Annihilators

Definition 1. The annihilator of any subset of a vector space ( need not be a subspace) is the set of all vectors in such that is identically zero for all in .

Thus and ( ). If is finite-dimensional and contains a non-zero vector, so that , then Section: Dual bases , Theorem 3 shows that .

Theorem 1. If is an -dimensional subspace of an -dimensional vector space , then is an -dimensional subspace of .

Proof. We leave it to the reader to verify that (in fact , for an arbitrary ) is always a subspace; we shall prove only the statement concerning the dimension of .

Let be a basis in whose first elements are in (and form therefore a basis for ); let be the dual basis in . We denote by the subspace (in ) spanned by ; clearly has dimension . We shall prove that .

If is any vector in , then is a linear combination of , and, for any , we have In other words, is in for ; it follows that is contained in , Suppose, on the other hand, that is any element of . Since , being in , is a linear combination of the basis vectors , we may write Since, by assumption, is in , we have, for every , in other words, is a linear combination of . This proves that is in , and consequently that and the theorem follows. ◻

Theorem 2. If is a subspace in a finite-dimensional vector space , then ( ) .

Proof. Observe that we use here the convention, established at the end of Section: Reflexivity , that identifies and . By definition, is the set of all vectors such that for all in . Since, by the definition of , for all in and all in , it follows that . The desired conclusion now follows from a dimension argument. Let be -dimensional; then the dimension of is , and that of is . Hence , as was to be proved. ◻

EXERCISES

Exercise 1. Define a non-zero linear functional on such that if and , then .

Exercise 2. The vectors , , and form a basis of . If is the dual basis, and if , find , , and .

Exercise 3. Prove that if is a linear functional on an -dimensional vector space , then the set of all those vectors for which is a subspace of ; what is the dimension of that subspace?

Exercise 4. If whenever is a vector in , then is a linear functional on ; find a basis of the subspace consisting of all those vectors for which .

Exercise 5. Prove that if , and if are linear functionals on an -dimensional vector space , then there exists a non-zero vector in such that for . What does this result say about the solutions of linear equations?

Exercise 6. Suppose that and that are linear functionals on an -dimensional vector space . Under what conditions on the scalars is it true that there exists a vector in such that for ? What does this result say about the solutions of linear equations?

Exercise 7. If is an -dimensional vector space over a finite field, and if , then the number of -dimensional subspaces of is the same as the number of -dimensional subspaces.

Exercise 8.

Prove that if is any subset of a finite-dimensional vector space, then coincides with the subspace spanned by .
If and are subsets of a vector space, and if , then .
If and are subspaces of a finite-dimensional vector space, then and . (Hint: make repeated use of (b) and of Section: Annihilators , Theorem 2.)
Is the conclusion of (c) valid for not necessarily finite-dimensional vector spaces?

Exercise 9. This exercise is concerned with vector spaces that need not be finite-dimensional; most of its parts (but not all) depend on the sort of transfinite reasoning that is needed to prove that every vector space has a basis (cf. Section: Bases , Ex. 11).

Suppose that and are scalar-valued functions defined on a set ; if and are scalars write for the function defined by for all in . The set of all such functions is a vector space with respect to this definition of the linear operations, and the same is true of the set of all finitely non-zero functions. (A function on is finitely non-zero if the set of those elements of for which is finite.)
Every vector space is isomorphic to the set of all finitely non-zero functions on some set.
If is a vector space with basis , and if is a scalar-valued function defined on the set , then there exists a unique linear functional on such that for all in .
Use (a), (b), and (c) to conclude that every vector space is isomorphic to a subspace of .
Which vector spaces are isomorphic to their own duals?
If is a linearly independent subset of a vector space , then there exists a basis of containing . (Compare this result with the theorem of Section: Bases .)
If is a set and if is an element of , write for the scalar-valued function defined on by writing or according as or . Let be the set of all functions together with the function defined by for all in . Prove that if is infinite, then is a linearly independent subset of the vector space of all scalar-valued functions on .
The natural correspondence from to is defined for all vector spaces (not only for the finite-dimensional ones); if is in , define the corresponding element of by writing for all in . Prove that if is reflexive (i.e., if every in can be obtained in this manner by a suitable choice of ), then is finite-dimensional. (Hint: represent as the set of all scalar-valued functions on some set, and then use (g), (f), and (c) to construct an element of that is not induced by an element of .)

Warning: the assertion that a vector space is reflexive if and only if it is finite-dimensional would shock most of the experts in the subject. The reason is that the customary and fruitful generalization of the concept of reflexivity to infinite-dimensional spaces is not the simple-minded one given in (h).

EXERCISES

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