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Unformatted text preview: 3.4. THE DUAL OF A VECTOR SPACE AND MATRICES 177 Lemma 3.4.4. Let V be a ﬁnite dimensional vector space over a ﬁeld K .
For each nonzero v 2 V , there is a linear functional f 2 V such that
f .v/ ¤ 0. Proof. We know that any linearly independent subset of V is contained in
a basis. If v is a nonzero vector in V , then fv g is linearly independent.
Therefore, there is a basis B of V with v 2 B . Let f be any function from
B into K with f .v/ ¤ 0. By Proposition 3.3.32, f extends to a linear
functional on V .
I Theorem 3.4.5. If V is a ﬁnite dimensional vector space, then Ã W V
V is a linear isomorphism. ! Proof. We already know that Ã is linear.
If v is a nonzero vector in V , then there is an f 2 V such that
f .v/ ¤ 0, by Lemma 3.4.4. Thus Ã.v/.f / D f .v/ ¤ 0, and Ã.v/ ¤
0. Thus Ã is injective. Applying Proposition 3.4.2(c) twice, we have
dim.V / D dim.V / D dim.V /. Therefore any injective linear map
from V to V is necessarily surjective, by Proposition 3.3.35.
I
Finite dimensionality is essential for this theorem. For an inﬁnite dimensional vector space, Ã W V ! .V / is injective, but not surjective. Duals of subspaces and quotients
Let V be a ﬁnite dimensional vector space over K . For any subset
S Â V , let S ı denote the set of f 2 V such that hv; f i D 0 for all
v 2 S . Likewise, for A Â V , let Aı denote the set of v 2 V such that
hv; f i D 0 for all f 2 A. (We identify V with V .) S ı is called the
annihilator of S in V .
Lemma 3.4.6. Let S and T be subsets of V , and W a subspace of V .
(a) S ı is a subspace of V .
(b) If S Â T , then T ı Â S ı and S ıı Â T ıı
(c) T Â T ıı .
(d) W D W ıı .
(e) S ı D span.S /ı and S ıı D span.S /. ...
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This note was uploaded on 12/15/2011 for the course MAC 1105 taught by Professor Everage during the Fall '08 term at FSU.
 Fall '08
 EVERAGE
 Algebra, Matrices, Vector Space

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