# Relationship of Basis to Dual Basis

1. Nov 26, 2011

### marschmellow

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?

Thanks.

2. Nov 26, 2011

### HallsofIvy

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 $v_1, v_2, v_3, ..., v_n$ for V, you can define n functions by $f_1(x_1)= 1$, $f_1(x_i)= 0$ for $i\ne 1$, $f_2(x_2)= 1$, $f_2(x_i)= 0$ for $i\ne 2$ and, generally, $f_j(x_i)= n$ if i= j, $f_j(x_i)= 0$ if $i\ne j$. 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.