Frame of A Vector Space - The Dual Frame

The Dual Frame

The frame condition is both sufficient and necessary for allowing the construction of a dual or conjugate frame, relative the original frame, . The duality of this frame implies that


\sum_{k} \langle \mathbf{v} | \mathbf{\tilde{e}}_{k} \rangle \mathbf{e}_{k} =
\sum_{k} \langle \mathbf{v} | \mathbf{e}_{k} \rangle \mathbf{\tilde{e}}_{k} = \mathbf{v}

is satisfied for all . In order to construct the dual frame, we first need the linear mapping: defined as


\mathbf{S} \mathbf{v} = \sum_{k} \langle \mathbf{v} | \mathbf{e}_{k} \rangle \mathbf{e}_{k}

From this definition of and linearity in the first argument of the inner product, it now follows that


\langle \mathbf{S} \mathbf{v} | \mathbf{v} \rangle =
\sum_{k} |\langle \mathbf{v} | \mathbf{e}_{k} \rangle|^{2}

which can be inserted into the frame condition to get

A \| \mathbf{v} \|^{2} \leq
\langle \mathbf{S} \mathbf{v} | \mathbf{v} \rangle \leq B \| \mathbf{v} \|^{2}
\text{ for all }\mathbf{v} \in V

The properties of can be summarised as follows:

  1. is self-adjoint, positive definite, and has positive upper and lower bounds. This leads to
  2. the inverse of exists and it, too, is self-adjoint, positive definite, and has positive upper and lower bounds.

The dual frame is defined by mapping each element of the frame with :


\tilde{\mathbf{e}}_{k} = \mathbf{S}^{-1} \mathbf{e}_{k}

To see that this make sense, let be arbitrary and set


\mathbf{u} =
\sum_{k} \langle \mathbf{v} | \mathbf{e}_{k} \rangle \tilde{\mathbf{e}}_{k}

It is then the case that


\mathbf{u} =
\sum_{k} \langle \mathbf{v} | \mathbf{e}_{k} \rangle ( \mathbf{S}^{-1} \mathbf{e}_{k} ) =
\mathbf{S}^{-1} \left ( \sum_{k} \langle \mathbf{v} | \mathbf{e}_{k} \rangle \mathbf{e}_{k} \right ) =
\mathbf{S}^{-1} \mathbf{S} \mathbf{v} = \mathbf{v}

which proves that


\mathbf{v} = \sum_{k} \langle \mathbf{v} | \mathbf{e}_{k} \rangle \tilde{\mathbf{e}}_{k}

Alternatively, we can set


\mathbf{u} = \sum_{k} \langle \mathbf{v} | \tilde{\mathbf{e}}_{k} \rangle \mathbf{e}_{k}

By inserting the above definition of and applying known properties of and its inverse, we get


\mathbf{u} =
\sum_{k} \langle \mathbf{v} | \mathbf{S}^{-1} \mathbf{e}_{k} \rangle \mathbf{e}_{k} =
\sum_{k} \langle \mathbf{S}^{-1} \mathbf{v} | \mathbf{e}_{k} \rangle \mathbf{e}_{k} =
\mathbf{S} (\mathbf{S}^{-1} \mathbf{v}) = \mathbf{v}

which shows that


\mathbf{v} = \sum_{k} \langle \mathbf{v} | \tilde{\mathbf{e}}_{k} \rangle \mathbf{e}_{k}

This derivation of the dual frame is a summary of section 3 in the article by Duffin and Schaeffer. They use the term conjugate frame for what here is called dual frame.

Read more about this topic:  Frame Of A Vector Space

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