Let F be a subspace of a real vector space V and let(adsbygoogle = window.adsbygoogle || []).push({});

G \subset V_C

i.e. a subspace of its complexification. Define the real subspace of G by

G_R := G \cap V.

There is a symplectic form w[u,v]. The annihilator subspace F^perp of V is defined by

F^perp = {v \in V : w[u,v] = 0 for all u \in F}

The annihilator subspace G^perp of V_C is defined by

G^perp = {v \in V_C : w[u,v] = 0 for all u \in G}.

Then the following results hold:

(F^perp)_C = (F_C)^perp,

It is easy to see that (F^perp)_C subset of (F_C)^perp. Could somone prove it the other way? I'm thinking F_C is larger than F so the annihilator condition is more restrictive?

thanks

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# Homework Help: Complexification of subspace

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