Page 152 - DMTH502_LINEAR_ALGEBRA

Linear Algebra

Notes Corollary: Let V be a finite-dimensional vector space over the field F. Each basis for V* is the
dual of some basis for V.

Proof: Let *= f 1 ,..., f n be a basis for V*. By Theorem 2 of unit 10 there is a basis L 1 ,...,L for
n
V** such that

L f . ...(5)
i i ij
Using the corollary above, for each i there is a vector , in V such that
L (f) = f()
i i
for every f in V*, i.e., such that L L . It follows immediately that ,..., is a basis for V and
i i 1 n
that * is the dual of this basis.

In view of Theorem 1, we usually identify with L and say that V ‘is’ the dual space of V* or
that the spaces V, V* are naturally in duality with one another. Each is the dual space of the other.
In the last corollary we have an illustration of how that can be useful. Here is a further illustration.
If E is a subset of V*, then the annihilator E is (technically) a subset of V**. If we choose to
0
0
identify V and V** as in Theorem (1), then E is a subspace of V, namely, the set of all in V such
that f = 0 for all f in E. In a corollary of Theorem 3 of unit 10 we noted that each subspace W
is determined by its annihilator W°. How is it determined? The answer is that W is the subspace
annihilated by all f in W°, that is, the intersection of the null spaces of all f’s in W°. In our present
notation for annihilators, the answer may be phrased very simply: W = (W°)°.

Theorem 2: If S is any subset of a finite-dimensional vector space V, then (S°)° is the subspace
spanned by S.

Proof: Let W be the subspace spanned by S. Clearly W° = S°. Therefore, what we are to prove is
that W = W°°. We have given one proof. Here is another. By Theorem 3 of unit 10.

dimW dimW dimV
...(6)
dimW dimW dimV *

and since dimV dimV * we have
dimW dimW .

Since W is a subspace of W°°, we see that W = W°°.
The results of this section hold for arbitrary vector spaces; however the proofs require the use of
the so-called Axiom of Choice. Here we shall not tackle annihilators for general vector spaces.
But, there are two results about linear functionals on arbitrary vector spaces which are so
fundamental that we should include them.
Let V be a vector space. We want to define hyperspaces in V. Unless V is finite-dimensional, we
cannot do that with the dimension of the hyperspace. But, we can express the idea that a space N
falls just one dimension short of filling out V, in the following way:
1. N is a proper subspace of V;
2. If W is a subspace of V which contains N, then either W = N or W = V.
Conditions (1) and (2) together say that N is a proper subspace and there is no larger proper
subspace, in short, N is a maximal proper subspace.

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