3 1 4 5 9 57 5 3 10 1 A Using Wolfram Alpha Check That M 21 1 5 0 5 Is A Symplectic Matrix And That 3 1 0 2 1 (43.01 KiB) Viewed 63 times
3 1 -4.5 -9.57 5 3 -10 1. (a) using wolfram alpha, check that M = -21 1.5 0.5 is a symplectic matrix, and that -3 1 0 -2 -3 2 = No, that M has it satisfies the various theorems from this pdf (use wolfram alpha to see Mª determinant 1, and that theorem 3, corollary 2, and theorem 8 and theorem 9 apply)
A B be a 2n x 2n matrix with the four n x n blocks A, B, C, D. Then C D Theorem 3 Let M = ΙΑ Β is a symplectic matrix if and only if it satisfies the two equivalent conditions C D (1) A¹C, B¹ D symmetric and A¹D - CtB = I (2) AB, CDt symmetric and ADt - BCT = I
Theorem 8 Let M = Sp(2n) Then A is an eigenvector of M if and only if X-¹ is an eigenvector for M and the multiplicities of X and X-1 agree. If ±1 is an eigenvalue of M then it occurs with even multiplicity.
Theorem 9 If Mv = \v, Mv = X'v, with \\' ‡ 1, then wo(v, v') = 0 n
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