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Annihilators : Linear Algebra Notes
Satya Mandal
September 21, 2005
Let F be a filed and V be vector space over F with dim(V ) = n < โ. As usual V โ^ will denote the dual space of V and V โโ^ = (V โ)โ.
- Definition 0.1 For a subset S โ V define the annihilator
ann(S) = {f โ V โ^ : f (v) = 0 f or allv โ S}.
We also use the notation S^0 = ann(S). We may also, temporarily, use the notation
ann(V โ^ : S) = ann(S)
to underscore the fact that ann(V โ^ : S) is a subspece of V โ.
- First note that for S โ V, the annihilator ann(S) โ V โ^ is a subspace of the dual space.
- Also note for S โ V, if W = Span(S) then
ann(S) = ann(W ).
- Now let S โ V โ. Then according to above definition, annihi- lator of S is ann(S) = ann(V โโ^ : S) is a subspace of V โโ. In this case, there is another natural annihilator of S as a subspace of V as follows:
ann(V : S) = {v โ V : f (v) = 0 f or all f โ S}.
It is possible to mix up two annihilator of S. The first one is a subspace of the double dual V โโ^ and the second one is the subspace of V. We will justify that these two annihilators are same via the natural identification of V and V โโ.
- Let L : V โโผ V โโ^ be the natural isomorphism.
- Let S โ V โ. Then
L(ann(V : S)) = ann(V โโ^ : S).
Proof. Let v โ ann(V : S). Then f (v) = 0 for all f โ S. Hence L(v)(f ) = f (v) = 0 for all f โ S. So, L(v) โ ann(V โโ^ : S). Therefore
L(ann(V : S)) โ ann(V โโ^ : S).
Now let G โ ann(V โโ^ : S). Since L is an isomorphism, we have G = L(v) for some v โ V. We need to show that v โ ann(V : S). But for f โ S we have 0 = G(f ) = L(v)(f ) = f (v). Therefore v โ ann(V : S). So,
ann(V โโ^ : S) โ L(ann(V : S)).
Hence the proof is complete.
- Now, for a subspace W โ V, the annihilator W 0 = ann(V : W ) has two annihilators. They are ann(V : W 0 ) and ann(V โโ^ : W 0 ). It follows from above that
L(ann(V : W 0 )) = ann(V โโ^ : W 0 ).
Since, L is a linear isomorphism, we also have
dim(ann(V : W 0 )) = dim(ann(V โโ^ : W 0 )).
Lemma 0.1 Suppose V is vector space of finite dimension, dim V = n, over F. Let f, g โ V โ^ be two linear functionals. let Nf be the null space of f and Ng be the null space of g. Then, Nf โ Ng if and only if g = cf for some c โ F.
Proof. (โ:) Obvious. (โ:) If g = 0 then g = cf with c = 0. So, we assume that g 6 = 0. So, dim(Ng) = nโ 1. Since Nf โ Ng, we have f 6 = 0 and dim(Nf ) = nโ 1. Therefore, Nf = Ng = N (say). Now, pick e /โ N. Since dim(V ) = n, it follows that V = N + Fe. Also, f (e) 6 = 0 and g(e) 6 = 0. Write c = g(e)/f (e). Claim that
g = cf.
First note, g(e) = cf (e). Now, for x โ V, we have x = y + ฮปe for some y โ N and ฮป โ F. Therefore
g(x) = g(y) + ฮปg(e) = ฮปg(e) = ฮปcf (e) = c(f (y + ฮปe)) = cf (x).
So, the proof is complete.
Following is Theorem 20, page 110.
Theorem 0.2 Suppose V is vector space of finite dimension, dim V = n, over F. Let g, f 1 ,... , fr โ V โ^ be linear functionals. Let N be the null space of g and Ni be the null space of fi. Then, N 1 โฉ N 2 โฉ ยท ยท ยท โฉ Nr โ N if and only if g =
โr i=1 cifi^ for some ci โ F.
Proof. (โ:) Obvious. (โ:) We use induction on r to prove this part. Case r = 1, is the above Lemma 0.1. Now, we assume the validity of the thoerem for r โ 1 functionals and prove it for r. Write V โฒ^ = Nr. Let gโฒ^ = G|V โฒ^ , f 1 โฒ = (f 1 )|V โฒ^ ,... , f (^) rโฒโ 1 = (frโ 1 )|V โฒ^ ,
be the restrictions of the respective functionals to V โฒ. Induction ap- plies for these functionals and it follows that
gโฒ^ =
rโโ 1
i=
cif (^) iโฒ
for some ci โ F. This means, for all x โ V โฒ^ = Nr, we have
g(x) =
rโโ 1
i=
cifi(x).
This means, for all x โ V โฒ^ = Nr, we have
g(x) =
rโโ 1
i=
cifi(x).
Write
h = g โ
rโโ 1
i=
cifi.
Then Nr = V โฒ^ โ N ull โ Space(h). By the case r = 1 (or by Lemma 0.1) It follows that h = crfr for some cr โ F. Hence
g =
rโโ 1
i=
cifi + h =
โ^ r
i=
cifi
and the proof is complete.
