歼灭者

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

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

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

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

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

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

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

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

EXERCISES

Exercise 1. Define a non-zero linear functional $y$ on $ℂ^{3}$ 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 $ℂ^{3}$ . 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 $𝒱$ , then the set of all those vectors $x$ for which $[x, y] = 0$ is a subspace of $𝒱$ ; what is the dimension of that subspace?

Exercise 4. If $y (x) = ξ_{1} + ξ_{2} + ξ_{3}$ whenever $x = (ξ_{1}, ξ_{2}, ξ_{3})$ is a vector in $ℂ^{3}$ , then $y$ is a linear functional on $ℂ^{3}$ ; 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 $y_{1}, \dots, y_{m}$ are linear functionals on an $n$ -dimensional vector space $𝒱$ , then there exists a non-zero vector $x$ in $𝒱$ such that $[x, y_{j}] = 0$ for $j = 1, \dots, m$ . What does this result say about the solutions of linear equations?

Exercise 6. Suppose that $m < n$ and that $y_{1}, \dots, y_{m}$ are linear functionals on an $n$ -dimensional vector space $𝒱$ . Under what conditions on the scalars $α_{1}, \dots, α_{m}$ is it true that there exists a vector $x$ in $𝒱$ such that $[x, y_{j}] = α_{j}$ for $j = 1, \dots, m$ ? What does this result say about the solutions of linear equations?

Exercise 7. If $𝒱$ 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 $𝒱$ is the same as the number of $(n - m)$ -dimensional subspaces.

Exercise 8.

Prove that if $𝒮$ is any subset of a finite-dimensional vector space, then $𝒮^{00}$ coincides with the subspace spanned by $𝒮$ .
If $𝒮$ and $𝒯$ are subsets of a vector space, and if $𝒮 \subset 𝒯$ , then $𝒯^{0} \subset 𝒮^{0}$ .
If $ℳ$ and $𝒩$ are subspaces of a finite-dimensional vector space, then $(ℳ \cap 𝒩)^{0} = ℳ^{0} + 𝒩^{0}$ and $(ℳ + 𝒩)^{0} = ℳ^{0} \cap 𝒩^{0}$ . (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 $f$ and $g$ are scalar-valued functions defined on a set $𝒳$ ; if $α$ and $β$ are scalars write $h = α f + β g$ for the function defined by $h (x) = α f (x) + β g (x)$ for all $x$ in $𝒳$ . 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 $𝒳$ is finitely non-zero if the set of those elements $x$ of $𝒳$ for which $f (x) \neq 0$ is finite.)
Every vector space is isomorphic to the set of all finitely non-zero functions on some set.
If $𝒱$ is a vector space with basis $𝒳$ , and if $f$ is a scalar-valued function defined on the set $𝒳$ , then there exists a unique linear functional $y$ on $𝒱$ such that $[x, y] = f (x)$ for all $x$ in $𝒳$ .
Use (a), (b), and (c) to conclude that every vector space $𝒱$ is isomorphic to a subspace of .
Which vector spaces are isomorphic to their own duals?
If $𝒴$ is a linearly independent subset of a vector space $𝒱$ , then there exists a basis of $𝒱$ containing $𝒴$ . (Compare this result with the theorem of Section: Bases .)
If $𝒳$ is a set and if $y$ is an element of $𝒳$ , write $f_{y}$ for the scalar-valued function defined on $𝒳$ by writing $f_{y} (x) = 1$ or $0$ according as $x = y$ or $x \neq y$ . Let $𝒴$ be the set of all functions $f_{y}$ together with the function $g$ defined by $g (x) = 1$ for all $x$ in $𝒳$ . Prove that if $𝒳$ is infinite, then $𝒴$ is a linearly independent subset of the vector space of all scalar-valued functions on $𝒳$ .
The natural correspondence from $𝒱$ to is defined for all vector spaces (not only for the finite-dimensional ones); if $x$ is in $𝒱$ , define the corresponding element $z_{0}$ of by writing $z_{0} (y) = [x_{0}, y]$ for all $y$ in . Prove that if $𝒱$ is reflexive (i.e., if every $x$ in $𝒱$ can be obtained in this manner by a suitable choice of $x_{0}$ ), then $𝒱$ 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 $𝒱$ .)

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).