# Math Help - null-space and annihilators

1. ## null-space and annihilators

I tried to work the following two rules, if someone can please confirm to me that they are indeed correct and guide how can I prove/counter-prove them

V - n dim vector space
V* dual space of V i.e. Hom(V,F)

Let S be a sub-set of V. A(S) is annihilator of S (thus is a sub-space of V*)
Let F be a sub-set of V*. N(F) is null-space of F (which means all v in V, such that for all f in F f(v)=0). N(F) can be shown to be a sub-space of V.

L(X) is linear span of any sub-set of any vector space.

Under the above notations, I want to establish/dis-prove the following:

1. N(A(S)) = L(S)

2. A(N(F)) = L(F)

Will really appreciate help here. I am just getting too confused here.

2. Originally Posted by aman_cc
I tried to work the following two rules, if someone can please confirm to me that they are indeed correct and guide how can I prove/counter-prove them

V - n dim vector space
V* dual space of V i.e. Hom(V,F)

Let S be a sub-set of V. A(S) is annihilator of S (thus is a sub-space of V*)
Let F be a sub-set of V*. N(F) is null-space of F (which means all v in V, such that for all f in F f(v)=0). N(F) can be shown to be a sub-space of V.

L(X) is linear span of any sub-set of any vector space.

Under the above notations, I want to establish/dis-prove the following:

1. N(A(S)) = L(S)

2. A(N(F)) = L(F)
For 1., N(A(S)) is a subspace containing S, so it contains L(S). For the reverse inclusion, suppose that x∉L(S). Let $\{e_1,\ldots,e_k\}$ be a basis for S. Then the set $\{e_1,\ldots,e_k,x\}$ is linearly independent, so we can extend it to a basis $\{e_1,\ldots,e_n\}$ for V, where $e_{k+1} = x$. Define an element f of V* by $f\bigl(\textstyle\sum \alpha_ie_i\bigr) = \alpha_{k+1}$. Then f∈A(S). But f(x)≠0, so x∉N(A(S)).

2. is also true, and it should be provable in a similar way.