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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 independent. 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....
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 Fall '08
 EVERAGE
 Algebra, Vector Space

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