dimension theorem for symplectic complement (proof)

We denote by V the dual spaceMathworldPlanetmathPlanetmath of V, i.e., linear mappings from V to . Moreover, we assume known that dimV=dimV for any vector spaceMathworldPlanetmath V.

We begin by showing that the mapping S:VV*, aω(a,) is an linear isomorphism. First, linearity is clear, and since ω is non-degenerate, kerS={0}, so S is injective. To show that S is surjective, we apply the http://planetmath.org/node/2238rank-nullity theoremMathworldPlanetmath to S, which yields dimV=dimimgS. We now have imgSV* and dimimgS=dimV. (The first assertion follows directly from the definition of S.) Hence imgS=V (see this page (http://planetmath.org/VectorSubspace)), and S is a surjection. We have shown that S is a linear isomorphism.

Let us next define the mapping T:VW*, aω(a,). Applying the http://planetmath.org/node/2238rank-nullity theorem to T yields

dimV = dimkerT+dimimgT. (1)

Now kerT=Wω and imgT=W*. To see the latter assertion, first note that from the definition of T, we have imgTW*. Since S is a linear isomorphism, we also have imgTW*. Then, since dimW=dimW*, the result follows from equation 1.

Title dimensionPlanetmathPlanetmathPlanetmath theorem for symplectic complement (proof)
Canonical name DimensionTheoremForSymplecticComplementproof
Date of creation 2013-03-22 13:32:52
Last modified on 2013-03-22 13:32:52
Owner matte (1858)
Last modified by matte (1858)
Numerical id 5
Author matte (1858)
Entry type Proof
Classification msc 15A04