Definition 1. The annihilator \(\mathcal{S}^{0}\) of any subset \(\mathcal{S}\) of a vector space \(\mathcal{V}\) ( \(\mathcal{S}\) need not be a subspace) is the set of all vectors \(y\) in \(\mathcal{V}^{\prime}\) such that \([x, y]\) is identically zero for all \(x\) in \(\mathcal{S}\) .

Thus \(\mathcal{O}^{0}=\mathcal{V}^{\prime}\) and \(\mathcal{V}^{0}=\mathcal{O}\) ( \(\subset \mathcal{V}^{\prime}\) ). If \(\mathcal{V}\) is finite-dimensional and \(\mathcal{S}\) contains a non-zero vector, so that \(\mathcal{S} \neq \mathcal{O}\) , then Section: Dual bases , Theorem 3 shows that \(\mathcal{S}^{0} \neq \mathcal{V}^{\prime}\) .

Theorem 1. If \(\mathcal{M}\) is an \(m\) -dimensional subspace of an \(n\) -dimensional vector space \(\mathcal{V}\) , then \(\mathcal{M}^{0}\) is an \((n-m)\) -dimensional subspace of \(\mathcal{V}^{\prime}\) .

Proof. We leave it to the reader to verify that \(\mathcal{M}^{0}\) (in fact \(\mathcal{S}^{0}\) , for an arbitrary \(\mathcal{S}\) ) is always a subspace; we shall prove only the statement concerning the dimension of \(\mathcal{M}^{0}\) .

Let \(\mathcal{X}=\{x_{1}, \ldots, x_{n}\}\) be a basis in \(\mathcal{V}\) whose first \(m\) elements are in \(\mathcal{M}\) (and form therefore a basis for \(\mathcal{M}\) ); let \(\mathcal{X}^{\prime}=\{y_{1}, \ldots, y_{n}\}\) be the dual basis in \(\mathcal{V}^{\prime}\) . We denote by \(\mathcal{N}\) the subspace (in \(\mathcal{V}^{\prime}\) ) spanned by \(y_{m+1}, \ldots, y_{n}\) ; clearly \(\mathcal{N}\) has dimension \(n-m\) . We shall prove that \(\mathcal{M}^{0}=\mathcal{N}\) .

If \(x\) is any vector in \(\mathcal{M}\) , then \(x\) is a linear combination of \(x_{1}, \ldots, x_{m}\) , \[x=\sum_{i=1}^{m} \xi_{i} x_{i},\] and, for any \(j=m+1, \ldots, n\) , we have \[[x, y_{j}]=\sum_{i=1}^{m} \xi_{i}[x_{i}, y_{j}]=0.\] In other words, \(y_{j}\) is in \(\mathcal{M}^{0}\) for \(j=m+1, \ldots, n\) ; it follows that \(\mathcal{N}\) is contained in \(\mathcal{M}^{0}\) , \[\mathcal{N} \subset \mathcal{M}^{0} .\] Suppose, on the other hand, that \(y\) is any element of \(\mathcal{M}^{0}\) . Since \(y\) , being in \(\mathcal{V}^{\prime}\) , is a linear combination of the basis vectors \(y_{1}, \ldots, y_{n}\) , we may write \[y=\sum_{j=1}^{n} \eta_{j} y_{j}.\] Since, by assumption, \(y\) is in \(\mathcal{M}^{0}\) , we have, for every \(i=1, \ldots, m\) , \[0=t[x_{i}, y]=\sum_{j=1}^{n} \eta_{j}[x_{i}, y_{j}]=\eta_{i}\] in other words, \(y\) is a linear combination of \(y_{m+1}, \ldots, y_{n}\) . This proves that \(y\) is in \(\mathcal{N}\) , and consequently that \[\mathcal{M}^{0} \subset \mathcal{N}\] and the theorem follows. ◻

Theorem 2. If \(\mathcal{M}\) is a subspace in a finite-dimensional vector space \(\mathcal{V}\) , then \(\mathcal{M}^{00}\) ( \(=(\mathcal{M}^{0})^{0}\) ) \(=\mathcal{M}\) .

Proof. Observe that we use here the convention, established at the end of Section: Reflexivity , that identifies \(\mathcal{V}\) and \(\mathcal{V}^{\prime \prime}\) . By definition, \(\mathcal{M}^{00}\) is the set of all vectors \(x\) such that \([x, y]=0\) for all \(y\) in \(\mathcal{M}^{0}\) . Since, by the definition of \(\mathcal{M}^{0}\) , \([x, y]=0\) for all \(x\) in \(\mathcal{M}\) and all \(y\) in \(\mathcal{M}^{0}\) , it follows that \(\mathcal{M} \subset \mathcal{M}^{00}\) . The desired conclusion now follows from a dimension argument. Let \(\mathcal{M}\) be \(m\) -dimensional; then the dimension of \(\mathcal{M}^{0}\) is \(n-m\) , and that of \(\mathcal{M}^{00}\) is \(n-(n-m)=m\) . Hence \(\mathcal{M}=\mathcal{M}^{00}\) , as was to be proved. ◻

EXERCISES

Exercise 1. Define a non-zero linear functional \(y\) on \(\mathbb{C}^3\) such that if \(x_1 = (1, 1, 1)\) and \(x_2 = (1, 1, -1)\) , then \([x_1, y] = [x_2, y] = 0\) .

Exercise 2. The vectors \(x_1 = (1, 1, 1)\) , \(x_2 = (1, 1, -1)\) , and \(x_3 = (1, -1, -1)\) form a basis of \(\mathbb{C}^3\) . If \(\{y_1, y_2, y_3\}\) is the dual basis, and if \(x = (0, 1, 0)\) , find \([x, y_1]\) , \([x, y_2]\) , and \([x, y_3]\) .

Exercise 3. Prove that if \(y\) is a linear functional on an \(n\) -dimensional vector space \(\mathcal{V}\) , then the set of all those vectors \(x\) for which \([x, y] = 0\) is a subspace of \(\mathcal{V}\) ; what is the dimension of that subspace?

Exercise 4. If \(y(x) = \xi_1 + \xi_2 + \xi_3\) whenever \(x = (\xi_1, \xi_2, \xi_3)\) is a vector in \(\mathbb{C}^3\) , then \(y\) is a linear functional on \(\mathbb{C}^3\) ; find a basis of the subspace consisting of all those vectors \(x\) for which \([x, y] = 0\) .

Exercise 5. Prove that if \(m < n\) , and if \(y_1, \ldots, y_m\) are linear functionals on an \(n\) -dimensional vector space \(\mathcal{V}\) , then there exists a non-zero vector \(x\) in \(\mathcal{V}\) such that \([x, y_j] = 0\) for \(j = 1, \ldots, m\) . What does this result say about the solutions of linear equations?

Exercise 6. Suppose that \(m < n\) and that \(y_1, \ldots, y_m\) are linear functionals on an \(n\) -dimensional vector space \(\mathcal{V}\) . Under what conditions on the scalars \(\alpha_1, \ldots, \alpha_m\) is it true that there exists a vector \(x\) in \(\mathcal{V}\) such that \([x, y_j] = \alpha_j\) for \(j = 1, \ldots, m\) ? What does this result say about the solutions of linear equations?

Exercise 7. If \(\mathcal{V}\) is an \(n\) -dimensional vector space over a finite field, and if \(0 \leq m \leq n\) , then the number of \(m\) -dimensional subspaces of \(\mathcal{V}\) is the same as the number of \((n - m)\) -dimensional subspaces.

Exercise 8. 

  1. Prove that if \(\mathcal{S}\) is any subset of a finite-dimensional vector space, then \(\mathcal{S}^{00}\) coincides with the subspace spanned by \(\mathcal{S}\) .
  2. If \(\mathcal{S}\) and \(\mathcal{T}\) are subsets of a vector space, and if \(\mathcal{S} \subset \mathcal{T}\) , then \(\mathcal{T}^{0} \subset \mathcal{S}^{0}\) .
  3. If \(\mathcal{M}\) and \(\mathcal{N}\) are subspaces of a finite-dimensional vector space, then \((\mathcal{M} \cap \mathcal{N})^{0} = \mathcal{M}^{0} + \mathcal{N}^{0}\) and \((\mathcal{M} + \mathcal{N})^{0} = \mathcal{M}^{0} \cap \mathcal{N}^{0}\) . (Hint: make repeated use of (b) and of Section: Annihilators , Theorem 2.)
  4. 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).

  1. Suppose that \(f\) and \(g\) are scalar-valued functions defined on a set \(\mathcal{X}\) ; if \(\alpha\) and \(\beta\) are scalars write \(h = \alpha f + \beta g\) for the function defined by \(h(x) = \alpha f(x) + \beta g(x)\) for all \(x\) in \(\mathcal{X}\) . 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 \(f\) on \(\mathcal{X}\) is finitely non-zero if the set of those elements \(x\) of \(\mathcal{X}\) for which \(f(x) \neq 0\) is finite.)
  2. Every vector space is isomorphic to the set of all finitely non-zero functions on some set.
  3. If \(\mathcal{V}\) is a vector space with basis \(\mathcal{X}\) , and if \(f\) is a scalar-valued function defined on the set \(\mathcal{X}\) , then there exists a unique linear functional \(y\) on \(\mathcal{V}\) such that \([x, y] = f(x)\) for all \(x\) in \(\mathcal{X}\) .
  4. Use (a), (b), and (c) to conclude that every vector space \(\mathcal{V}\) is isomorphic to a subspace of \(\mathcal{V}^\prime\) .
  5. Which vector spaces are isomorphic to their own duals?
  6. If \(\mathcal{Y}\) is a linearly independent subset of a vector space \(\mathcal{V}\) , then there exists a basis of \(\mathcal{V}\) containing \(\mathcal{Y}\) . (Compare this result with the theorem of Section: Bases .)
  7. If \(\mathcal{X}\) is a set and if \(y\) is an element of \(\mathcal{X}\) , write \(f_y\) for the scalar-valued function defined on \(\mathcal{X}\) by writing \(f_y(x) = 1\) or \(0\) according as \(x = y\) or \(x \neq y\) . Let \(\mathcal{Y}\) be the set of all functions \(f_y\) together with the function \(g\) defined by \(g(x) = 1\) for all \(x\) in \(\mathcal{X}\) . Prove that if \(\mathcal{X}\) is infinite, then \(\mathcal{Y}\) is a linearly independent subset of the vector space of all scalar-valued functions on \(\mathcal{X}\) .
  8. The natural correspondence from \(\mathcal{V}\) to \(\mathcal{V}^{\prime\prime}\) is defined for all vector spaces (not only for the finite-dimensional ones); if \(x\) is in \(\mathcal{V}\) , define the corresponding element \(z_0\) of \(\mathcal{V}^{\prime\prime}\) by writing \(z_0(y) = [x_0, y]\) for all \(y\) in \(\mathcal{V}^\prime\) . Prove that if \(\mathcal{V}\) is reflexive (i.e., if every \(x\) in \(\mathcal{V}\) can be obtained in this manner by a suitable choice of \(x_0\) ), then \(\mathcal{V}\) is finite-dimensional. (Hint: represent \(\mathcal{V}^\prime\) as the set of all scalar-valued functions on some set, and then use (g), (f), and (c) to construct an element of \(\mathcal{V}^{\prime\prime}\) that is not induced by an element of \(\mathcal{V}\) .)

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).