Definitions: Two linear operators T and U on a finite dimensional vector space V

Question

Definitions: Two linear operators T and U on a finite dimensional vector space V are called simultaneously diagnolizable if there exists an ordered basis for V such that both [T] and [U] are diagonol matricies. Similarly, A,B Mnxn(F) are called simultaneously diagnolizable if there exists an invertible matrix Q Mnxn(F) such that both Q-1AQ and Q-1BQ are diagonol matricies.

(a) Prove that if T and U are simultaneously diagnolizable operators, then T and U commute (i.e. TU=UT)

(b) Show that if A and B are simultaneously diagnoliazable matricies, then A and B commute.

Explanation / Answer

. Let be an eigenvalue of T and x x 2 E. Then T (Ux) = (T U)x = (UT )x = U(T x) = U(x) = Ux; so Ux 2 E, proving that E is U-invariant. Let 1; : : : ; k denote the eigenvalues of T . Since U is diagonalizable by assumption and each eigenspace of T is U-invariant, there is a basis j for every Ej , j = 1; : : : ; k, consisting of eigenvectors of U. Let =Skj=1 j . Then is linearly independent by the theorem on pages 89 90. Moreover, the number of vectors in j is the geometric multiplic- ity of j , and thus also the algebraic multiplicity of j since T is diagonalizable (see part (1) of the theorem on page 91). Thus the number of vectors in is the sum of the algebraic multiplicities of the eigenvalues of T , i.e., the dimension of V . Since is a linearly independent set of n vectors, it is a basis for V . Each vector in is an eigenvector of T since j is a basis for Ej and each vector in is an eigenvector of U by construction. Therefore [T ]and [U]are both diagonal matrices. Let = fx1; : : : ; xng and suppose [T ] = diag(1; : : : ; n) and [U] =diag(1; : : : ; n). Then [T U] = [T ][U] = diag(11; : : : ; nn) = diag(11; : : : ; nn) = [U][T ] = [UT ]: Since T U and UT have the same matrix representations with respect to , it fol- lows that T Uxj = UT xj for every j. And since linear operators are completely determined by the images of basis vectors, it follows that T U = UT .

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