Re: Duals and anihilators

given u in U, it is enough to show that φ(u) is in Ann(Ann(U)) (this proof only works in finite-dimensional vector spaces, by the way, as in general, U is not isomorphic to U**).

but if f is in Ann(U), this means f(u) = 0 for all u in U. hence for any such f, φ(u)(f) = f(u) = 0, that is φ(u) is contained in Ann(Ann(U)), which means that φ(U) = U** is contained in Ann(Ann(U)).

but clearly Ann(Ann(U)) is a subset of U**, so the two sets are equal (the finite-dimensionality comes in by way of asserting φ is ONTO U**).

*************

now suppose we take G in Ann(Ann(U)). since Ann(Ann(U)) = U**, and φ is an isomorphism, φ^{-1} is well-defined. letting u = φ^{-1}(G), we have:

φ(u) = φ(φ^{-1}(G)) = id_{U**}(G) = G.

Re: Duals and anihilators

(Yes, I had forgotten to say that V is finite dimensionnal. We haven't gotten to infinite-dimensionnal vector spaces yet.)

Quote:

Originally Posted by

**Deveno** but clearly Ann(Ann(U)) is a subset of U**

I'm really confused about this... Ann(U) is the set of all f in V* such that f(u) = 0 for all u in U. So Ann(U) is a subset of V*. The same way, Ann(Ann(U)) would be a subset of V**. More precisely, I can't see how a function from U* to the field (an element of U**) could be equal to a function from V* to the field (an element of Ann(Ann(U))). Maybe I got my definitions wrong?

Re: Duals and anihilators

the isomorphism φ is actually induced from a similar isomorphism (defined the same way) from V to V**.

i understand your confusion. let's look at a concrete example. suppose V = R^{3}. suppose U = span(S), where S = {(1,0,1),(1,1,0)}.

now Ann(U) = {f in V*: f(u) = 0, for all u in U}. let's find a basis for Ann(U). let {p_{j} in V*: p_{j}(e_{i}) = δ_{ij}}. this is a basis for V*.

so, for example, p_{2}(x,y,z) = y. now suppose f = a_{1}p_{1}+a_{2}p_{2}+a_{3}p_{3} is in Ann(U).

this means f(1,0,1) = 0, and f(1,1,0) = 0. so from the first equation we get: a_{1} + a_{3} = 0, and from the second we get: a_{1} + a_{2} = 0.

combining these two, we get: a_{2} = a_{3}. so f is of the form: f = -cp_{1} + cp_{2} + cp_{3}, for c in R. this is a 1-dimensional subspace of V*, with basis {-p_{1}+p_{2}+p_{3}}.

now Ann(Ann(U)) = {G in V**: G(f) = 0, for all f in Ann(U)}. this in turn means for any G in Ann(Ann(U)), -G(p_{1})+G(p_{2})+G(p_{3}) = 0.

now let's find a basis for Ann(Ann(U)). to do this, we need a basis for V**. let's use B = {E_{1},E_{2},E_{3}} where E_{k}(p_{j}) = δ_{jk}.

so if G = b_{1}E_{1}+b_{2}E_{2}+b_{3}E_{3}, from -G(p_{1})+G(p_{2})+G(p_{3}) = 0

we get -b_{1}+b_{2}+b_{3} = 0. we thus have 2 free parameters, say s and t, and G = s(E_{1}+E_{3}) + t(E_{1}+E_{2}),

that is, a basis for Ann(Ann(U)) is {E_{1}+E_{2},E_{1}+E_{3}}. see the similarity with U?

ok, so suppose we consider the isomorphism ψ:V-->V** given by [ψ(v)](f) = f(v), for f in V* let's see what we get for v = (1,1,0), and v = (1,0,1).

given f = a_{1}p_{1}+a_{2}p_{2}+a_{3}p_{3} in V* we have:

ψ(1,1,0)(f) = f(1,1,0) = a_{1}+a_{2}

ψ(1,0,1)(f) = f(1,0,1) = a_{1}+a_{3}

but (E_{1}+E_{2})(f) = E_{1}(a_{1}p_{1}+a_{2}p_{2}+a_{3}p_{3}) + E_{2}(a_{1}p_{1}+a_{2}p_{2}+a_{3}p_{3}) = a_{1}+a_{2}

and (E_{1}+E_{3})(f) = E_{1}(a_{1}p_{1}+a_{2}p_{2}+a_{3}p_{3}) + E_{3}(a_{1}p_{1}+a_{2}p_{2}+a_{3}p_{3}) = a_{1}+a_{3}

that is: ψ(1,1,0) = E_{1}+E_{2}, ψ(1,0,1) = E_{1}+E_{3}.

so ψ does indeed map U to Ann(Ann(U)).

in particular, if f = -cp_{1}+cp_{2}+cp_{3} (that is, if f is in Ann(U)), we see:

ψ(x,y,z)(f) = c(y+z-x). if we extend the basis {(1,1,0),(1,0,1)} to the basis {(1,1,0),(1,0,1),(-1,1,1)} for R^{3}, and consider f ≠ 0 in Ann(U) on these:

ψ(v)(f) = ψ[d_{1}(1,1,0)+d_{2}(1,0,1)+d_{3}(-1,1,1)](f) = f(d_{1}+d_{2}-d_{3},d_{1}+d_{3},d_{2}+d_{3})

= 3cd_{3}, which is 0 if and only if d_{3} = 0, that is, if v is in U, so ψ(v) is in Ann(Ann(U)) if and only if v is in U.

Re: Duals and anihilators

Thanks, it's all clear now.