Periodicity in Sequences Defined by Linear Recurrence by Engstrom H. T.
By Engstrom H. T.
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Additional info for Periodicity in Sequences Defined by Linear Recurrence Relations
Erdos says that every noninvertible matrix is a product of idempotent matrices. , nonzero, nonidentity) idempotent matrix and N is nilpotent, does E + N have two distinct eigenvalues? ) Of course, the canonical forms describe all idempotent and nilpotent matrices to within similarity, in a simple and beautiful way. But real-life examples won’t always appear this way. ) For 2 × 2 matrices, nontrivial idempotent matrices are precisely those with rank 1 and trace 1. , nonzero) nilpotent matrices are those of rank 1 and trace 0.
Then the matrices Ai and Bi must be of the same size and similar for i = 1, 2, . . , k. Proof Inasmuch as A and B are similar, they must have the same characteristic polynomial, say p(x) = (x − λ1 )m1 (x − λ2 )m2 · · · (x − λk )mk . Moreover, from our eigenvalue hypotheses, Ai and Bi are then mi × mi matrices. Let P be an invertible matrix with P −1 AP = B. As a k × k blocked matrix (with diagonal blocks matching those of A and B), write P = (Pij ). We show that the off-diagonal blocks of P are zero.
As a k × k blocked matrix (with diagonal blocks matching those of A and B), write P = (Pij ). We show that the off-diagonal blocks of P are zero. Fix indices i, j with i = j. From P −1 AP = B we have AP = PB, whence Ai Pij = Pij Bj . 1 (taking C = 0), this implies Pij = 0 because Ai and Bj have no common eigenvalue. Thus, P = diag(P11 , P22 , . . , Pkk ) is block diagonal. From the way in which block diagonal matrices multiply, we now see that each Pii is invertible and Pii−1 Ai Pii = Bi . Thus, Ai and Bi are similar, as our proposition claims.