Thread: Duals and anihilators

Hi!

I have this problem in my book that I really have trouble with:

Given a vector space V and a subspace U of V. Prove that phi of U [phi is the natural isomorphism from U to U** given by phi(u)(f) = f(u)] is equal (not just isomorphic!) to the double annihilator of U. Given G in the double annihilator, find u in U such that G=phi(u)

Any help would be greatly appreciated!!

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**).

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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.

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

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?

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 = R³. 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, p₂(x,y,z) = y. now suppose f = a₁p₁+a₂p₂+a₃p₃ is in Ann(U).

this means f(1,0,1) = 0, and f(1,1,0) = 0. so from the first equation we get: a₁ + a₃ = 0, and from the second we get: a₁ + a₂ = 0.

combining these two, we get: a₂ = a₃. so f is of the form: f = -cp₁ + cp₂ + cp₃, for c in R. this is a 1-dimensional subspace of V*, with basis {-p₁+p₂+p₃}.

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(p₁)+G(p₂)+G(p₃) = 0.

now let's find a basis for Ann(Ann(U)). to do this, we need a basis for V**. let's use B = {E₁,E₂,E₃} where E_k(p_j) = δ_jk.

so if G = b₁E₁+b₂E₂+b₃E₃, from -G(p₁)+G(p₂)+G(p₃) = 0

we get -b₁+b₂+b₃ = 0. we thus have 2 free parameters, say s and t, and G = s(E₁+E₃) + t(E₁+E₂),

that is, a basis for Ann(Ann(U)) is {E₁+E₂,E₁+E₃}. 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 = a₁p₁+a₂p₂+a₃p₃ in V* we have:

ψ(1,1,0)(f) = f(1,1,0) = a₁+a₂
ψ(1,0,1)(f) = f(1,0,1) = a₁+a₃

but (E₁+E₂)(f) = E₁(a₁p₁+a₂p₂+a₃p₃) + E₂(a₁p₁+a₂p₂+a₃p₃) = a₁+a₂

and (E₁+E₃)(f) = E₁(a₁p₁+a₂p₂+a₃p₃) + E₃(a₁p₁+a₂p₂+a₃p₃) = a₁+a₃

that is: ψ(1,1,0) = E₁+E₂, ψ(1,0,1) = E₁+E₃.

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

in particular, if f = -cp₁+cp₂+cp₃ (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 R³, and consider f ≠ 0 in Ann(U) on these:

ψ(v)(f) = ψ[d₁(1,1,0)+d₂(1,0,1)+d₃(-1,1,1)](f) = f(d₁+d₂-d₃,d₁+d₃,d₂+d₃)

= 3cd₃, which is 0 if and only if d₃ = 0, that is, if v is in U, so ψ(v) is in Ann(Ann(U)) if and only if v is in U.

Thanks, it's all clear now.