Question

P.9.22

YI=U. THIS IS KNOWI as we prispev P.9.22 Suppose that A,B EM, commute and A has distinct eigenvalues. Use Theorem 8.5.1 to show that B is diagonalizable. Moreover, show that there is an invertible S e M, such that s-'AS and S-BS are both diagonal matrices. - . .. anather proof of the assertion in the preceding problem

Theorem 8.5.1

Answer 1

2.92 Given that ABE Mn commente ie AB=BA and a has distinct eigenvalues We know that the eigenvector corresponding to distinct eigenvalues are lineroly independent Thees A has n linearly independent eigen vectors soy Vir Vai- Von Again A square matsin A of order n is diagonalisable if and only if it has n the arty independent eigenvectors. Hence A is diagonalisable Again AB = BA, so by the theorem 8.5.1 the eigenvectors V,V, ...,Vn are also eigenvectors of B. => B has n linearly independent eigen vectors Hence B is diagonalisable. Past to let arg, .., dn be the eigenvalues of A corresponding to eigenrectors V ,..," and &, dar..gan be the eigenvalues of B corresponding to eigenvectores vivon, Un. Let Ps= [vi, vai...n] PSEMA Then § is non singular since Vinag Vn care thearty independent colemns of s. s is invertible .

Nerea AS = Arvi, Val--., NJ = [ Avis Avg) --., Avn] = [1, V., Az Yig., A Ann] = [vi, Y. ..., Vn] diag (2, 93) 9n] . = diay [903. 10) -.., in] → s'as = diag [2,122,---, an] A is diagoncetti - s'As is a diagonal matrin Again BS = B [ Vis Vas..., mn] = [BV, , B Vg, ---, BV] = [Q, V, y V2 ) --, on Vn] = [vis Vasco., Vn] diag [ordon.on) = s diag [dosohas.., On] => s'bs = diag [dy, dy,--., an] . =) 5'BS is a diagonal matoin