Eigenvalues, Eigenvectors, and Diagonalization

1 Eigenvalues, Eigenvectors, and Diagonal-izationMath 240 EigenvaluesandEigenvectorsDiagonalizatio nEigenvalues, eigenvectors , and DiagonalizationMath 240 Calculus IIIS ummer 2013, Session IIWednesday, July 24, 2013 Eigenvalues, Eigenvectors, and Diagonal-izationMath 240 EigenvaluesandEigenvectorsDiagonalizatio nAgenda1. Eigenvalues and Eigenvectors2. DiagonalizationEigenvalues, eigenvectors , and Diagonal-izationMath 240 EigenvaluesandEigenvectorsDiagonalizatio nIntroductionNext week, we will apply linear algebra to solving that is particularly easy to solve isy = has the solutiony=ceat, wherecis any real (or complex) in terms of linear transformations,y=ceatisthe solution to the vector equationT(y) =ay,(1)whereT:Ck(I) Ck 1(I)isT(y) =y.

2 We are going tostudy equation (1) in a more general , Eigenvectors, and Diagonal-izationMath 240 EigenvaluesandEigenvectorsDiagonalizatio nDefinitionDefinitionLetAbe ann value of for whichAv= vhasnontrivialsolutionsvare an eigenvector ofis not an eigenvector ofxyxyAAAA vvvvvvv v Figure: A geometrical description of eigenvectors , Eigenvectors, and Diagonal-izationMath 240 EigenvaluesandEigenvectorsDiagonalizatio nExampleExampleIfAis the matrixA=[1 1 3 5],then the vectorv= (1,3)is an eigenvector forAbecauseAv=[1 1 3 5][13]=[412]= corresponding eigenvalue is = that ifAv= vandcis any scalar, thenA(cv) =cAv=c( v) = (cv).

3 Consequently, ifvis an eigenvector ofA, then so iscvfor anynonzero , Eigenvectors, and Diagonal-izationMath 240 EigenvaluesandEigenvectorsDiagonalizatio nFinding eigenvaluesThe eigenvector/ eigenvalue equation can be rewritten as(A I)v= eigenvalues ofAare the values of for which the aboveequation has nontrivial are nontrivialsolutions if and only ifdet (A I) = a givenn nmatrixA, the polynomialp( ) = det(A I)is called thecharacteristic polynomialofA, and the equationp( ) = 0is called thecharacteristic eigenvalues ofAare the roots of its , Eigenvectors, and Diagonal-izationMath 240 EigenvaluesandEigenvectorsDiagonalizatio nFinding eigenvectorsIf is a root of the characteristic polynomial, then the nonzeroelements ofnullspace (A I)will be eigenvectors nonzero linear combinations of eigenvectors for a singleeigenvalue are still eigenvectors , we ll find a set of linearlyindependent eigenvectors for each , Eigenvectors, and Diagonal-izationMath 240 EigenvaluesandEigenvectorsDiagonalizatio nExampleFind all of the eigenvalues and eigenvectors ofA=[5 48 7].

4 Compute the characteristic polynomialdet(A I) = 5 48 7 = 2+ 2 roots are = 3and = 1. These are the = 3, we have the eigenvector(1,2).If = 1, thenA I=[4 48 8],which gives us the eigenvector(1,1). Eigenvalues, Eigenvector s,and Diagonal-izationMath 240 EigenvaluesandEigenvectorsDiagonalizatio nRepeated eigenvaluesFind all of the eigenvalues and eigenvectors ofA= 5 12 6 3 10 6 3 12 8 .Compute the characteristic polynomial ( 2)2( + 1).DefinitionIfAis a matrix with characteristic polynomialp( ), themultiplicity of a root ofpis called thealgebraic multiplicityof the eigenvalue .ExampleIn the example above, the eigenvalue = 2has algebraicmultiplicity 2, while = 1has algebraic multiplicity , Eigenvectors, and Diagonal-izationMath 240 EigenvaluesandEigenvectorsDiagonalizatio nRepeated eigenvaluesThe eigenvalue = 2gives us two linearly independenteigenvectors( 4,1,0)and(2,0,1).

5 When = 1, we obtain the single eigenvector( 1,1,1).DefinitionThe number of linearly independent eigenvectors correspondingto a single eigenvalue is itsgeometric , the eigenvalue = 2has geometric multiplicity 2, while = 1has geometric multiplicity geometric multiplicity of an eigenvalue is less than or equalto its algebraic matrix that has an eigenvalue whose geometric multiplicity isless than its algebraic multiplicity is , Eigenvectors, and Diagonal-izationMath 240 EigenvaluesandEigenvectorsDiagonalizatio nA defective matrixFind all of the eigenvalues and eigenvectors ofA=[1 10 1].The characteristic polynomial is( 1)2, so we have a singleeigenvalue = 1with algebraic multiplicity matrixA I=[0 10 0]has a one-dimensional null space spanned by the vector(1,0).

6 Thus, the geometric multiplicity of this eigenvalue is , Eigenvectors, and Diagonal-izationMath 240 EigenvaluesandEigenvectorsDiagonalizatio nComplex eigenvaluesFind all of the eigenvalues and eigenvectors ofA=[ 2 63 4].The characteristic polynomial is 2 2 + 10. Its roots are 1= 1 + 3iand 2= 1= 1 eigenvector corresponding to 1is( 1 +i,1).TheoremLetAbe a square matrix with real elements. If is a complexeigenvalue ofAwith eigenvectorv, then is an eigenvalue ofAwith eigenvector corresponding to 2= 1is( 1 i,1). Eigenvalues, Eigenvectors, and Diagonal-izationMath 240 EigenvaluesandEigenvectorsDiagonalizatio nSegueIf ann nmatrixAis nondefective, then a set of linearlyindependent eigenvectors forAwill form a basis weexpress the linear transformationT(x) =Axas a matrixtransformation relative to this basis, it will look like 10 n.

7 The following example will demonstrate the utility of such , Eigenvectors, and Diagonal-izationMath 240 EigenvaluesandEigenvectorsDiagonalizatio nDifferential equation exampleDetermine all solutions to the linear system of differentialequationsx =[x 1x 2]=[5x1 4x28x1 7x2]=[5 48 7][x1x2]= know that the coefficient matrix has eigenvalues 1= 1and 2= 3with corresponding eigenvectorsv1= (1,1)andv2= (1,2), the basis{v1,v2}, we write thelinear transformationT(x) =Axin the matrix representation[1 00 3]. Eigenvalues, Eigenvectors, and Diagonal-izationMath 240 EigenvaluesandEigenvectorsDiagonalizatio nDifferential equation exampleNow consider the new linear systemy =[y 1y 2]=[1 00 3][y1y2]= has the obvious solutiony1=c1etandy2=c2e 3t,for any is this relevant tox =Ax?

8 A[v1v2]=[Av1Av2]=[v1 3v2]=[v1v2] [v1v2]. Sincey =ByandAS=SB, we have(Sy) =Sy =SBy=ASy=A(Sy).Thus, a solution tox =Axis given byx=Sy=[1 11 2][c1etc2e 3t]=[c1et+c2e 3tc1et+ 2c2e 3t].