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Wednesday, July 17, 2013

Dual Annihilators (11.3.3)

Dummit and Foote Abstract Algebra, section 11.3, exercise 3:

MathJax TeX Test Page Let SV for V some finite dimensional space. Define Ann(S)={vV | f(v)=0 for all fS}.

(a) Show Ann(S) is a subspace of V.
(b) Let W1,W2V be subspaces. Show Ann(W1+W2)=Ann(W1)Ann(W2) and Ann(W1W2)=Ann(W1)+Ann(W2).
(c) Prove W1=W2Ann(W1)=Ann(W2).
(d) Prove Ann(S)=Ann(span S).
(e) Assume V has for basis {v1,...,vn}. Prove that if S={v1,...,vk} then Ann(S) has basis {vk+1,...,vn}.
(f) Assume V is finite dimensional. Prove that if WV is a subspace then dim Ann(W)=dim Vdim W.

Proof: We shall approach this problem from another angle to obtain the final result first, and then recover the rest as implications. But first, (a):v1,v2Ann(S)f(v1+αv2)=f(v1)+αf(v2)=0     fS Let S be a finite set of linearly independent linear functionals v1,...,vk, perhaps a basis for an arbitrary subspace of V. Define a linear transformationF:VFkF(v)=(v1(v),...,vk(v))Under this terminology, we can see Ann(S)=ker F. Letting A be the k×n reduced row echelon matrix of F from basis e1,...,en (of V) to e1,...,ek (of Fk) we observe the linear functionals manifest as linearly independent rows, so that the kernel is of dimension nk=dim Vdim span S. This is (f), since (d) is immediately evident. (e) then follows easily, as the argument for the basis is clear and dimensional restrictions show it spans.

When SV, define the subspace Ann(S)={vV | v(s)=0 for all sS} to be the dual notion of the Ann defined above. A parallel argument by evaluations at vi gives dim Ann(S)=dim Vdim span S. When S is a subspace of V or of V we show Ann(Ann(S))=S since clearly holds, and dimension restrictions show , which now easily gives (c).

We now show Ann(W1+W2)=Ann(W1)Ann(W2) when W1,W2 are subspaces of V (or V). () Let vAnn(W1)Ann(W2). We have (w1+w2)(v)=w1(v)+w2(v)=0 (respectively, v(w1+w2)=v(w1)+v(w2)=0). () Let vAnn(W1+W2). We have w1(v)=(w1+0)(v)=0 and similarly w2(v)=0 (respectively, v(w1)=v(w1+0)=0 and v(w2)=0). Now, to finish, we haveAnn(W1W2)=Ann(W1)+Ann(W2)Ann(Ann(W1W2))=Ann(Ann(W1)+Ann(W2))W1W2=Ann(Ann(W1))Ann(Ann(W2))=W1W2 

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