# 0.1 A matrix times a vector

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If a transformation is made on the signal with a non-orthogonal basis system, then Parseval's theorem does not hold and the concept of energy does not move back and forth between domains. We can get around someof these restrictions by using frames rather than bases.

## Frames and tight frames

In order to look at a more general expansion system than a basis and to generalize the ideas of orthogonality and of energybeing calculated in the original expansion system or the transformed system, the concept of frame is defined. A frame decomposition or representation is generally more robust and flexible than abasis decomposition or representation but it requires more computation and memory [link] , [link] , [link] . Sometimes a frame is called a redundant basis or representing an underdetermined or underspecified set ofequations.

If a set of vectors, ${\mathbf{f}}_{\mathbf{k}}$ , span a vector space (or subspace) but are not necessarily independent nor orthogonal, bounds on the energy in the transform can still be defined. A set of vectors that span a vector space is called a frame if two constants, $A$ and $B$ exist such that

${0

and the two constants are called the frame bounds for the system. This can be written

${0

where

$\mathbf{c}=\mathbf{F}\mathbf{x}$

If the ${\mathbf{f}}_{\mathbf{k}}$ are linearly independent but not orthogonal, then the frame is a non-orthogonal basis. If the ${\mathbf{f}}_{\mathbf{k}}$ are not independent the frame is called redundant since there are more than the minimum number of expansion vectors thata basis would have. If the frame bounds are equal, $A=B$ , the system is called a tight frame and it has many of features of an orthogonal basis. If the bounds are equal to each other and to one, $A=B=1$ , then the frame is a basis and is tight. It is, therefore, an orthogonal basis.

So a frame is a generalization of a basis and a tight frame is a generalization of an orthogonal basis. If , $A=B$ , the frame is tight and we have a scaled Parseval's theorem:

${A||\mathbf{x}||}^{2}=\sum _{k}{|\left({\mathbf{f}}_{\mathbf{k}},\mathbf{x}\right)|}^{2}$

If $A=B>1$ , then the number of expansion vectors are more than needed for a basis and $A$ is a measure of the redundancy of the system (for normalized frame vectors). For example, if there are three frame vectors in a two dimensional vector space, $A=3/2$ .

A finite dimensional matrix version of the redundant case would have $\mathbf{F}$ in [link] with more columns than rows but with full row rank. For example

$\left[\begin{array}{ccc}{a}_{00}& {a}_{01}& {a}_{02}\\ {a}_{10}& {a}_{11}& {a}_{12}\end{array}\right]\left[\begin{array}{c}{x}_{0}\\ {x}_{1}\\ {x}_{2}\end{array}\right]=\left[\begin{array}{c}{b}_{0}\\ {b}_{1}\end{array}\right]$

has three frame vectors as the columns of $\mathbf{A}$ but in a two dimensional space.

The prototypical example is called the Mercedes-Benz tight frame where three frame vectors that are ${120}^{\circ }$ apart are used in a two-dimensional plane and look like the Mercedes car hood ornament. These three frame vectors must be asfar apart from each other as possible to be tight, hence the ${120}^{\circ }$ separation. But, they can be rotated any amount and remain tight [link] , [link] and, therefore, are not unique.

$\left[\begin{array}{ccc}1& -0.5& -0.5\\ 0& 0.866& -0.866\end{array}\right]\left[\begin{array}{c}{x}_{0}\\ {x}_{1}\\ {x}_{2}\end{array}\right]=\left[\begin{array}{c}{b}_{0}\\ {b}_{1}\end{array}\right]$

In the next section, we will use the pseudo-inverse of $\mathbf{A}$ to find the optimal $\mathbf{x}$ for a given $\mathbf{b}$ .

So the frame bounds $A$ and $B$ in [link] are an indication of the redundancy of the expansion system ${f}_{k}$ and to how close they are to being orthogonal or tight. Indeed, [link] is a sort of approximate Parseval's theorem [link] , [link] , [link] , [link] , [link] , [link] , [link] , [link] .

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