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Dual (or biorthogonal) bases

02/04/2011

While I was reading the paper, Life Beyond Bases: The Advent of Frames (Part I), I noticed a new and important word dual or biorthogonal bases.

Where does dual base come from?

  • A simple example

Consider b_1,b_2 are two independent vectors in R^2. Then b_1, b_2 is a base. Now consider an arbitrary vector x, it can be written as

x=c_1 b_1 +c_2 b_2

My question is how to compute c_1, c_2. The conventional answer is easy. Let c=[c_1,c_2]^T, B=[b_1,b_2], then x=Bc, so the coefficients are c=B^{-1}x. There is no difficulty to get the coefficients. But can we go a little further?

  • Go a little further

We already know x=Bc and c=B^{-1}x. Hence of course x=Bc=BB^{-1}x. Write B^{-1}=\tilde{B}=[\tilde{b_1}, \tilde{b_2}]^T. So we have

x=(\tilde{b_1}^Tx)b_1+(\tilde{b_2}^Tx)b_2=c_1b_1+c_2b_2

  • Summary

Given a base b_1,b_2 and an arbitrary vector x, then

x=c_1b_1+c_2b_2=(\tilde{b_1}^Tx)b_1+(\tilde{b_2}^Tx)b_2

where [\tilde{b_1}, \tilde{b_2}] is the dual base. Note \tilde{b_1}^Tx is the projection length of x on the basis $\tilde{b_1}$. And \tilde{b_i}^Tb_j=0, \tilde{b_i}^Tb_i=1. x can also be written as

x=c_1\tilde{b_1}+c_2\tilde{b_2}=(b_1^Tx)\tilde{b_1}+(b_2^Tx)\tilde{b_2}

We know [b_1,b_2][\tilde{b_1},\tilde{b_2}]^T=I. For a orthogonal base, the dual base is itself. Then

x=(b_1^Tx)b_1+(b_1^Tx)b_2

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