[The Wikipedia page on vector space frames](http://en.wikipedia.org/wiki/Frame_of_a_vector_space) gives a construction to find a dual frame for a given frame. Specifically, given a set of vectors $\{ e\_k \}$ in a Hilbert space $\mathcal{H}$ such that for all $v\in\mathcal{H}$ and some $ A \leq B<\infty$ we have
$$A||v||^2\leq\sum\_k|\langle e\_k|v\rangle|^2\leq B ||v||^2,$$
the construction gives (among other things) a set of linear functionals $\phi\_k:\mathcal{H}\rightarrow\mathbb{C}$ such that for all $v\in\mathcal{H}$
$$v=\sum\_k e\_k \cdot\phi\_k(v).$$


My question is, how unique are these functionals? (or alternatively, their dual images in $\mathcal{H}$.) The construction gives a natural way to find these dual images, which depends on the inverse of the map $S(v)=\sum\_k\langle v|e\_k\rangle e\_k$ and thus on the inner product used, but that doesn't mean the corresponding functionals will depend on the inner product.


If I have, as in one standard example, three noncollinear vectors in a two-dimensional space, then I would naively expect to have "one real degree of freedom" in choosing the coordinates. How does this show up? If there are other sets of functionals, what do their dual images in $\mathcal{H}$ look like? Is the one picked by $S$ special? if so, how? Does any of this depend on the (in)finiteness of the space dimension?



Unicity of a vector space frame's dual frame
Consider the finite-dimensional case. Let $d$ be the dimension and $n$ the number of frame elements. Take $\Phi$ to denote the $d\times n$ matrix whose columns are the frame elements. Then $\Psi$ is a dual frame for $\Phi$ precisely when $\Phi\Psi^\*=I$. In particular, the $i$th row of $\Psi$ must have inner product $1$ with the $i$th row of $\Phi$ and be orthogonal to all other rows of $\Phi$.


Now suppose $d=2$, $n=3$, and $\Phi$ has rank $2$ (e.g., the columns are noncollinear). Then the rows of $\Phi$ are linearly independent, and so the first column of $\Psi^\*$ is a member of some shifted version of the $1$-dimensional nullspace of $\Phi$ (the second column has its own shifted version). In the real case, this produces your one real degree of freedom for each coordinate.


The dual frame that is constructed in the Wikipedia page is called the canonical dual frame. In matrix parlance, this is just the Moore–Penrose pseudoinverse: $\Psi=(\Phi\Phi^\*)^{-1}\Phi$. This dual frame is special because


$$
\|\Psi^\* x\|\_2\leq\|y\|\_2
\quad
\mbox{whenever}
\quad
x=\Phi y,
$$


that is, canonical dual frame coefficients have minimal $2$-norm. For applications, the canonical dual frame is used because it's optimally robust to additive noise.


As you'd expect, in the infinite-dimensional case, a frame will typically have many dual frames other than its canonical dual, but the canonical dual still has the same notion of optimality. More information on the finite frames case can be found [in this book](http://books.google.com/books?isbn=0817683739).


