The document discusses equiangular tight frames (ETFs) in the context of finite dimensional real or complex spaces, providing definitions, properties, and theorems related to frames, dual frames, and various types of frame operators. Key concepts include frame analysis and synthesis, tight frames, coherence, and grassmannian frames, along with their mathematical foundations. The text lists original results and theorems, focusing on the implications of frame redundancy and the reconstruction of vectors from frame coefficients.

1. RESULTS ON EQUIANGULAR TIGHT FRAMES
SHAILESH KUMAR
Contents
1. Introduction 1
1.1. Unit norm frame 4
1.2. Dual frame and pseudo inverse 6
1.3. Analysis and Synthesis 7
1.4. Tight frames 8
1.5. Coherence and Grassmannian frames 9
1.6. Equiangular tight frames 9
2. More Properties of Frames and Dictionaries 9
3. Vandermonde Matrices 11
4. Grassmannian frames 11
4.1. Existence of Grassmannian frames 11
4.2. Optimal Grassmannian frames 12
5. Full Spark Frames 17
5.1. Vandermonde matrices 17
6. Subspace angles 17
6.1. Principal angles 18
6.2. One dimensional subspaces 19
6.3. Optimal Grassmannian Frame 20
Appendix A. Linear Algebra 21
References 21
Index 22
1. Introduction
We will be concerned with finite dimensional real or complex spaces RM
or CM
.
Most of the results will be specified in terms of field F which is either R or C.
Thus, FM
means either RM
or CM
. Some results will be specific to either RM
or
CM
. A general introduction to frame theory can be found in [3]. While [3] builds
the theory over general Hilbert spaces, our treatment will be restricted to finite
dimensional spaces.
In the sequel the results marked with are potentially original. Many of them
are very simple. It is possible that some of them might have appeared in some prior
work in some other context. The new results are listed here for quick reference
• theorem 21
• theorem 31
• theorem 35
1

2. 2 SHAILESH KUMAR
• theorem 36
Definition 1 A finite frame ΦN
M in a real or complex M-dimensional Hilbert
space FM
is a sequence of N ≥ M vectors {φk}N
k=1, φk ∈ FM
, satisfying the
following condition
α x 2
2 ≤
N
k=1
| x, φk |2
≤ β x 2
2, ∀ x ∈ FM
(1.1)
where α, β are positive constants 0 < α ≤ β called the lower and upper frame
bounds.
In this work, we will only work with finite dimensional spaces and finite frames.
Thus, whenever we say a frame, we mean a finite frame.
Definition 2 The M × N matrix Φ = φ1 . . . φN with the frame vectors
as the columns is known as the frame synthesis operator.
In the sequel we will use the frame symbol ΦM
N and its synthesis operator Φ inter-
changeably.
Definition 3 The frame analysis operator is a mapping from FM
to FN
defined as
y ΦH
x. (1.2)
yi = x, φi . The entries in y are known as the frame coefficients which are
the inner product of the frame vectors with x.
Clearly
y 2
2 =
N
k=1
| x, φk |2
= ΦH
x 2
2 = xH
ΦΦH
x. (1.3)
We can rewrite (1.1) as
α x 2
2 ≤ ΦH
x 2
2 ≤ β x 2
2, ∀ x ∈ FM
(1.4)
Definition 4 The frame operator S associated with the frame {φk}N
k=1 is
defined by
Sx
N
k=1
x, φk φk. (1.5)
We note that
N
k=1
x, φk φk = ΦΦH
x. (1.6)

3. ETF 3
Thus,
S = ΦΦH
∈ FM×M
(1.7)
Also
xH
ΦΦH
x = xH
Sx. (1.8)
Theorem 1 A frame spans FM
.
If the frame didn’t span FM
, then the lower frame bound would be 0. This means
that both Φ and S are full rank.
Theorem 2 The frame operator is positive definite.
Proof. We note that the frame bounds can be rewritten as
α x 2
2 ≤ xH
Sx ≤ β x 2
2, ∀ x ∈ FM
. (1.9)
Since α > 0, hence xH
Sx > 0 ∀ x = 0.
Theorem 3 The frame analysis operator is injective.
Proof. Let x1 and x2 be two vectors in FN
such that ΦH
x1 = ΦH
x2. Then ΦH
(x1 −
x2) = 0. Thus, ΦH
(x1 − x2) 2
2 = 0. This can happen only if x1 − x2 = 0 and the
two vectors are identical. This shows that ΦH
is injective.
Corollary 4 The frame analysis operator is bijective between FM
and its range
given by R(ΦH
).
Theorem 5 The upper and lower frame bounds of a frame are given by the
largest and smallest eigen values of the frame operator S = ΦΦH
respectively.
Alternatively they are the largest and smallest (non-zero) singular values of Φ.
Proof. This can be seen from the frame bounds inequality (1.9).
Definition 5 The operator G = ΦH
Φ is called the Gram matrix associated
with Φ. This is also known as the Grammian.
The (i, j)-th entry in G is the inner product of φi and φj. The (i, i)-th entry is
obviously the norm-squared of i-th frame vector given by φi
2
2 = φH
i φi.
Theorem 6 The non-zero eigen values of the frame operator S and the Gram
matrix G are identical.

4. 4 SHAILESH KUMAR
Proof. Let λ be a (non-zero) eigen value of S = ΦΦH
. Then, there exists x = 0
such that
ΦΦH
x = λx.
Multiplying both sides by ΦH
and defining y = ΦH
x, we obtain
ΦH
Φy = λy.
Since x = 0 =⇒ y = 0, hence y is an eigen vector of G with the same eigen
value.
We note that
ΦH
x, y = x, Φy . (1.10)
Theorem 7 The rank of G = ΦH
Φ is M. Exactly M eigen values of G are
positive. Rest are zero.
Proof. Let x ∈ N(Φ). Then
Φx = 0 =⇒ ΦH
Φx = 0 =⇒ N(Φ) ⊆ N(ΦH
Φ).
Let x ∈ N(ΦH
Φ). Then,
ΦH
Φx = 0 =⇒ xH
ΦH
Φx = 0 =⇒ Φx 2
2 = 0 =⇒ x ∈ N(Φ).
Thus N(Φ) = N(ΦH
Φ). Since rank of Φ is M, hence rank(G) = M. Thus exactly
M of eigen values of G are positive.
Definition 6 The redundancy of a frame is defined as ρ = N
M .
An orthonormal basis (ONB) for FM
has exactly M vectors. The redundancy of a
frame tells us about relative number of extra vectors in a frame viz. an ONB.
Definition 7 When the frame vectors {φk}N
k=1 are linearly independent, then
the frame is called a Riesz basis.
Theorem 8 If a finite frame is a Riesz basis, then M = N.
This is obvious since any set of more than M vectors in FM
is linearly dependent
and any set of less than M vectors cannot span FM
.
1.1. Unit norm frame.
Definition 8 A frame is called unit norm frame (UNF) if each frame vector
satisfies φk 2 = 1.
A unit norm frame is also known as a dictionary.

5. ETF 5
The diagonal entries of the Grammian for a unit norm frame are 1. Naturally
for UNF, the trace of the Grammian is given by
tr(G) = N. (1.11)
Theorem 9 The redundancy of a unit norm frame is bounded by
α ≤
N
M
≤ β. (1.12)
Proof. Recall that S = ΦΦH
∈ FM×M
. There are exactly M eigen values of S. All
of these are positive since S is positive definite. Thus tr(S) =
M
k=1 λi is bounded
by
αM = λminM ≤
M
k=1
λi = tr(ΦΦH
) ≤ λmaxM = βM.
But
tr(ΦΦH
) = tr(ΦH
Φ) = tr(G) = N.
We get
αM ≤ N ≤ βM.
Dividing by M, we obtain the result.
Theorem 10 [1] For a UNF Φ the sum of non-zero eigen values of G = ΦH
Φ
is N. Let the eigen values of G be ordered in by decreasing size as
λ1 ≥ · · · ≥ λN .
Then
M
k=1
λk = N. (1.13)
Further, the sum of square of eigen values is bounded by
M
k=1
λ2
k ≥
N2
M
. (1.14)
The lower bound is achieved only if λk = N
M for all M positive eigen values.
Proof. Recall that rank(G) = M. Thus, only M of these eigen values are non-zero.
We have
M
k=1
λk = tr(G) =
N
k=1
| φk, φk | =
N
k=1
1 = N.
Now set, ek = λk − N
M . Then
M
k=1
ek =
M
k=1
λk −
N
M
= N − M
N
M
= 0.

6. 6 SHAILESH KUMAR
Thus,
M
k=1
λ2
k =
M
k=1
N
M
+ ek
2
=
M
k=1
N2
M2
+
2N
M
M
k=1
ek +
M
k=1
e2
k
=
N2
M
+
M
k=1
e2
k ≥
N2
M
.
The lower bound is achieved only if ek = 0, thus λk = N
M for k = 1, . . . , M.
1.2. Dual frame and pseudo inverse. The reconstruction of x from its frame
coefficients ΦH
x is calculated with a pseudo-inverse of ΦH
.
Theorem 11 If {φk}N
k=1 is a frame but not a Riesz basis, then the correspond-
ing frame analysis operator ΦH
admits an infinite number of left inverses.
Proof. We know that N(Φ) = R(ΦH
)⊥
. If {φk}N
k=1 is a frame but not a Riesz basis,
then {φk}N
k=1 is linearly dependent. Thus, there exists an a ∈ N(Φ) with a = 0.
Since the restriction of ΦH
to R(ΦH
) is bijective thus invertible, hence ΦH
admits
a left inverse. Since the restriction of the left inverse to R(ΦH
)⊥
may be arbitrary,
hence there is an infinite number of left inverses.
The Moore-Penrose pseudo-inverse is defined as the left inverse that is zero on
R(ΦH
)⊥
.
Theorem 12 If {φk}N
k=1 is a frame, then S = ΦΦH
is invertible and the
pseudo inverse satisfies
ΦH†
= (ΦΦH
)−1
Φ. (1.15)
Alternatively, the right pseudo-inverse for the frame synthesis operator Φ is
given by
Φ†
= ΦH
(ΦΦH
)−1
. (1.16)
Proof. Since S is positive definite, it is invertible. Next, we see that
ΦH†
ΦH
x = (ΦΦH
)−1
ΦΦH
x = x. (1.17)
Thus, ΦH†
is a left inverse. For any a ∈ R(ΦH
)⊥
= N(Φ), we have Φa = 0. Thus,
ΦH†
a = (ΦΦH
)−1
Φa = 0.
Thus, it is the pseudo inverse.
Similarly, we note that
ΦΦ†
= ΦΦH
(ΦΦH
)−1
= I. (1.18)
Thus, Φ†
is the right pseudo-inverse of Φ.

7. ETF 7
Definition 9 Let {φk}N
k=1 be a frame. The vectors of a dual frame are
defined by
φk = (ΦΦH
)−1
φk, 1 ≤ k ≤ N. (1.19)
The dual frame consists of {φk}N
k=1. The dual frame synthesis operator
is defined by
Φ φ1 . . . φN = (ΦΦH
)−1
Φ = S−1
Φ. (1.20)
The dual frame analysis operator is given by ΦH
.
Theorem 13 The dual frame operator S = S−1
.
Proof.
ΦΦH
= (ΦΦH
)−1
Φ (ΦΦH
)−1
Φ
H
= (ΦΦH
)−1
ΦΦH
(ΦΦH
)−1
= (ΦΦH
)−1
.
Thus S = S−1
.
Theorem 14 Let 0 < α ≤ β be the frame bounds of Φ. Then, the frame
bounds of Φ are given by 1
β and 1
α . Specifically,
1
β
x 2
2 ≤
N
k=1
| x, φk |2
≤
1
α
x 2
2, ∀ x ∈ FM
(1.21)
Proof. Note that
N
k=1
| x, φk |2
= ΦH
x 2
2 = xH
Sx = xH
S−1
x.
Since S is positive definite, hence S−1
is positive definite too. The eigen values of
S−1
are the inverse of eigen values of S. Thus if α is the minimum eigen value of S
then 1
α is the maximum eigen value of S−1
. Similarly for β. The result follows.
1.3. Analysis and Synthesis.
Theorem 15 We have
x = ΦΦH
x = ΦΦH
x. (1.22)
This can also be expressed as
x =
N
k=1
x, φk φk =
N
k=1
x, φk φk. (1.23)
Proof.
ΦΦH
x = (ΦΦH
)−1
ΦΦH
x = x.

8. 8 SHAILESH KUMAR
Similarly
ΦΦH
x = ΦΦH
(ΦΦH
)−1
x = x.
1.4. Tight frames.
Definition 10 When the frame bounds α and β are equal, a frame is said to
be tight.
Definition 11 A tight frame is called unit norm tight frame (UNTF) if
each frame vector satisfies φk 2 = 1.
Theorem 16 The redundancy of a unit norm tight frame(UNTF) is given by
α = β =
N
M
. (1.24)
We can rewrite (1.1) as
N
M
x 2
2 = ΦH
x 2
2 ∀ x ∈ FM
. (1.25)
Note that α and β are upper and lower bounds of the eigen values of S = ΦΦH
.
In a tight frame, they are same, thus all eigen values are identical.
Theorem 17 Let Φ be a tight frame, then S = αI where α is the one and only
eigen value of S.
Proof. Let S = V ΛV H
be the eigen value decomposition of S. Since α = β, hence
all eigen values of S are equal. Thus Λ = αI. Thus S = αV IV H
= αI.
It is obvious that Φ = S−1
Φ = 1
α Φ. Thus, the reconstruction formula looks like
x =
N
k=1
x, φk φk =
1
α
N
k=1
x, φk φk. (1.26)
Compare this with the reconstruction formula for an orthonormal basis. Except
for an extra factor of 1
α which is the inverse of the frame redundancy factor, the
reconstruction formula is identical to ONB.
Theorem 18 Rows of the frame synthesis operator Φ of a tight frame ΦN
M are
orthogonal. The squared norm of each row is α. Specifically, the norm of the
rows of a UNTF is N
M .

9. ETF 9
Proof. We have S = ΦΦH
= αI. Clearly the inner product of distinct rows is 0
and squared norm of each row is α.
1.5. Coherence and Grassmannian frames. A good introduction to Grassman-
nian frames can be found in [4].
Definition 12 Let {φk}N
k=1 be a unit norm frame with the frame synthesis
operator Φ. The coherence or the maximum correlation of the frame is
defined as
µ(Φ) max
k=l
| φk, φl |. (1.27)
Whenever the frame is obvious from the context, we may denote the coherence
simply by µ.
Definition 13 Fix N ≥ M. Let Ω denote the set of all unit norm frames over
FM
consisting of N vectors. A frame Φ ∈ Ω is called a Grassmannian frame
if
µ(Φ) = inf
X∈Ω
{µ(X)} (1.28)
i.e. the coherence of Φ is the smallest possible over all frames in Ω.
One can see that the idea of Grassmannian frames follows the mini-max principle
(minimize the maximum correlation).
1.6. Equiangular tight frames. In orthonormal bases, any two basis vectors are
orthogonal to each other. This notion is generalized in frames through the idea of
equiangular tight frames.
Definition 14 A unit norm tight frame {φk}N
k=1 is called equiangular if
| φk, φl | = p whenever k = l (1.29)
i.e. the angle between the lines spanned by any two frame vectors is same.
In other words, the absolute value of the off-diagonal entries in the Gram matrix
is also the same. We note that 0 ≤ p < 1 where 0 is achieved only when Φ is an
orthonormal basis. Further µ(Φ) = p. We also note that
| φk, φl | = | φk, −φl |.
Thus, a sign change in frame vectors doesn’t make any difference.
2. More Properties of Frames and Dictionaries
We have already defined the coherence or maximum correlation of a dictionary.

10. 10 SHAILESH KUMAR
Definition 15 The spark of a frame is the size of the smallest linearly de-
pendent subset of frame vectors.
spark(Φ) = min{ x 0 : Φx = 0, x = 0}. (2.1)
Definition 16 Let {φk}N
k=1 be a unit norm frame with the frame synthesis
operator Φ. The minimum correlation of the frame is defined as
µmin(Φ) min
k=l
| φk, φl |. (2.2)
Theorem 19 The coherence of a frame ΦN
M is invariant to unitary transfor-
mation.
Proof. Let Ψ be an orthonormal basis for FM
. Let ΨΦ be the frame constructed
by the orthonormal transformation. Then
Ψφk, Ψφl = φk, ΨH
Ψφl = φk, φl
since the inner products are invariant to unitary transformation. It is obvious that
max
k=l
| Ψφk, Ψφl | = max
k=l
| φk, φl |.
Theorem 20 The Grammian of a frame ΦN
M is invariant to unitary transfor-
mation.
Proof. Let Ψ be an orthonormal basis for FM
. Let ΨΦ be the frame constructed
by the orthonormal transformation. Then
(ΨΦ)H
(ΨΦ) = ΦH
ΨH
ΨΦ = ΦH
Φ.
This result is simply due to the invariance of inner products over orthonormal
transformation.
Theorem 21 Let ΦN
M be a frame in FM
. Let Γ be a diagonal matrix whose
diagonal entries are chosen from {1, −1} when F = R or {eiθ
: 0 ≤ θ < 2π}
when F = C. Let α, β be frame bounds of Φ. Then α, β are also frame bounds
of ΦΓ. Moreover:
(1) If Φ is a unit norm frame, then so is ΦΓ.
(2) If Φ is a tight frame, then so is ΦΓ.
(3) If Φ is a unit norm tight frame, then so is ΦΓ.
(4) If Φ is an equiangular tight frame then so is ΦΓ.

11. ETF 11
Proof. Let the vectors in ΦΓ be γkφk where γk is drawn from {1, −1} or {eiθ
: 0 ≤
θ < 2π}. Clearly |γk| = 1.
Now,
(ΦΓ)H
x 2
2 =
N
k=1
| x, γkφk |2
=
N
k=1
|γk|2
| x, φk |2
=
N
k=1
| x, φk |2
= ΦH
x 2
2.
Thus, the frame bounds are identical.
For other results we note that:
(1) If φk is unit norm vector then so is γkφk.
(2) Since frame bounds are same, hence tight frame property is preserved.
(3) Application of the above two statements.
(4) The absolute value of the inner product is identical:
| φk, φl | = | γkφk, γlφl |.
Hence equiangular property is preserved also.
3. Vandermonde Matrices
A Vandermonde matrix has the following form.
V =





1 1 . . . 1
γ1 γ2 . . . γN
...
...
...
...
γM−1
1 γM−1
2 . . . γM−1
N





(3.1)
A square Vandermonde matrix (N = M) has the following determinant:
det(V ) =
1≤i<j≤M
(γi − γj). (3.2)
When N ≥ M, then every M × M submatrix of V is also Vandermonde and the
submatrices are non-singular as long as the bases {γk}N
k=1 are distinct.
Theorem 22 The only Vandermonde matrices that are equal norm and tight
have bases in the complex unit circle. Among these, the frames with the small-
est worst case coherence have bases that are equally spaced in the complex unit
circle, provided N ≥ 2M.
Proof. Since the matrix is a tight frame, hence it is full rank. Its rows have equal
norms.
4. Grassmannian frames
4.1. Existence of Grassmannian frames. Let SM−1
= {x ∈ FM
: x 2 =
1}denote the unit sphere in FM
. A unit norm frame is constructed by picking up
vectors from SM−1
.

12. 12 SHAILESH KUMAR
We demonstrate that maximum correlation of a sequence of vectors is a contin-
uous function. We define the function
f :FM
× · · · × FM
→ [0, 1]
f(x1, . . . , xN ) = µ({xk}N
k=1).
Define a norm over X = FM
× · · · × FM
(N times) given by
{xk}N
k=1 X
=
N
k=1
xk 2.
It is easy to check that the norm so defined satisfies all properties of a norm.
Theorem 23 [1] The maximum correlation function from X to [0, 1] is con-
tinuous.
Proof. We only provide the proof outline here. Fix any point {xk}N
k=1 in X. We
need to show that for any > 0 there exists a δ > 0 such that |µ({xk}N
k=1) −
µ({yk}N
k=1)| < whenever {xk}N
k=1 − {yk}N
k=1 X < δ.
Theorem 24 for N ≥ M, a Grassmannian frame exists.
Proof. Note that SM−1
is a compact subset of FM
. This implies S = {SM−1
×
· · · × SM−1
} is a compact subset of X. Since, µ is continuous over X, hence it is
bounded over S and a minimum as well maximum value exists. Thus, unit norm
sequences {xk}N
k=1 exist in S with minimum coherence. What is left to show that
they also span FM
. If {xk}N
k=1 doesn’t span FM
, then we can easily replace one of
the vectors in the sequence with a unit vector from the orthogonal complement of
{xk}N
k=1 in FM
. This process doesn’t increase coherence. We can keep doing it till
the sequence spans FM
. Thus there are indeed some sequences which satisfy all the
criteria for a frame with minimum coherence. This completes the proof.
Theorem 25 Let ΦN
M be a Grassmannian frame and let Ψ be an orthonormal
basis (unitary matrix) for FM
. Then, ΨΦ is also a Grassmannian frame.
Proof. Since (due to theorem 19) coherence is invariant to unitary transformation,
hence ΦN
M and ΨΦN
M have identical coherence. Thus, ΨΦN
M is also a Grassmannian
frame.
4.2. Optimal Grassmannian frames.

13. ETF 13
Theorem 26 Let ΦN
M with N ≥ M be a unit norm frame. Then
µ(Φ) ≥
N − M
M(N − 1)
. (4.1)
The equality holds if and only if Φ is equiangular tight frame with frame bounds
α = β = N
M .
(4.1) gives a lower bound to the coherence of any unit norm frame. Thus, a Grass-
mannian frame can achieve this bound if and only if it is an equiangular tight frame
with the specified frame bound. Such a Grassmannian frame is known as optimal
Grassmannian frame.
Proof. Let Φ be a Grassmannian frame and G = ΦH
Φ ∈ CN×N
be its Grassman-
nian. G is symmetric and positive semi-definite. Its eigen-values are non-negative.
Let the eigen values of G be ordered in by decreasing size as
λ1 ≥ · · · ≥ λN .
From theorem 10, we have
M
k=1 λk = N and
M
k=1 λ2
k ≥ N2
M . The lower bound is
achieved only if λk = N
M for all M positive eigen values. Let G = g1 . . . gN
in terms of its column vectors. Consider the matrix G2
. Its eigen values are
λ2
1 ≥ · · · ≥ λ2
N . Since G2
= GH
G. Thus,
M
k=1
λ2
k = tr(G2
) =
N
k=1
gH
k gk.
But
gH
k gk = gk
2
2
N
l=1
| φk, φl |2
.
We get
N2
M
≤
M
k=1
λ2
k =
N
k=1
N
l=1
| φk, φl |2
. (4.2)
Since G is Hermitian, hence | φk, φl | = | φl, φk |. We can rewrite
N2
M
≤
k=l
| φk, φl |2
+
k<l
| φk, φl |2
+
k>l
| φk, φl |2
= N + 2
k<l
| φk, φl |2
≤ N + 2
N(N − 1)
2
max
k=l
| φk, φl |2
.
Rearranging this, we obtain
µ2
(Φ) ≥
N − M
M(N − 1)
.

14. 14 SHAILESH KUMAR
We now show that equality holds only if Φ is an equiangular tight frame. We note
that µ2
(Φ) = N−M
M(N−1) forces that
N2
M
=
M
k=1
λ2
k =
N
k=1
N
l=1
| φk, φl |2
.
This can happen only if λk = N
M for k = 1, . . . , M. Immediately, since the largest
and smallest eigen values of the frame operator ΦΦH
are equal, hence Φ is a tight
frame. To see that Φ is also equiangular, we proceed as follows. Recall from above
that
N + 2
k<l
| φk, φl |2
=
N2
M
when Φ is tight. Hence
k<l
| φk, φl |2
=
N(N − M)
2M
.
Since we are assuming that max
k=l
| φk, φl |2
= N−M
M(N−1) , hence for any k = l,
| φk, φl |2
=
N − M
M(N − 1)
− εk,l
where εk,l ≥ 0. Summing over k < l, we get
N(N − M)
2M
=
k<l
N − M
M(N − 1)
− εk,l
=
N(N − 1)
2
N − M
M(N − 1)
−
k<l
εk,l
=
N(N − M)
2M
−
k<l
εk,l.
Therefore, k<l εk,l = 0 and since εk,l is non-negative, hence each of it has to be
zero. Thus, Φ is equiangular and
| φk, φl |2
=
N − M
M(N − 1)
.
We have shown that if the lower bound for coherence is satisfied, then Φ has to be
an ETF. We now show the converse.
Let Φ be an equiangular tight frame. Then the frame bounds are α = β = N
M .
Further, there is µ ∈ [0, 1), such that | φk, φl |2
= µ2
for k = l. Since Φ is tight,
hence λk = N
M for k = 1, . . . , M and zero otherwise. We have
N2
M
=
M
k=1
λ2
k =
N
k=1
N
l=1
| φk, φl |2
= N + N(N − 1)µ2
.
Rearranging, we get the lower bound for µ.
Theorem 27 For real matrices, if N > M(M+1)
2 , then Φ can’t be equiangular.

15. ETF 15
Proof. Let ΦN
M be an equiangular frame. Let Pk : RM
→ RM
be the projection of
x on to the line spanned by φk, i.e., Pkx = x, φk φk (or Pk = φkφT
k ). Let V be the
vector space of symmetric linear mappings RM
→ RM
. Then dim(V ) = M(M+1)
2 .
We define an inner product over V as C, D V = tr(CD). Since Φ is equiangular,
there exists µ ∈ [0, 1] such that φk, φl = ±µ for k = l.
If µ = 1, then we have N = 2 since the vectors in Φ are assumed to be distinct
and unit norm. Clearly for M ≥ 2, N = 2 < 3 ≤ M(M+1)
2 .
µ = 0 is possible only if all vectors are orthogonal to each other. This gives us
an orthonormal basis with N = M ≤ M(M+1)
2 .
We now consider µ ∈ (0, 1). Now,
Pk, Pl V = φk, φl =
1 if k = l
µ2
if k = l
Then, the Grammian G of the set {P1, . . . , PN } ⊂ V is
[G]k,l ==
1 if k = l
µ2
if k = l
Determinant of G is given by
det(G) = (1 + (N − 1)µ2
)(1 − µ2
)N−1
= 0.
Therefore G is invertible and full rank. We now have
N = rank(G) = dim(span{P1, . . . , PN }) ≤ dim(V ) =
M(M + 1)
2
.
We have shown that if Φ is ETF, then N ≤ M(M+1)
2 . By contraposition, the result
is proved.
Theorem 28 For complex matrices, if N > M2
, then Φ can’t be equiangular.
Proof. The proof follows theorem 27. Main differences are as follows. We let
V to be the real vector space of Hermitian linear mappings CM
→ CM
. Then,
dim(V ) = M2
. The projection operators for φk are defined as Pk = φkφH
k . The
inner product definition remains the same. Rest of the argument is identical.
Theorem 29 For optimal Grassmannian frame, the minimum and maximum
correlations are identical.
Proof. The reason is that all off-diagonal entries in G = ΦH
Φ have same magnitude.

16. 16 SHAILESH KUMAR
Theorem 30 [4] Let M, N ∈ N with N ≥ M. Assume R is a Hermitian
N × N matrix with entries Rk,k = 1 and
Rk,l =



± N−M
M(N−1) if F = R
±i N−M
M(N−1) if F = C
, (4.3)
for k, l = 1, . . . , N; k = l. If the eigenvalues λ1, . . . , λN of R are such that
λ1 = · · · = λM = N
M and other eigen values are 0, then there exists a frame
ΦN
M that achieves the bound (4.1).
Proof. Since R is Hermitian, it has a spectral factorization of the form R = WΛWH
.
Without loss of generality, we can assume that the non-zero eigen values of R are
contained in the first M diagonal entries of Λ. Let W be formed by the first M
columns of W. Then, it is clear that R = W N
M IWH
. Alternatively
R =
N
M
W
N
M
W
H
.
Let φk = N
M {Wk,l}M
l=1. In other words, φk is the k-th row (transposed) of
N
M W . Clearly,
φk, φl = φH
l φk = φT
k
¯φl = Rl,k.
Hence, {φk}N
k=1 is equiangular. {φk}N
k=1 is also tight since the non-zero eigen
values of R are identical. Hence by theorem 26, {φk}N
k=1 achieves the optimal
Grassmannian frame coherence bound.
Theorem 31 Let ΦN
M be an optimal Grassmannian frame with coherence
µ. Then for every vector unit norm vector x ∈ FM
, there exists at least one
vector φ in ΦN
M such that
| x, φ | ≥ µ. (4.4)
Proof. Since ΦN
M spans FM
, hence x is a linear combination of M or less vectors
from ΦN
M . If x is a scalar multiple of one of the vectors in ΦN
M (say φk) , then
| x, φk | = 1 > µ.
We now consider the case where x is a linear combination of more than one
vectors in ΦN
M . Thus, it is different from all vectors in ΦN
M .
For the sake of contradiction, assume that | x, φ | < µ for every vector in ΦN
M .
Now construct a new frame XN
M by replacing one of the vectors (say φ1) in ΦN
M
with x. Clearly coherence of XN
M is also µ (the absolute inner product of any pair
of vectors in XN
M excluding x). Thus, XN
M achieves the lower bound in (4.1). But
this is possible only if XN
M is equiangular, i.e. | x, φk | = µ for k = 2, . . . , N. We
reach a contradiction. Hence, there exists at least one vector φ in ΦN
M such that
| x, φ | ≥ µ.

17. ETF 17
5. Full Spark Frames
This section considers the design of full spark frames. i.e. unit norm frames ΦM
N
such that every set of M frame vectors in the frame is linearly independent and
spans FM
.
A frame with larger spark can provide unique sparse representations for signals
with larger sparsity levels.
Definition 17 A frame ΦM
N is called full spark if spark(Φ) = M + 1.
For a full-spark frame, every M × M submatrix is invertible.
5.1. Vandermonde matrices.
Theorem 32 A Vandermonde matrix is full spark if and only if its bases are
distinct.
6. Subspace angles
Consider an arbitrary vector x ∈ FM
. The angle between x and standard basis
vectors is given by
cos θi =
|xi|
x 2
where we ignore the sign / phase of xi. We consider the angle between x and the
vector closest to it in the standard basis. The minimum angle is given by
cos θmin = max
1≤i≤M
|xi|
x 2
.
For a general orthonormal basis Ψ, we have
cos θmin = max
1≤i≤M
|ψH
i x|
x 2
.
If x is one of the basis vectors in Ψ, then θmin is zero degree. How large can
θmin be? Since, the angle calculation is independent of the norm of x, hence let us
assume x to be a unit norm vector.
We will call this angle as the principal angle and the corresponding vector as
the principal vector.
Theorem 33 The largest principal angle between any vector x and an or-
thonormal basis is given by cos θ = 1√
M
.
Proof. Let x = Ψv where v 2 = 1. Since Ψ is unitary, x 2 = v 2 = 1. Also
v = ΨH
x, thus vi = ψH
i x. The principal angle is given by
cos θmin = max
1≤i≤M
|ψH
i x| = max
1≤i≤M
|vi|.

18. 18 SHAILESH KUMAR
Choose v = 1√
M
[1, . . . , 1]T
∈ FM
. Clearly, the angle between x = Ψv and each of
the basis vectors is cos θ = 1√
M
. Suppose, there is a unit norm vector y = Ψu, such
that y is further away from all vectors in Ψ compared to x. Then,
max
1≤i≤M
|ui| <
1
√
M
.
But this means that |ui| < 1√
M
, resulting in u 2 < 1 which is a contradiction.
The minimum angle can be generalized for a unit norm frame as
cos θmin = max
1≤k≤N
|φH
k x|
x 2
.
Let Φ be an optimal Grassmannian frame (i.e. it is also a unit norm equiangular
tight frame). Let x be an arbitrary unit norm vector. What is the minimum angle
of x with the columns of Φ?
To start with, if x is a vector in the sequence Φ, then the minimum angle is 0◦
.
TODO complete this.
6.1. Principal angles.
Definition 18 [2] Let U and V be given subspaces of FM
. Assume that
p = dim(U) ≥ dim(V) = q ≥ 1.
The smallest principal angle θ1(U, V) = θ1 ∈ [0, π/2] between U and V is
defined by
cos θ1 = sup
u∈U,v∈V
uH
v = uH
1 v1, u 2 = 1, v 2 = 1. (6.1)
For k = 2, . . . , q, the k-th principal angle θk is defined by
cos θk = sup
u∈Uk−1,v∈Vk−1
uH
v = uH
k vk, u 2 = 1, v 2 = 1. (6.2)
where Uk(Vk) is orthogonal complement of span{u1, . . . , uk} (span{v1, . . . , vk})
in U (V) respectively. The vectors {u1, . . . , uk} and {v1, . . . , vk} are called
principal vectors.
If we have U and V as matrices whose ranges are U and V respectively, then
in order to find the principal angles, we construct orthonormal bases QU and QV .
We then compute the inner product matrix G = QH
U QV . The SVD of G gives the
principal angles. In particular, the smallest principal angle is given by cos θ1 = σ1,
the largest singular value. We state the result in the following theorem.
Theorem 34 Let U and V be subspaces as in definition 18. Let QU and QV
be orthonormal bases for U and V respectively. Put G = QH
U QV . Let the SVD
of this p × q matrix be
G = Y ΣZH
, Σ = diag(σ1, . . . , σq).

19. ETF 19
We assume that σ1 ≥ · · · ≥ σq. Then the principal angles and principal vectors
associated with this pair of subspaces are given by
cos θk = σk, U = QU Y, V = UV Z. (6.3)
Proof. Note that QH
U QU = Ip and QH
V QV = Iq. Also, Y H
Y = Y Y H
= Ip and
ZH
Z = ZZH
= Iq.
The singular values and singular vectors of a matrix G are characterized by
σk = max
y 2=1, z 2=1
(yH
Gz) = yH
k Gzk,
subject to
yH
yj = zH
zj = 0, j = 1, . . . , k − 1.
Let u = QU y ∈ U and v = QV z ∈ V, then it follows that u 2 = y 2, v 2 = z 2
and
yH
yi = uH
ui, zH
zi = vH
vi.
Since yH
Gz = yH
QH
U QV z = uH
v, hence we can rewrite
σk = max
u 2=1, v 2=1
(uH
v) = uH
k vk,
subject to
uH
uj = vH
vj = 0, j = 1, . . . , k − 1.
This is exactly the definition of the principal angles and vectors which concludes
the proof.
6.2. One dimensional subspaces. The principal angle between the subspaces
spanned by two vectors x1
, x2
∈ FM
is given by
cos θx =
| x1
, x2
|
x1
2 x2
2
. (6.4)
Let Φ be a unit norm tight frame. Then with y1
= ΦH
x1
and y2
= ΦH
x2
, the
angle between y1
and y2
in FN
is given by
cos θy =
| y1
, y2
|
y1
2 y2
2
. (6.5)
The frame bounds give us
N
M
x 2
2 = ΦH
x 2
2.
Thus
ΦH
x 2 =
N
M
x 2.
Thus, we have
y1
2 y2
2 =
N
M
x1
2
N
M
x2
2 =
N
M
x1
2 x2
2.
Further,
y1
, y2
= ΦH
x1
, ΦH
x2
= x2H
ΦΦH
x1
= x2H
Sx1
= x2H N
M
x2H
x1
.
Thus
y1
, y2
=
N
M
x1
, x2
.

20. 20 SHAILESH KUMAR
This gives us
cos θy =
| y1
, y2
|
y1
2 y2
2
. =
N
M | x1
, x2
|
N
M x1
2 x2
2
= cos θx.
This result is summarized in following theorem.
Theorem 35 Let Φ be a unit norm tight frame. Then the principal angle
between the one dimensional subspaces spanned by two vectors x1
and x2
in FM
is same as the principal angle between the one dimensional subspaces spanned
by the vectors y1
= ΦH
x1
and y2
= ΦH
x2
in FN
.
6.3. Optimal Grassmannian Frame. Recall that an optimal Grassmannian frame
is an equiangular tight frame (with unit norm columns). The correlation between
any two vectors of the frame is given by µ = N−M
M(N−1) .
Let ΦN
M be an optimal Grassmannian frame. It is obvious that the angle between
any two one dimensional subspaces spanned the vectors in ΦN
M is given by cos θ1 =
N−M
M(N−1) .
Following discussion focuses on subspace angles between independent subspaces
created by choosing atoms from ΦN
M .
Let us consider two subspaces as U = span{φk, φl} and V = span{φm} where
k, l, m are different.
If U and V are dependent (happens when φm ∈ U), then the smallest principal
angle is 0◦
.
Let us now assume that U and V are independent. What is the smallest principal
angle between U and V? We restrict the following discussion to real optimal Grass-
mannian frames. We have qk, ql = ±µ. If qk, ql = −µ, then lets just replace ql by
−ql to ensure that qk, ql = µ. This change doesn’t impact U. Also see theorem 21.
Similarly, let’s change qm by −qm if necessary to ensure that qk, qm = µ.
Consider two vectors qa = qk + ql and qb = qk − ql. We have
qk ± ql
2
2 = qk
2
2 + ql
2
2 ± 2 qk, ql = 2(1 ± µ).
Let qa and qb be obtained by normalizing qa and qb respectively.
If ql, qm = µ, then qa, qm = 2µ and qb, qm = 0. We have
qa, qm =
2
1 + µ
µ.
On the other hand if ql, qm = −µ, then qa, qm = 0 and qb, qm = 2µ. We have
qb, qm =
2
1 − µ
µ.
In both cases, we have found a vector in U whose inner product with qm is larger
than µ. If µ 1, then in both cases, the inner product is approximated by
√
2µ.
This result is summarized in following theorem.

21. ETF 21
Theorem 36 Let ΦN
M be a real optimal Grassmannian frame with coherence
µ where µ 1. Let us consider two subspaces as U = span{φk, φl} and
V = span{φm} where k, l, m are different. Further, assume that U and V V V
are independent. Then the smallest principal angle between U and V V V is
approximated by
cos θ1 ≈
√
2µ.
The question now is, can we generalize this result for more vectors in the subspaces
U and V?
Appendix A. Linear Algebra
Some results from linear algebra are collected here for reference.
Theorem 37 Let Hn be an n × n matrix with 1 on the main diagonal and p
elsewhere. Let Cn be a matrix identical to Hn except that its (1, 1)-th element
equals p. Then
det(Hn) = (1 + (n − 1)p)(1 − p)n−1
(A.1)
and
det(Cn) = p(1 − p)n−1
. (A.2)
References
[1] John J Benedetto and Joseph D Kolesar. Geometric properties of grassmannian
frames for and. EURASIP Journal on Advances in Signal Processing, 2006(1):
049850, 2006.
[2] ke Bj¨orck and Gene H Golub. Numerical methods for computing angles between
linear subspaces. Mathematics of computation, 27(123):579–594, 1973.
[3] Stephane Mallat. A wavelet tour of signal processing: the sparse way. Access
Online via Elsevier, 2008.
[4] Thomas Strohmer and Robert W Heath. Grassmannian frames with appli-
cations to coding and communication. Applied and computational harmonic
analysis, 14(3):257–275, 2003.

22. Index
Coherence, 9
Dictionary, 4
Dual frame, 7
Equiangular tight frame, 9
Frame, 2
Frame analysis operator, 2
Frame coefficients, 2
Frame operator, 2
Frame redundancy, 4
Frame synthesis operator, 2
Full spark frame, 17
Gram matrix, 3
Grammian, 3
Grassmannian frame, 9
Maximum correlation, 9
Minimum correlation, 10
Principal angle, 18
Principal vector, 18
Redundancy, 4
Riesz basis, 4
Smallest principal angle, 18
Spark, 10
Tight frame, 8
Unit norm frame, 4
Unit norm tight frame, 8
22