Eigenvalues and Eigenvectors §5.2 Diagonalization

1 PreviewDiagonalizationExamplesExplicit DiagonalizationEigenvalues and Eigenvectors DiagonalizationSatya Mandal, KUSummer 2017 Satya Mandal, KUEigenvalues and Eigenvectors DiagonalizationPreviewDiagonalizationExa mplesExplicit DiagonalizationGoalsSupposeAis square matrix of necessary and sufficient condition when there isan invertible matrixPsuch thatP 1 APis a Mandal, KUEigenvalues and Eigenvectors DiagonalizationPreviewDiagonalizationExa mplesExplicit DiagonalizationDefinitionsITwo square matricesA,Bare said to be similar, if thereis an invertible matrixP, such thatA=P square matrixAsaid to be diagonalizable, if there is aninvertible matrixP, such thatP 1 APis a diagonalmatrix. That means, ifAis similar to a diagonal matrix,we say thatAis Mandal, KUEigenvalues and Eigenvectors DiagonalizationPreviewDiagonalizationExa mplesExplicit DiagonalizationTheorem ,Bare two similar matrices.

2 Then,AandBhavesame | I A|=| I P 1BP|=| (P 1P) P 1BP|=|P 1( I B)P|=|P 1|| I B||P|=|P| 1| I B||P|=| I B|So,AandBhas same characteristic polynomials. So, theyhave same Eigenvalues . The proof is Mandal, KUEigenvalues and Eigenvectors DiagonalizationPreviewDiagonalizationExa mplesExplicit DiagonalizationTheorem : DiagonalizabilityWe ask, when a square matrix is diagonalizable?Theorem A square matrixA, of ordern, is diagonalizableif and only ifAhasnlinearly independent are two statements to prove. First, 1AP=D,and henceAP=PDwherePis an invertible matrix andDis a diagonal ,D= 10 00 2 0 00 n ,P=(p1p2 pn)Satya Mandal, KUEigenvalues and Eigenvectors DiagonalizationPreviewDiagonalizationExa mplesExplicit DiagonalizationContinuedISinceAP=PA, we haveA(p1p2 pn)=(p1p2 pn) 10 00 2 0 00 n.

3 Or(Ap1Ap2 Apn)=( 1p1 2p2 npn)Satya Mandal, KUEigenvalues and Eigenvectors DiagonalizationPreviewDiagonalizationExa mplesExplicit DiagonalizationContinuedISo,Api= ipifori= 1,2, ,nSincePis invertible,pi6=0and hencepiis aneigenvector ofA, for .IAlso,rank(P) = , its columns{p1,p2,..,pn}arelinearly , it is established that ifAis diagonalizable, thenAhasnlinearly independent Mandal, KUEigenvalues and Eigenvectors DiagonalizationPreviewDiagonalizationExa mplesExplicit DiagonalizationContinuedINow, we prove the converse. So, we assumeAbas hasnlinearly independent Eigenvectors :{p1,p2,..,pn}ISo,Ap1= 1p1,Ap2= 2p2, ,Apn= npnfor some Mandal, KUEigenvalues and Eigenvectors DiagonalizationPreviewDiagonalizationExa mplesExplicit DiagonalizationContinuedIWrite,P=(p1p2 pn)andD= 10 00 2 0 00 n.

4 IIt follows from the equationsApi= ipithatAP= ,P 1AP=Dis proof is Mandal, KUEigenvalues and Eigenvectors DiagonalizationPreviewDiagonalizationExa mplesExplicit DiagonalizationSteps for DiagonalizingSupposeAis a square matrix of not havenlinearly independent Eigenvectors ,thenAis not possible, findnlinearly independent eigenvectorsp1,p2, ,pnforAwith corresponding Eigenvalues 1, 2,.., , writeP=(p1p2 pn)andD= 10 00 2 0 00 n IWe haveD=P 1 APis a diagonal Mandal, KUEigenvalues and Eigenvectors DiagonalizationPreviewDiagonalizationExa mplesExplicit DiagonalizationCorollary a vectors space andx1,x2,..,xnbe vectors inV. Then,x1,x2,..,xnare linearly dependent if and only ifthere is an integerm nsuch that (1)x1,x2.

5 ,xmarelinearly dependent and (2)xm span(x1,x2,..,xm 1). ,x2,..,xnare linearly dependent. ByTheorem , one of these vectors is a linear combination ofthe rest. By relabeling, we can assumexnis a linearcombination ofx1,x2,..,xn 1. Letm= min{k:xk span(x1,x2,..,xk 1)}Satya Mandal, KUEigenvalues and Eigenvectors DiagonalizationPreviewDiagonalizationExa mplesExplicit DiagonalizationContinuedIfx1,x2,..,xm 1are linearly dependent, then we could applyTheorem again, which would lead to a contradiction,thatmis minimum. So,x1,x2,..,xm 1are linearlyindependent. This establishes one way , suppose there is anm nsuch that (1) and (2)holds. Then,xm=c1x1+ +cm 1xm 1for somec1,..,cm 1 RSo,c1x1+ +cm 1xm 1+ ( 1)xm=0which is a nontrivial linear combination.

6 So,x1,x2,..,xm,..,xnare linearly Mandal, KUEigenvalues and Eigenvectors DiagonalizationPreviewDiagonalizationExa mplesExplicit DiagonalizationTheorem : With Distinct EigenvaluesLetAbe a square matrixA, of ordern. SupposeAhasndistinct Eigenvalues . ThenIthe corresponding Eigenvectors are linearly independentIandAis second statement follows from the first, by So, we prove the first statement 1, 2,.., nbe distinct Eigenvalues , fori= 1,2,..,nwe haveAxi= ixiwherexi6= 0are Mandal, KUEigenvalues and Eigenvectors DiagonalizationPreviewDiagonalizationExa mplesExplicit DiagonalizationContinuedIWe need to prove thatx1,x2,..,xnare linearlyindependent. We prove by contra-positive , assume they are linearly Corollary there is anm<nsuch thatx1,x2.

7 ,xmare mutually linearly independent andxm+1is in can be written as a linear combination of{x1,x2,..,xm}.So,xm+1=c1x1+c2x2+ +cmxm(1)Here, at least oneci6= , if needed, wecan assumec16= Mandal, KUEigenvalues and Eigenvectors DiagonalizationPreviewDiagonalizationExa mplesExplicit DiagonalizationContinuedIMultiply (1) byAon the left:Axm+1=c1Ax1+c2Ax2+ +cmAxm(2)Now, useAxi= ixi,: m+1xm+1= 1c1x1+ 2c2x2+ + mcmxm(3)IAlso, multiply (1) by m+1,we have m+1xm+1= m+1c1x1+ m+1c2x2+ + m+1cmxm(4)Satya Mandal, KUEigenvalues and Eigenvectors DiagonalizationPreviewDiagonalizationExa mplesExplicit DiagonalizationContinuedISubtract (3) from (4):( m+1 1)c1x1+( m+1 2)c2x2+ +( m+1 m)cmxm= these vectors are linearly independent, and hence( m+1 i)ci= 0fori= 1,2, , 0 we get m+1 1= 0 or m+1= that is are distinct.

8 So, we conclude thatx1,x2,..,xnare linearly independent. The proof Mandal, KUEigenvalues and Eigenvectors DiagonalizationPreviewDiagonalizationExa mplesExplicit DiagonalizationExample 23 10 1 200 3 andP= 11 50 1 100 2 .Verify thatAis diagonalizable, by computingP :We do it in a two Use TI to computeP 1= 11 30 .So,P 1AP= 20 00 1 000 3 .So, it is verified thatP 1 APis a diagonal Mandal, KUEigenvalues and Eigenvectors DiagonalizationPreviewDiagonalizationExa mplesExplicit DiagonalizationExample (31 9 3).Show thatAis not :Use Theorem and show thatAdoes not have2 linearly independent Eigenvectors . To do this, we have findand count the dimensions of all the eigenspacesE( ).We doit in a few , find all the Eigenvalues .

9 To do this, we solvedet( I A) = 3 19 + 3 = 2= , = 0 is the only eigenvalue Mandal, KUEigenvalues and Eigenvectors DiagonalizationPreviewDiagonalizationExa mplesExplicit DiagonalizationContinuedINow we compute the eigenspaceE(0) of the eigenvalue = haveE(0) is solution space of(0I A)(xy)=(00)or( 3 193)(xy)=(00)Using TI (or by hand), a parametric solution of thissystem is given byx= .5ty= (0) ={(t, 3t) :t R}=R1, 3).So, the (sum of) dimension(s) of the eigenspace(s)= dimE(0) = 1< not Mandal, KUEigenvalues and Eigenvectors DiagonalizationPreviewDiagonalizationExa mplesExplicit DiagonalizationExample 1110 3100 3 .Show thatAis not :Use Theorem and show thatAdoes not have3 linearly independent find the Eigenvalues , we solvedet( I A) = 1 1 10 + 3 100 + 3 = ( 1)( +3)2= , = 1, 3 are the only Eigenvalues Mandal, KUEigenvalues and Eigenvectors DiagonalizationPreviewDiagonalizationExa mplesExplicit DiagonalizationContinuedIWe haveE(1) is solution space of(I A) xyz = 000 Or 0 1 104 1004 xyz = 000 (As an alternative approach, avoid solving this system.

10 The (column) rank of the coefficient matrix is 2. So,dim(E(1)) =nullity= 3 rank= 3 2 = Mandal, KUEigenvalues and Eigenvectors DiagonalizationPreviewDiagonalizationExa mplesExplicit DiagonalizationContinuedINow we compute the dimension dimE( 3).E( 3) is thesolution space of( 3I A) xyz = 000 or 4 1 100 1000 xyz = 000 The rank of the coefficient matrix is 2 (use TI, if youneed). So,dim(E( 3)) =nullity= 3 rank= 3 2 = Mandal, KUEigenvalues and Eigenvectors DiagonalizationPreviewDiagonalizationExa mplesExplicit DiagonalizationContinuedISo, the sum of dimensions of the eigenspaces= dimE(1) + dimE( 3) = 2< not Mandal, KUEigenvalues and Eigenvectors DiagonalizationPreviewDiagonalizationExa mplesExplicit DiagonalizationExample A= 17 113 20 2100 Find its eigenvaluesand determine (use Theorem ), ifAis diagonalizable.