Frame of a vector space

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This article is about a generalization of bases to linearly dependent sets of vectors. For a linearly independent set of vectors, see k-frame.

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. 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.

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 c_{k}. If the set \{ \mathbf{e}_{k} \} does not span V, then such coefficients do not exist for every such \mathbf{v}. If \{ \mathbf{e}_{k} \} spans V and also is linearly independent, this set forms a basis of V, and the coefficients c_{k} 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, a strategy is to remove vectors from the set until it becomes linearly independent and forms a basis. There are some problems with this plan:

  1. By removing vectors arbitrarily 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 c_{k} without removing elements in \{\mathbf{e}_{k}\}. The coefficients c_{k} will no longer be uniquely determined by \mathbf{v}. Therefore, the vector \mathbf{v} can be represented as a linear combination of \{\mathbf{e}_{k}\} in more than one way.

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 c_{k} 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 respectively. Note that the frame bounds A and B are not unique because numbers less than A and greater than B are also considered frame bounds. The optimal upper bound is the infimum of all upper bounds. Likewise, the optimal lower bound is the supremum of all lower bounds. The optimal bounds are in fact frame bounds as well.

We say that a frame is overcomplete or redundant if it is not a basis. Intuitively, we can think of a frame as an "overcomplete" basis.

It can be shown that the frame condition entails the existence of a set of dual frame vectors \{ \mathbf{\tilde{e}}_{k} \} with the property that

\mathbf{v} = 
\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}

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

Relation to bases[edit]

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} with the dot product, 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[edit]

Tight frames[edit]

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

Uniform frames[edit]

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

Dual frames[edit]

The frame condition is both sufficient and necessary for allowing the construction of a dual or conjugate frame,  \{ \tilde{\mathbf{e}}_{k} \}, relative to the original frame,  \{ \mathbf{e}_{k} \}. The duality of this frame implies that

\mathbf{v} = 
\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}

is satisfied for all \mathbf{v} \in V. In order to construct a 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} .

The linear mapping \mathbf{S} is called the frame operator of the frame \{\mathbf{e}_{k}\}.

From this definition of \mathbf{S} and linearity in the first argument of the inner product, it 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. The frame operator \mathbf{S} is self-adjoint, positive definite, and has positive upper and lower bounds.
  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 makes 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} .

The numbers \langle \mathbf{v} , \tilde{\mathbf{e}}_{k} \rangle are called frame coefficients. This derivation of a 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 a dual frame.

Note that the dual frame \{\tilde{\mathbf{e}}_{k}\} is called the canonical dual of \{\mathbf{e}_{k}\} because it acts similarly as a dual to its basis.

When the frame \{\mathbf{e}_{k}\} is overcomplete, a vector \mathbf{v} can be written as a linear combination of \{\mathbf{e}_{k}\} in more than one way. That is, there exist different choice of coefficients \{c_{k}\} such that \mathbf{v} = \sum_{k} c_{k} \mathbf{e}_{k}. This allows us some freedom for the choice of coefficients \{c_{k}\} other than \langle \mathbf{v} , \tilde{\mathbf{e}}_{k} \rangle. It is necessary that the frame \{\mathbf{e}_{k}\} is overcomplete for other such coefficients \{c_{k}\} to exist. If so, then there exist frames \{\mathbf{g}_{k}\} \neq \{\tilde{\mathbf{e}}_{k}\} for which

\mathbf{v} = \sum_{k} \langle \mathbf{v} , \mathbf{g}_{k} \rangle \mathbf{e}_{k}

for all \mathbf{v} \in V. We call \{\mathbf{g}_{k}\} a dual frame of \{\mathbf{e}_{k}\}.

See also[edit]