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- Thread starter marschmellow
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HallsofIvy

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More generally, given basis vectors [itex]v_1, v_2, v_3, ..., v_n[/itex] for V, you can define n functions by [itex]f_1(x_1)= 1[/itex], [itex]f_1(x_i)= 0[/itex] for [itex]i\ne 1[/itex], [itex]f_2(x_2)= 1[/itex], [itex]f_2(x_i)= 0[/itex] for [itex]i\ne 2[/itex] and, generally, [itex]f_j(x_i)= n[/itex] if i= j, [itex]f_j(x_i)= 0[/itex] if [itex]i\ne j[/itex]. Show that those functions form a basis for the dual of V.

In finite dimensional vector spaces, if two vectors spaces have the same dimensionl, they are isomorphic so the dual of a finite dimensional vector space is isomorphic to the vector space. In infinite dimensions, that is not true. However, the "dual of the dual" of a vector space is always isomorphic to the vector space.

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