Lecture 34: Similar Matrices Math 2270 - University of Utah

Math2270-Lecture34 AS,whereAisthematrixoftheeigenvaluesofA, , ,9,10,12,131 SimilarMatricesTwomatricesAandBaresimila rifthereexistsaninvertiblematrixMsuchtha tB=M ,obviously,Aissimilartoitself,andifB=M AMthenA=MBAi- ,soifAissimilartoB, ,notethatifAM BM1andB=ii CPi9thenA=Ai (M CM2)M1(PvI2M1) C(M2M1).So,ifAissimilartoB,andBissimilar toC, Similar isreflexive,symmetric,andtransitive, regivenasetofobjects,andanequivalencerel ationontheseobjects,andyou ,andthey ,we ,we sright,bytheendoftheday,you , ,ifB=Ai ,thenM xisaneigenvectorofB=M ) wehaveMBA1 x=B(ii x)=So,A1 xisaneigenvectorforB,witheigenvalue).Not ethatasMisinvertiblesois,ofcourse,A11,an dsoA1 , , ,bothAandBaresimilartoA, , ,thezeromatrix1 O00hastherepeatedeigenvalue0, (010alsohastherepeatedeigenvalue0, c2 cclC//2fcjc( (Od3 cCQ0 cI c+3 TheJordanFormNotonlycanthematrixA=(3)not bediagonalized,butit ,forexample,thematrix/510( ,thenumberonthediagonal, ,ifamatrixAhassindependenteigenvectors,t henitissimilartoamatrixJthathassJordanbl ocksonitsdiagonal:1 JiM AM=TheseJo))

2 Eigenvalues and Eigenvectors of Similar Ma trices Two similar matrices have the same eigenvalues, even though they will usually have different eigenvectors. Said more precisely, if B = Ai’AJ.I and x is an eigenvector of A, then M’x is an eigenvector of B = M’AM. The proof is quick. Suppose Ax )x. Then as A = A1BM’ we have MBA1’x = B ...

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