Frame of a vector space


Frame of a vector space

In linear algebra, a frame of a vector space V with an inner product can be seen as a generalization of the idea of a basis to sets which may be linearly dependent. The key issue related to the construction of a frame appears when we have a sequence of vectors \{\mathbf{e}_{k}\}, with each \mathbf{e}_{k} \in V and we want to express an arbitrary element \mathbf{v} \in V as a linear combination of the vectors \{\mathbf{e}_{k}\}:

 \mathbf{v} = \sum_{k} c_{k} \mathbf{e}_{k}

and want to determine the coefficients ck. If the set \{ \mathbf{e}_{k} \} does not span V, then these coefficients cannot be determined for all such \mathbf{v}. If \{ \mathbf{e}_{k} \} spans V and also is linearly independent, this set forms a basis of V, and the coefficients ck are uniquely determined by \mathbf{v}: they are the coordinates of \mathbf{v} relative to this basis. If, however, \{\mathbf{e}_{k}\} spans V but is not linearly independent, the question of how to determine the coefficients becomes less apparent, in particular if V is of infinite dimension.

Given that \{\mathbf{e}_{k}\} spans V and is linearly dependent, it may appear obvious that we should remove vectors from the set until it becomes linearly independent and forms a basis. There are some problems with this strategy:

  1. By removing vectors randomly from the set, it may lose its possibility to span V before it becomes linearly independent.
  2. Even if it is possible to devise a specific way to remove vectors from the set until it becomes a basis, this approach may become infeasible in practice if the set is large or infinite.
  3. In some applications, it may be an advantage to use more vectors than necessary to represent \mathbf{v}. This means that we want to find the coefficients ck without removing elements in \{\mathbf{e}_{k}\}.

In 1952, Duffin and Schaeffer gave a solution to this problem, by describing a condition on the set \{ \mathbf{e}_{k} \} that makes it possible to compute the coefficients ck in a simple way. More precisely, a frame is a set \{ \mathbf{e}_{k} \} of elements of V which satisfy the so-called frame condition:

There exist two real numbers, A and B such that 0 < A \leq B < \infty and
A \| \mathbf{v} \|^{2} \leq \sum_{k} |\langle \mathbf{v} | \mathbf{e}_{k} \rangle|^{2} \leq B \| \mathbf{v} \|^{2}
\text{ for all }\mathbf{v} \in V.
This means that the constants A and B can be chosen independently of v: they only depend on the set \{\mathbf{e}_{k}\}.

The numbers A and B are called lower and upper frame bounds.

It can be shown that the frame condition is both necessary and sufficient to form a frame a set of dual frame vectors \{ \mathbf{\tilde{e}}_{k} \} with the following property:


\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}

for any \mathbf{v} \in V. This implies that a frame together with its dual frame has the same properties as a basis and its dual basis in terms of reconstructing a vector from scalar products.

Contents

Relation to bases

If the set \{\mathbf{e}_{k}\} is a frame of V, it spans V. Otherwise there would exist at least one non-zero \mathbf{v} \in V which would be orthogonal to all \mathbf{e}_{k}. If we insert \mathbf{v} into the frame condition, we obtain


A \| \mathbf{v} \|^{2} \leq 0 \leq B \| \mathbf{v} \|^{2} ;

therefore A \leq 0, which is a violation of the initial assumptions on the lower frame bound.

If a set of vectors spans V, this is not a sufficient condition for calling the set a frame. As an example, consider  V = \mathbb{R}^{2} and the infinite set \{\mathbf{e}_{k}\} given by

\left\{ (1,0) , \, (0,1), \, \left(0,\frac{1}{\sqrt{2}}\right) , \, \left(0,\frac{1}{\sqrt{3}}\right), \ldots \right\}.

This set spans V but since \sum_k |\langle \mathbf{e}_k|(0,1)\rangle|^2 = 0 + 1 + \frac{1}{2} + \frac{1}{3} +\cdots = \infty we cannot choose B < \infty. Consequently, the set \{\mathbf{e}_{k}\} is not a frame.

Types of frames

Tight frames

A frame is tight if the frame bounds A and B are equal. This means that the frame obeys a generalized Parseval's identity. If A = B = 1, then a frame is either called normalized or Parseval. However, some of the literature refers to a frame for which \forall k\ \|\mathbf{e}_k\| = c where c is a constant independent of k (see uniform below) as a normalized frame.

Uniform frames

A frame is uniform if each element has the same norm: \forall k\ \|\mathbf{e}_k\| = c where c is a constant independent of k. A uniform normalized tight frame with c = 1 is an orthonormal basis.

The dual frame

The frame condition is both sufficient and necessary for allowing the construction of a dual or conjugate frame,  \{ \tilde{\mathbf{e}}_{k} \}, relative the original frame,  \{ \mathbf{e}_{k} \}. 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 \mathbf{v} \in V. In order to construct the dual frame, we first need the linear mapping: \mathbf{S} : V \rightarrow V defined as


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

From this definition of \mathbf{S} 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 \mathbf{S} can be summarised as follows:

  1. \mathbf{S} is self-adjoint, positive definite, and has positive upper and lower bounds. This leads to
  2. the inverse \mathbf{S}^{-1} of \mathbf{S} 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 \mathbf{S}^{-1}:


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

To see that this make sense, let \mathbf{v} \in V 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 \tilde{\mathbf{e}}_{k} and applying known properties of \mathbf{S} 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.

History

Frames were introduced by Duffin and Schaeffer in their study on nonharmonic Fourier series. They remained obscure until Mallat, Daubechies, and others used them to analyze wavelets in the 1980s. Some practical uses of frames today include robust coding and design and analysis of filter banks.

See also

References

  • Ole Christensen (2003). An Introduction to Frames and Riesz Bases. Birkhäuser. 
  • R. J. Duffin and A. C. Schaeffer (1952). "A class of nonharmonic Fourier series". Trans. Amer. Math. Soc. (Transactions of the American Mathematical Society, Vol. 72, No. 2) 72 (2): 341–366. doi:10.2307/1990760. JSTOR 1990760. 

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