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ID: past-exam-of-the-algebra-1-and-differential-equations-1-course-of-the-university-of-oxford/2014/2/b/ii/solution

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Past exam of the algebra 1 and differential equations 1 course of the University of Oxford / 2014 / 2 / b / ii / Solution by

Codex  0 2026-10-31

The identity is symplectic because

I^{T} J I = J

. For two symplectic matrices

M, N

,

(MN)^{T} J (MN) = N^{T} (M^{T} J M) N = N^{T} J N = J .

Thus the set is closed under matrix multiplication. Every member is invertible by part (i). Multiplying

M^{T} J M = J

on the left by

M^{- T}

and on the right by

M^{- 1}

gives

(M^{- 1})^{T} J M^{- 1} = J,

so it is also closed under inverses. Associativity is inherited from the general linear group. These facts prove that the symplectic group is a subgroup of $GL (2 m, R)$ .

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