College Algebra Exam Review 164

College Algebra Exam Review 164 - /v j .v i / D n X j D 1...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
174 3. PRODUCTS OF GROUPS for A;B;C 2 Hom K .V;W / and v 2 V . The reader is invited to check the remaining details in Exercise 3.4.1 . An important special instance of the preceeding construction is the vector space dual to V , Hom K .V;K/ , which we also denote by V ± . A linear map from V into the one dimensional vector space of scalars K is called a linear functional on V . V ± is the space of all linear functionals on V . Let us summarize our observations: Proposition 3.4.1. Let V be a vector space over a field K . (a) For any vector space W , Hom K .V;W / is a vector space. (b) In particular, V ± D Hom K .V;K/ is a vector space. Suppose now that V is finite dimensional with ordered basis B D .v 1 ;v 2 ;:::;v n / . Every element v 2 V has a unique expansion v D P n i D 1 ˛ i v i . For 1 ± j ± n define v ± j 2 V ± by v ± j . P n i D 1 ˛ i v i / D ˛ j . The functional v ± j is the unique element of V ± satisfying v ± j .v i / D ı i;j for 1 ± i ± n . 3 I claim that B ± D .v ± 1 ;v ± 2 ;:::;v ± n / is a a basis of V ± . In fact, for any f 2 V ± , consider the functional Q f D P n j D 1 f.v j /v ± j . We have Q f .v i / D n X j D 1 f.v j
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: /v j .v i / D n X j D 1 f.v j / i;j D f.v i /: Thus f.v i / D Q f .v i / for each element v i 2 B . It follows from Proposition 3.3.32 that f D Q f . This means that B spans V . Next we check the linear independence of B . Suppose P n j D 1 j v j D (the zero functional in V ). Applying both sides to a basis vector v i , we get D n X j D 1 j v j .v i / D n X j D 1 j i;j D i : Thus all the coefcients i are zero, which shows that B is linearly inde-pendent. B is called the basis of V dual to B . We showed above that for f 2 V , the expansion of f in terms of the basis B is f D n X j D 1 f.v j /v j : 3 Here i;j is the so called Kronecker delta, dened by i;j D 1 if i D j and i;j D otherwise....
View Full Document

Ask a homework question - tutors are online