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ak416
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Can someone prove this to me? I know that if you have a finite dimensional vector space V with a dual space V*, then every ordered basis for V* is the dual basis for some basis for V (this follows from a theorem). But if you're just given an arbitrary vector space V. Let's say the Space of R^n over the field R, then how could the elements of that space be linear functions of another space (and thus V is the dual space of another space)? And what would that space be? All i can guess is that V can be thought of as set of constant functions...
The two main theorems I am supplied with in this section are
Thm: Suppose V is a finite dimensional vector space with an ordered basis B = {x_1,...,x_n}. Let f_i (1<= i<=n) be the ith coordinate function wrt B, and let B* = {f_1,...,f_n}. Then B* is an ordered basis for V*, and, for any f element of V*, we have f = Sum i=1 to n[f(x_i)f_i
Thm: Let V be a finite dimensional vector space, and and define U: V->V** by U(x) = x' where x': V* -> F is defined by x'(f) = f(x) (F is a field). Then U is an isomorphism.
So again, I would like a proof of the statement "Every vector space is a dual of some other vector space" and an example using let's say R^n as the dual space. Thanks.
The two main theorems I am supplied with in this section are
Thm: Suppose V is a finite dimensional vector space with an ordered basis B = {x_1,...,x_n}. Let f_i (1<= i<=n) be the ith coordinate function wrt B, and let B* = {f_1,...,f_n}. Then B* is an ordered basis for V*, and, for any f element of V*, we have f = Sum i=1 to n[f(x_i)f_i
Thm: Let V be a finite dimensional vector space, and and define U: V->V** by U(x) = x' where x': V* -> F is defined by x'(f) = f(x) (F is a field). Then U is an isomorphism.
So again, I would like a proof of the statement "Every vector space is a dual of some other vector space" and an example using let's say R^n as the dual space. Thanks.