The Wikipedia page on vector space frames 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?


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.


You are essentially asking about the structure of the set $D_X$ of all dual frames $Y$ of a given frame $X$. As mentioned by Dustin G. Mixon, there is always one preferred dual frame, namely the canonical dual frame $X'$. Now regarding uniqueness: $D_X=(X')$ if and only if $X$ is a Riesz basis.

If $H$ has finite dimension, then a frame $X$ is a Riesz basis if and only if $|X|=\dim H$. So you will get infinitely many duals as soon as $|X|\geq \dim H+1$, and only the canonical dual when $|X|=\dim H$.

In general, the set $D_X$ has a natural structure of affine space via the consideration of idempotents. Here is a detailed outline.

Analysis/frame/synthesis operator: for any family $(x_j , j\in J)$, we consider the so-called analysis operator $\theta_X:H\longrightarrow \mathbb{C}^J$ sending $x$ to $((x,x_j))_{j\in J}$. By definition, $X$ is a frame if $\theta_X$ is bounded above and below from $H$ to $\ell^2(J)$. That is, $\theta_X$ is bounded and the positive so-called frame operator $\theta_X^*\theta_X$ is invertible (with spectrum in $[A,B]$, the constants of the definition). Denoting by $(e_j,j\in J)$ the canonical basis of $\ell^2(J)$, note that the adjoint (called the synthesis operator) satisfies $\theta_X^*e_j=x_j$.

A useful projection: when $X$ is a frame, the projection $p_X$ onto the range of $\theta_X$ is given by $p_X=\theta_X(\theta_X^*\theta_X)^{-1}\theta_X^*$.

Riesz basis: a frame $X$ is a Riesz basis if and only if $\theta_X^*$ is injective, if and only if $\theta_X$ is surjective, if and only if $\theta_X$ is invertible ($\theta_X^*$ yielding the equivalence between $X$ and the canonical basis of $\ell^2(J)$). That's equivalent to $p_X=1$.

Dual frame: a frame $Y$ is called a dual frame of a frame $X$ if $\theta_Y^*\theta_X=\mbox{Id}_H=1$. That is $v=\sum_j (x_j,v)y_j$ for every $v\in H$. That's equivalent to $\theta_X^*\theta_Y=\mbox{Id}_H=1$. That is $v=\sum_j (y_j,v)x_j$ for every $v\in H$.

Canonical dual fame: given a frame $X$, the canonical dual $X'$ of $X$ is given by the formula $\theta_{X'}^*=(\theta_X^*\theta_X)^{-1}\theta_X^*$. It is indeed trivial to check that $\theta_{X'}^*\theta_X=1$.

Idempotent characterization of duality: it is easy to see that, given two frames $X,Y$, they are alternate duals of each other if and only if $\theta_X\theta_Y^*$ is idempotent in $B(\ell^2(J))$.

Affine structure: given a frame $X$, the mapping $Y\longmapsto \theta_X\theta_Y^*$ is a bijection from the set $D_X$ of all duals of $X$ to the set of all idempotents $q\in B(\ell^2(J))$ with range equal to that of $\theta_X$. The latter is simply $$ D_X\simeq p_X+p_X B(\ell^2(J))(1-p_X) $$ where $p_X$ corresponds to the canonical dual $X'$. Since $p_X=0$ is impossible ($\theta_X^*p_X=\theta_X^*\neq 0$), it follows that $D_X$ is a singleton (unique dual) if and only if $p_X=1$ ($X$ Riesz basis). Otherwise, it is of course infinite and the canonical dual is optimal in the sense that it minimizes the norm of the idempotent $\theta_X\theta_Y^*$. It is indeed the only $Y$ for which the latter is $1$.


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