A Physical Interpretation of Tight Frames
P. G. Casazza
1
, M. Fickus
2
, J. Kovaˇ
cevi´
c
3
, M. T. Leon
4
, and J. C. Tremain
5
1
Mathematics Department, University of Missouri, Columbia, Missouri 65211
USA,
[email protected]
2
Department of Mathematics and Statistics, Air Force Institute of Technology,
WrightPatterson AFB, Ohio 45433 USA,
[email protected]
3
Department of Biomedical Engineering, Carnegie Mellon University, Pittsburgh,
Pennsylvania 15213, USA,
[email protected]
4
Mathematics Department, University of Missouri, Columbia, Missouri 65211
USA,
[email protected]
5
Mathematics Department, University of Missouri, Columbia, Missouri 65211
USA,
[email protected]
Summary.
We find finite tight frames when the lengths of the frame elements are
predetermined. In particular, we derive a “fundamental inequality” which completely
characterizes those sequences which arise as the lengths of a tight frame’s elements.
Furthermore, using concepts from classical physics, we show that this characteriza
tion has an intuitive physical interpretation.
1 Introduction
Let
H
N
be a finite
N
dimensional Hilbert space. A finite sequence
{
f
m
}
M
m
=1
of vectors is
A
tight
for
H
N
if there exists
A
≥
0 such that,
A f
2
=
M
m
=1

f, f
m

2
,
for all
f
∈
H
N
. An
A
tight frame
is an
A
tight sequence for which
A >
0. By
polarization,
{
f
m
}
M
m
=1
is
A
tight for
H
N
if and only if,
Af
=
M
m
=1
f, f
m
f
m
,
for all
f
∈
H
N
. Clearly, any orthonormal basis is a 1tight frame. However,
the converse is false. For example, the vertices of a tetrahedron, appropriately
centered and scaled, form a 1tight frame of four elements for
R
3
. Moreover,
while the elements of an orthonormal basis are of unit length
a priori
, there are
no explicit assumptions made about the lengths of a tight frame’s elements.
This raises the question,
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2
P. G. Casazza, M. Fickus, J. Kovaˇ
cevi´
c, M. T. Leon, and J. C. Tremain
Given positive integers
M
and
N
, for what sequences of nonnegative
numbers
{
a
m
}
M
m
=1
do there exist tight frames
{
f
m
}
M
m
=1
for
H
N
, such
that
f
m
=
a
m
for all
m
?
The answer to this question is the subject of this chapter. To be more precise,
we first derive a necessary condition upon the lengths.
Proposition 1.
If
{
f
m
}
M
m
=1
is
A
tight for
H
N
then,
max
m
=1
,...,M
f
m
2
≤
A
=
1
N
M
m
=1
f
m
2
.
(1)
Proof.
First note that for any
m
= 1
, . . . , M
,
f
m
4
=

f
m
, f
m

2
≤
M
m
=1

f
m
, f
m

2
=
A f
m
2
.
Thus,
f
m
2
≤
A
for all
m
= 1
, . . . , M
, yielding the inequality in (1). For the
equality, let
{
e
n
}
N
n
=1
be an orthonormal basis for
H
N
. By Parseval’s identity,
1
N
M
m
=1
f
m
2
=
1
N
M
m
=1
N
n
=1

f
m
, e
n

2
=
1
N
N
n
=1
A e
n
2
=
A.
Thus, the lengths
{
a
m
}
M
m
=1
of a tight frame of
M
elements for an
N

dimensional space must satisfy the
fundamental inequality
,
max
m
=1
,...,M
a
2
m
≤
1
N
M
m
=1
a
2
m
.
(2)
Remarkably, this easilyfound necessary condition will prove suﬃcient as
well. That is, we shall show that for any sequence
{
a
m
}
M
m
=1
which satisfies
(2), and for any
N
dimensional Hilbert space
H
N
, there exists a tight frame
{
f
m
}
M
m
=1
for
H
N
for which
f
m
=
a
m
for all
m
.
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 Spring '10
 RunyiYu
 Linear Algebra, Potential Energy, Hilbert space, Orthonormal basis

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