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

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