftwhenA=5421.1.etA=2ete3t2(e3tet)ete3t2e3tet2.etA=2et+et2(e3tet)ete3t2e3t+et3.etA=2e3tet2(ete3t)e3tet2ete3tcorrect4.etA=2e3t+et2(ete3t)e3tet2et+e3tExplanation:IfAcan be diagonalized byA=PDP1=Pd100d2P1,thenetA=PetDP1=Petd100etd2P1.NowAcan be diagonalized if we can findan eigenbasis ofR2of eigenvectorsv1,v2ofAcorresponding to eigenvaluesλ1,λ2, for then:A=Pλ100λ2P1,P= [v1v2].Butdet[AλI] =5λ421λ=λ2λ+ 3 = (λ3)(λ1) = 0,i.e.,λ1= 3 andλ2= 1.Correspondingeigenvectors arev1=21,v2=11,soP=2111,P1=1112.Consequently,etA=2111e3t00et1112=2e3tet2(ete3t)e3tet2ete3t.01510.0 pointsFind a matrixPso thatPd100d2P1,d1d2is a diagonalization of the matrixA=75421.P=51412.P=14153.P=15144.P=14155.P=51416.P=1514correctExplanation:To begin, we must find the eigenvectorsand eigenvalues ofA. To do this, we will usethe characteristic equation, det(AλI) = 0.
jones (dzj62) – HW10 – gilbert – (53525)9That is, we will look for the zeros of thecharacteristic polynomial.det(AλI) = (7λ)(2λ) + 20=λ2+ 5λ+ 6= (λ+ 2)(λ+ 3) = 0.SoD=λ100λ2=2003.Now to find the eigenvectors ofA, we willsolve for the nontrivial solution of the charac-teristic equation by row reducing the relatedaugmented matrices:[Aλ1I0] =7 + 25042 + 20=550440110000=u1=11,while[Aλ2I0] =7 + 35042 + 30=450450450000=u2=54.So,P= [u1u2] andA=PDP1is a diagonalization ofA.Consequently,D=2003,P=1514.01610.0 pointsThe eigenvalues of ann×nmatrixAarethe entries on the main diagonal ofA.True or False?1.FALSEcorrect2.TRUEExplanation:The eigenvalues of atriangular n×nma-trixAare the entries on the main diagonal ofA.This is not necessarily true for matricesthat are not triangular.

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture