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Relationship of Basis to Dual Basis

  1. Nov 26, 2011 #1
    If we're working in R^n and we consider the elements of a basis for R^n to be the column vectors of an nxn invertible matrix B, then what is the relationship between B and the matrix whose row vectors represent elements of the corresponding dual basis for R^n*? My guess, which Wikipedia helped me formulate, is that the row vectors of the inverse of B constitute the dual basis, but I'm still not sure. Also, if we're working in general finite-dimensional vector spaces, does the process of finding a dual basis become harder, or is it trivial once you know how to do it for R^n?

  2. jcsd
  3. Nov 26, 2011 #2


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    The "dual" of a vector space, V, is the set of linear functions from V to the real numbers. If you are representing the vectors, v, as column matrices, then you can represent the dual vectors, v*, as row matrices. The value of the function is the matrix product v*v.

    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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