Unformatted text preview: /v ± j .v i / D n X j D 1 f.v j /ı 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 ı i;j D ˛ i : Thus all the coefﬁcients ˛ 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 ı i;j is the so called “Kronecker delta”, deﬁned by ı i;j D 1 if i D j and ı i;j D otherwise....
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 Fall '08
 EVERAGE
 Linear Algebra, Algebra, Vector Space, Kronecker delta

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