On minimal scalings of scalable frames
Rachel Domagalski
Yeon Hyang Kim
Sivaram K. Narayan
Department of Mathematics
Central Michigan University
Mount Pleasant, MI 48859
Email: Domag1rj@cmich.edu
Department of Mathematics
Central Michigan University
Mount Pleasant, MI 48859
Email: kim4y@cmich.edu
Department of Mathematics
Central Michigan University
Mount Pleasant, MI 48859
Email: naray1sk@cmich.edu
Abstract—A tight frame in Rn is a redundant system which
has a reconstruction formula similar to that of an orthonormal
basis. For a unit-norm frame F = {fi }ki=1 , a scaling is a vector
c = (c(1), . . . , c(k)) ∈ Rk≥0 such that {c(i)fi }ki=1 is a tight
frame in Rn . If a scaling c exists, we say that F is a scalable
frame. A scaling c is a minimal scaling if {fi : c(i) > 0}
has no proper scalable subframes. In this paper, we present
the uniqueness of the orthogonal partitioning property of any
set of minimal scalings and provide a construction of scalable
frames by extending the standard orthonormal basis of Rn .
1. Introduction
A frame is a redundant set of vectors that span a finite
dimensional vector space. In recent years there is a new focus given to frames because they admit stable decomposition
and reconstruction algorithms. The study of frames began
in 1952 with their introduction by Duffin and Schaeffer [10]
and has since been expanded by Daubechies [9] and others
[8], [12], [13]. If {fi }ki=1 is an orthonormal basis for a finite
dimensional inner product spacePthen each vector f has a
k
unique representation as f =
i=1 hf, fi i fi . If a signal
is represented as a vector and transmitted by sending the
sequence of coefficients of its representation, then using an
orthonormal basis to analyze and later reconstruct the signal
is not always possible because the loss of any coefficient
during transmission means that the original signal cannot
be recovered. Redundancy is introduced in frames so that
it might be possible to reconstruct a signal if some of
the coefficients are lost. A tight frame is a special case
of a frame, which has a reconstruction formula similar
to that of an orthonormal basis. Because of this simple
formulation of reconstruction, tight frames are employed
in a variety of applications such as sampling, signal processing, filtering, smoothing, denoising, compression, image
processing, and in other areas. In the last couple of years
the theme of scalable frames was developed to maintain
erasure resilience or sparse expansion properties of frames
[2], [5], [14], [15]. In this paper, we further explore scalable
frames. For a frame F = {fi }ki=1 , a scaling is a vector
c = (c(1), . . . , c(k)) ∈ Rk≥0 such that {c(i)fi }ki=1 is a tight
frame in Rn . If a scaling c exists, we say that F is a scalable
c 2015 IEEE
978-1-4673-7353-1/15/$31.00 frame. We denote the scaled frame by cF . We present results
on the uniqueness of the orthogonal partitioning property
of minimal scalings and provide a construction of scalable
frames by extending the standard orthonormal basis of Rn .
2. Preliminaries
In this section we recall some basic properties of tight
frames and scalable frames in Rn . We present a few results
that will be used later in the paper. For the basic facts about
scalable frames we refer to [2], [3], [5], [14], [15].
Definition 2.1. A sequence {fi }ki=1 ⊆ Rn , is a frame for
Rn with frame bounds 0 < A ≤ B < ∞ if for all x ∈ Rn ,
A||x||2 ≤
k
X
|hx, fi i|2 ≤ B||x||2 .
(1)
i=1
Often it is useful to express frames both as sequences
as well as matrices. Therefore we abuse notation in the
following way: A frame F = {fi }ki=1 will be expressed as a
n × k matrix F whose column vectors are fi , i = 1, . . . , k .
Definition 2.2. A frame {fi }i∈I is said to be λ − tight if
λ = A = B in (1) and is said to be P arseval if A = B = 1.
A unit-norm frame is a frame such that each vector in the
frame has norm one.
We use diagram vectors to determine tight frames. In
R2 , any
vector
f can be expressed using polar coordinates,
f (1)
a cos θ
f =
=
, θ being the angle between the
f (2)
a sin θ
vector and the positive x-axis. The corresponding diagram
vector of f is
2
2
a cos 2θ
f (1) − f 2 (2)
f˜ =
= 2
.
2f (1)f (2)
a sin 2θ
Diagram vectors determine tight frames in the following
way:
Theorem 2.1 ( [13]). Let {fi }ki=1 be a sequence of vectors
k
in R2 , not all of which
Pk are˜ zero. Then {fi }i=1 is a tight
frame if and only if i=1 fi = 0.
The notion of diagram vectors was extended to vectors
in Rn in [5], and used to study properties of frames in [3],
[5], [6].


f (1)


Definition 2.3. For any vector f =  ...  ∈ Rn , we
f (n)
define the diagram vector of f , denoted as f˜, by


f 2 (1) − f 2 (2)


..


.
 2

f (n − 1) − f 2 (n)
1
˜
√

 ∈ Rn(n−1) ,
f=√
2nf (1)f (2) 
n−1




..


.
√
2nf (n − 1)f (n)
Proposition 2.3 ( [5]). Let {fi }ki=1 be a unit-norm frame
for Rn . There exist real numbers d(1), d(2), . . . , d(k) for
which {d(i)fi }ki=1 is a tight frame for Rn if and only if
there exist nonnegative numbers c(1), c(2), . . . , c(k) such
that {c(i)fi }ki=1 is a tight frame for Rn .
The proof of Proposition 2.3 follows from selecting
c(i) = |di | and Equation (2).
Lemma 2.4 ( [5]). Let {fi }ki=1 be a unit-norm frame for
Rn . If there exist nonnegative numbers c(1), c(2), . . . , c(k)
such that {c(i)fi }ki=1 is a λ-tight frame for some λ 6= 0,
then
c2 (1) + · · · + c2 (k) = λn.
Proof. If {c(i)fi }ki=1 is a λ-tight frame for Rn , then since
the frame operator for {c(i)fi }ki=1 is λIn , we have the trace
n X
k
X
where the difference of squares f 2 (i) − f 2 (j) and the product f (i)f (j) occur exactly once for i < j, i = 1, . . . , n − 1.
Theorem 2.2 ( [5]). Let {fi }ki=1 be a sequence of vectors in
Rn , not all of P
which are zero. Then {fi }ki=1 is a tight frame
k
if and only if i=1 f˜i = 0. Moreover, for any f, g ∈ Rn we
have
nhf, gi2 − ||f ||2 ||g||2 = (n − 1)hf˜, g̃i.
j=1 i=1
Since
{fi }ki=1
is a unit-norm frame, for each i,
fi2 (1) + · · · + fi2 (n) = 1,
which implies that
c2 (1) + · · · + c2 (k) = λn.
Proof. For any vectors f and g in Rn , we have
=
nhf, gi2 − kf k2 kgk2
!2
n
X
n
f (i)g(i) −
=
2n
i=1
f (i)g 2 (i) −
i=1
2n
f (i)
X
n
X
!
2
g (i)
i=1
n X
n
X
f 2 (i)g 2 (j)
i=1 j=1
Proof. Let {c(i)fi }ki=1 be a λ-tight frame. Suppose
√ there is
a unit vector f ∈ Rn such that |hf,
f
i|
≥
1/
n for all
i
√
i = 1, 2, · · · , k and |hf, fi i| > 1/ n for at least one i.
Since {c(i)fi }ki=1 is a λ-tight frame and kf k = 1, by the
assumption, we have
f (i)f (j)g(i)g(j)
1≤i<j≤n
+
X
f 2 (i) − f 2 (j)
g 2 (i) − g 2 (j)
1≤i<j≤n
=
We use Lemma 2.4 to provide a necessary condition for
tight frame scaling in Rn .
Theorem 2.5 ( [5]). Let {fi }ki=1 be a unit-norm frame for
Rn . If there exist positive numbers c(1), c(2), . . . , c(k) such
that {c(i)fi }ki=1 is a tight frame for Rn , then
√ there is no
unit vector f ∈ Rn such that√|hf, fi i| ≥ 1/ n for all i =
1, 2, · · · , k and |hf, fi i| > 1/ n for at least one i.
f (i)f (j)g(i)g(j)
1≤i<j≤n
n
X
2
=
!
2
i=1
X
+n
n
X
c2 (i)fi2 (j) = λn.
(n − 1)hf˜, g̃i.
λ=
k
X
k
c2 (i)|hf, fi i|2 >
i=1
2.1. Tight Frame Scaling in
Rn
which contradicts Lemma 2.4.
The authors characterized when a unit-norm frame
{fi }ki=1 for Rn can be scaled to a tight frame in [5]. In
this section, we present some useful results from [5], which
are related to scalable frames. First we note that for any real
number d and vector f ∈ Rn ,
e = d2 fe.
df
1X 2
c (i),
n i=1
(2)
The next proposition shows that we can always choose the
scaling constants for a unit-norm frame {fi }ki=1 ⊂ Rn to be
nonnegative numbers.
Next we use the Gramian of the associated diagram
vectors of a unit-norm frame to provide a necessary and
sufficient condition for tight frame scaling in Rn . For
{fi }ki=1 ⊂ Rn , the Gramian operator G is the k × k matrix
defined by
k
G = F ∗ F = (hfj , fi i)i,j=1 .
We note that the Gramian matrix is symmetric and positive
semidefinite.
Remark 1. Let A be a symmetric positive semidefinite
matrix and x ∈ Rn . Then xT Ax = 0 if and only if Ax = 0.
Moreover, if A is positive definite then xT Ax = 0 if and
only if x = 0.
D
Ek
e=
Let G
f˜j , f˜i
be the Gramian associated to
n i,j=1
ok
. We note that if {fi }ki=1 ⊂
the diagram vectors f˜i
i=1
n
R is a set of unit-norm vectors, then by Theorem 2.2, we
e ij = n|Gij |2 − 1. We have the following
have (n − 1)G
characterization of tight frame scaling using the Gramian
e.
G
{fi }ki=1
Proposition 2.6 ( [5]). Let
be a unit-norm frame
for Rn and c(1), · · · , c(k) be nonegative numbers. Then
{c(i)fi }ki=1 is a tight frame for Rn if and only if


c2 (1)


w =  ... 
c2 (k)
e.
belongs to the null space of G
k
n
Proof. The
Pk set {c(i)fi }i=1P⊂k R 2 is a˜ tight frame if and
only
if
i=1 (c(i)fi )˜ =
i=1
DP
E c (i)fi = 0 if and only
k
2
˜i , Pk c2 (i)f˜i = wT Gw
e = 0. From
if
c
(i)
f
i=1
i=1
e is positive semidefinite we conclude that
Remark 1, since G
e = 0.
Gw
For a given subset K in Rn , let conv(K) be the convex
hull of K . The following result on linear inequalities proves
to be useful.
Lemma 2.7 ( [7]). Let K be a compact subset of Rn . There
exists g ∈ Rn such that hg, f i > 0 for all f ∈ K if and
only if 0 ∈
/ conv(K).
A scaling c of a unit-norm frame F which as all positive
entries is called a strict scaling. We then have the main
theorem from [5] for strict tight frame scaling problem in
Rn .
Theorem 2.8 ( [5]). Let {fi }ki=1 be a unit-norm frame
D
Ek
e =
for Rn and let G
f˜j , f˜i
be the Gramian
ni,j=1
ok
e is
. Suppose G
associated to the diagram vectors f˜i
i=1
e
not invertible.
 Let { v1 , . . . , vl } be any basis for null(G)
v1 (i)


and ri :=  ...  for i = 1, . . . , k . Then there exist
vl (i)
c(i) > 0, i = 1, · · · , k so that {c(i)fi }ki=1 is a tight frame
if and only if 0 6∈ conv { r1 , . . . , rk }.
Proof. Suppose there exist c(i) > 0, i = 1, · · · , k so that
{c(i)fi }ki=1 is a tight frame. Then by Proposition 2.6, this


c2 (1)
. 
e = 0, where w = 
is equivalent to Gw
 .. . Since
c2 (k)
e ,
{ v1 , . . . , vl } is a basis for null(G)


hy, r1 i
l
X


..
w=
y(i)vi = 
,
.
i=1
hy, rk i
for some y ∈ Rl . Since hy, ri i = c2 (i) > 0 for i =
1, 2, · · · , k by considering K = { r1 , . . . , rk } in Lemma
2.7, this is equivalent to 0 6∈ conv { r1 , . . . , rk }. This
completes the proof.
There is an equivalent formulation of the tight frame
scaling problem as a matrix equation using the Gramian. Let
{fi }ki=1 be a unit-norm frame for Rn . There exist scalars
c(i) > 0, i = 1, · · · , k so that {c(i)fi }ki=1 is a tight frame
if and only if there exists a C = diag(c(1), . . . , c(k)) with
c(i) > 0 for all i so that F CC ∗ F ∗ = λI = (F C)(F C)∗ for
some λ > 0. Multiplying the equation on the left by F and
on the right by F ∗ yields F ∗ F CC ∗ F ∗ F = λF ∗ F . Since
G = F ∗ F , we have the following result.
Proposition 2.9 ( [5]). Let {fi }ki=1 be a unit-norm frame
for Rn and let G be the Gramian associated to {fi }ki=1 .
There exist scalars c(i) > 0, i = 1, · · · , k so that
{c(i)fi }ki=1 is a λ-tight frame if and only if there exist a
D := diag(d1 , . . . , dk ) with di > 0 for all i = 1, . . . , k so
that GDG = λG.
Using Theorem 2.8 we can explicitly determine when
there exists a solution to the strict tight frame scaling
problem. In particular, there exist algorithms to determine
the convex hull of a given set of points.
3. Scalability polytope and minimal scalings
In this section, we establish results on the structure of
the scalability polytope and some properties of minimal
scalings. To begin with, we present some definitions.
Let c = (c(1), . . . , c(k)) ∈ Rk . The support of c,
denoted by supp(c), is defined as {i : c(i) 6= 0}.
Definition 3.1. Let F = {fi }ki=1 be a frame in Rn . Define
P as
(
)
k
X
2
∗
P := (c(1), . . . , c(k)) : c(i) ≥ 0,
c (i)fi fi = In .
i=1
P is the set of all scalings and is called the scalability
polytope of F .
As stated in the previous definition, it is known that P
is a polytope. We have the following result about P :
Theorem 3.1 ( [2]). Let F = {fi }ki=1 be a frame in Rn
and P its scalability polytope. Then P is the convex hull of
the minimal scalings of F .
with ∅ ( J 0 ( J} .
3.1. Properties of the minimal scalings
A scaling of a unit-norm frame F is prime if the scaled
frame cF does not contain any proper, tight subframes and
non-prime otherwise. The following theorem was proved in
[3].
Theorem 3.2. [3] A scaling is non-prime if and only if it
is a convex combination of minimal scalings which can be
partitioned into two orthogonal subsets.
This motivated the consideration of orthogonal partitions
of subsets of minimal scalings. In this section we present
some of the results. We define the smallest orthogonal
partition of minimal scalings {vi }i∈I to be a partition
{ {vj }j∈J1 , . . . , {vj }j∈Ja }
We now state the theorem about unique orthogonal
partitioning property.
Theorem 3.5. Let {vi }i∈I be the set of minimal scalings
of a scalable frame F . Any scaling c can be orthogonally
partitioned as
X
X
c=
αj vj + . . . +
αj vj ,
(5)
j∈J1
j∈Ja
where { ∪j∈Ji supp(vj ) : i = 1, . . . , a } is a collection of
pairwise disjoint subsets of EC(cF ). If EC(cF ) is pairwise
disjoint, then {vj }j∈J1 ∪ . . . ∪ {vj }j∈Ja is the smallest orthogonal partition of {vi }i∈J1 ∪...∪Ja so that the orthogonal
decomposition in (5) is unique.
such that each subsets are mutually orthogonal (i.e.,
hvi , vj i = 0 if i ∈ Jk , j ∈ Jl , and j 6= k ) and each subset
cannot be further partitioned into orthogonal subsets.
We note the orthogonal decomposition given in (5) for
EC(cF ) is not unique in general.
Theorem 3.3. Let {vi }i∈I be the set of minimal scalings of
a scalable frame. The smallest orthogonal partition of any
subset of {vi }i∈I is unique.
3.2. Construction of a scalable frame by extending
the standard orthonormal basis
Proof. Let J ⊂ I . Suppose {vj }j∈J can be partitioned into
orthogonal subsets in two ways:
{vj }j∈J = {vj }j∈J1 ∪ . . . ∪ {vj }j∈Ja
(3)
= {vj }j∈K1 ∪ . . . ∪ {vj }j∈Kb ,
(4)
In this section, we provide a method to construct a tight
frame from the standard orthogonal basis vectors in Rn , by
adding some vectors and specifying the direction of one of
the vectors. In [14], the authors showed that it is not possible
to have such a tight frame with k = n + 1.
where each subset cannot be further partitioned into orthogonal subsets. It is enough to show that J1 ∩ K1 6= ∅ implies
J1 = K1 . Suppose that J1 6= K1 . Without loss of generality,
we assume that J1 \ K1 6= ∅, which implies that
Proposition 3.6 ( [14]). Let F = {fi }ki=1 ⊆ Rn be a frame
with no zero vectors and fi 6= ±fj for i 6= j . If {fj }j∈J
is strictly scalable with |J| = n, then {fj }j∈J∪{i} is not
strictly scalable for any i ∈ {1, . . . , k}.
J1 = (J1 \ K1 ) ∪ (J1 ∩ K1 ) .
This proposition states that if F = {f1 , . . . , fn , fn+1 }
is strictly scalable, the fn+1 must be collinear with a vector
in the orthonormal basis. Thus we need more conditions to
construct a tight frame containing the standard orthonormal
basis and a specified vector.
This is a contradiction to the assumption that J1 cannot
be partitioned further into orthogonal subsets. Since the
supports of the two partition in (3) are same, it is clear
that a = b.
Corollary 3.4. Let {vi }i∈I be the set of minimal scalings
of a scalable frame. If J ⊂ K ⊂ I and ∪i∈J supp(vi ) =
∪i∈K supp(vi ), then {vi }i∈J and {vi }i∈K have the same
smallest orthogonal partitions.
In the next result, we have a unique orthogonal partitioning property of any subset of the minimal scalings. In
order to state this result we need the notation of an empty
cover of the factor poset of a frame found in [1], [3]
Definition 3.2. Let F = { fi }i∈I be a finite frame in Rn .
We define its factor poset F(F ) ⊂ 2I to be the set
n
o
F(F ) = J ⊂ I : { fj }j∈J is a tight frame for Rn
partially ordered by inclusion. We assume ∅ ∈ F(F ). We
define the empty cover of F(F ), EC(F ), to be the set of
J ∈ F(F ) which covers ∅, that is,
EC(F ) = {J ∈ F(F ) : J 6= ∅ and @J 0 ∈ F(F )
Theorem 3.7. Let V = {v1 , . . . , vl } ⊆ R2 with |V | ≥ 2.
Let F = {e1 , e2 } ∪ V . Then F can be strictly scaled to
a tight frame when at least one vector, vi is in the I or
III quadrant (non-collinear with e1 or e2 ) and at least one
vector, vj , is in the II or IV quadrant (non-collinear with
e1 or e2 ), or all vectors are collinear with the orthonormal
basis.
In Rn , we first investigate when F := { e1 , . . . en , f, g }
is scalable, where the vector f has only two non zero
entries. The first condition is that the vector g must also have
only two non zero elements in the same entries as f since
{ e1 , . . . en } is scalable. Thus, without loss of generality, we
assume that
f = (a, b, 0, . . . , 0),
a2 + b2 = 1,
g = (s, t, 0, . . . , 0),
s2 + t2 = 1.
Then the diagram vector matrix of F is

1 −1 0
0 ···
1 0 −1 0 · · ·
.
..
..
..
..
.
.
.
.
.
.

0
0 ···
1 0

0 1 −1 0 · · ·
.
..
..
..
..
.
.
.
.
.
.
F̃ = 
0
0 ···
0 1
0 0
1 −1 · · ·

.
..
..
..
..
 ..
.
.
.
.

0 0
0
0
·
·
·

0 0
0
0 ···

..
..
..
..
.
.
.
.
···
0
0
..
.
a2 − b2
a2
..
.
−1
0
..
.
a2
b2
..
.
−1
0
..
.
b2
0
..
.
0
0
..
.
√

s2 − t2
2
s 
.. 

. 

s2 

t2 
.. 

. 
.
2
t

0 

.. 

√ . 
2nst 

0 

..
.
2nab
0
..
.
Thus if c is a scaling for the frame F , using Theorem 2.2
we have that
2
2
2
2
2
c (1) − c (2) + c (n + 1)(a − b )
+c2 (n + 2)(s2 − t2 ) = 0
c2 (1) − c2 (3) + c2 (n + 1)a2 + c2 (n + 2)s2 = 0
..
.
2
2
2
2
2
2
c (1) − c (n) + c (n + 1)a + c (n + 2)s = 0
c2 (2) − c2 (3) + c2 (n + 1)b2 + c2 (n + 2)t2 = 0
..
.
2
2
2
2
2
2
c (2) − c (n) + c (n + 1)b + c (n + 2)t = 0
√
√
c2 (n + 1) 2nab + c2 (n + 2) 2nst = 0
for Rn by adding f and g which have similar structure as
described in Theorem 3.8. We also note that Theorem 3.8
is related to Naimark Complement discussed in [4].
Acknowledgments
The authors thank the anonymous referees for their
valuable comments and suggestions. Results presented in
Section 3 are based on joint work with REU students Hong
Suh, and Xingyu Zhang, along with the graduate student
mentor Alice Chan [11].
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The choice of
2
t
2
a (at − bs)c (n + 2),
s
− b (at − bs)c2 (n + 2),
2
c (1) =
c2 (2) =
c2 (3) = . . . = c (n)
st
= 1 − ab
c2 (n + 2),
st 2
2
c (n + 1) = − ab c (n + 2)
[8] O. Christensen. Frames and bases: an introductory course. Birkhäuser
Boston, 2008.
[9] I. Daubechies. Ten lectures on wavelets. CBMS Series, SIAM, 1992.
[10] R.J. Duffin and A.C. Shaeffer. A class of nonharmonic Fourier series.
Proc. Amer. Math. Soc., 72:341-366, 1952.
[11] R. Domagalski, H. Suh, and X. Zhang. Tight Frame Structure and
Scalability. Central Michigan University 2014 Final Reports, 2014.
ensure that c is a scaling for F . A consequence is the
following result.
[12] C. Heil. A basis theory primer: Expanded edition (applied and
numerical harmonic analysis). Birkhäuser Boston, 2010.
Theorem 3.8. Let { e1 , . . . en } be the standard orthonormal
basis in Rn with n ≥ 2. Let f and g be two unit-norm
vectors in Rn . Then { e1 , . . . en , f, g } is scalable if f and g
have only two nonzero elements in the same entries, and the
product of the two non zero elements in f has the opposite
sign of the product of the two non zero elements in g .
[13] D. Han, K. Kornelson, D. Larson, and E. Weber. Frames for undergraduates. Student Mathematical Library, 40, American Mathematical
Society, Providence, RI, 2007.
Corollary 3.9. Let F = {e1 . . . , en , f, g} ⊆ Rn and a, b >
0 such that a2 + b2 = 1. If f = (a, b, 0, . . . , 0) and f =
(a, −b, 0, . . . , 0), then F is strictly scalable
√ with scalings
√
(c(1), . . . , c(n), 1, 1) such that c(1)
=
2b, c(2) = 2a,
√
and c(3) = c(4) = ... = c(n) = 2.
Remark 2. For n ≥ 2, using similar techniques,
{ e1 , . . . en−1 } in Rn can be extended to a tight frame
[14] G. Kutyniok, K. Okoudjou, and F. Philipp. Scalable frames and
convex geometry. Contemp. Math., 345, 2013.
[15] G. Kutyniok, K. A. Okoudjou, F. Philipp, and E. K. Tuley. Scalable
frames. Linear Algebra and its Applications, 438:2225-2238, 2013.