RESULTS ON EQUIANGULAR TIGHT FRAMES

SHAILESH KUMAR

Contents

1. Introduction 11.1. Unit norm frame 41.2. Dual frame and pseudo inverse 61.3. Analysis and Synthesis 71.4. Tight frames 81.5. Coherence and Grassmannian frames 91.6. Equiangular tight frames 92. More Properties of Frames and Dictionaries 93. Vandermonde Matrices 114. Grassmannian frames 114.1. Existence of Grassmannian frames 114.2. Optimal Grassmannian frames 125. Full Spark Frames 175.1. Vandermonde matrices 176. Subspace angles 176.1. Principal angles 186.2. One dimensional subspaces 196.3. Optimal Grassmannian Frame 20Appendix A. Linear Algebra 21References 21Index 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 orCM . A general introduction to frame theory can be found in [3]. While [3] buildsthe theory over general Hilbert spaces, our treatment will be restricted to finitedimensional spaces.

In the sequel the results marked with ~ are potentially original. Many of themare very simple. It is possible that some of them might have appeared in some priorwork in some other context. The new results are listed here for quick reference

• theorem 21• theorem 31• theorem 35

1

2 SHAILESH KUMAR

• theorem 36

Definition 1 A finite frame ΦNM in a real or complex M -dimensional Hilbertspace FM is a sequence of N ≥ M vectors φkNk=1, φk ∈ FM , satisfying thefollowing condition

α‖x‖22 ≤N∑k=1

|〈x, φk〉|2 ≤ β‖x‖22, ∀ x ∈ FM (1.1)

where α, β are positive constants 0 < α ≤ β called the lower and upper framebounds.

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 ΦMN and its synthesis operator Φ inter-changeably.

Definition 3 The frame analysis operator is a mapping from FM to FN

defined asy , ΦHx. (1.2)

yi = 〈x, φi〉. The entries in y are known as the frame coefficients which arethe inner product of the frame vectors with x.

Clearly

‖y‖22 =

N∑k=1

|〈x, φk〉|2 = ‖ΦHx‖22 = xHΦΦHx. (1.3)

We can rewrite (1.1) as

α‖x‖22 ≤ ‖ΦHx‖22 ≤ β‖x‖22, ∀ x ∈ FM (1.4)

Definition 4 The frame operator S associated with the frame φkNk=1 isdefined by

Sx ,N∑k=1

〈x, φk〉φk. (1.5)

We note thatN∑k=1

〈x, φk〉φk = ΦΦHx. (1.6)

ETF 3

Thus,

S = ΦΦH ∈ FM×M (1.7)

Also

xHΦΦHx = xHSx. (1.8)

Theorem 1 A frame spans FM .

If the frame didn’t span FM , then the lower frame bound would be 0. This meansthat 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‖22 ≤ xHSx ≤ β‖x‖22, ∀ x ∈ FM . (1.9)

Since α > 0, hence xHSx > 0 ∀ x 6= 0.

Theorem 3 The frame analysis operator is injective.

Proof. Let x1 and x2 be two vectors in FN such that ΦHx1 = ΦHx2. Then ΦH(x1−x2) = 0. Thus, ‖ΦH(x1 − x2)‖22 = 0. This can happen only if x1 − x2 = 0 and thetwo vectors are identical. This shows that ΦH is injective.

Corollary 4 The frame analysis operator is bijective between FM and its rangegiven by R(ΦH).

Theorem 5 The upper and lower frame bounds of a frame are given by thelargest 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 = ΦHΦ is called the Gram matrix associatedwith 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 isobviously the norm-squared of i-th frame vector given by ‖φi‖22 = φHi φi.

Theorem 6 The non-zero eigen values of the frame operator S and the Grammatrix G are identical.

4 SHAILESH KUMAR

Proof. Let λ be a (non-zero) eigen value of S = ΦΦH . Then, there exists x 6= 0such that

ΦΦHx = λx.

Multiplying both sides by ΦH and defining y = ΦHx, we obtain

ΦHΦy = λy.

Since x 6= 0 =⇒ y 6= 0, hence y is an eigen vector of G with the same eigenvalue.

We note that

〈ΦHx, y〉 = 〈x,Φy〉. (1.10)

Theorem 7 The rank of G = ΦHΦ is M. Exactly M eigen values of G arepositive. Rest are zero.

Proof. Let x ∈ N (Φ). Then

Φx = 0 =⇒ ΦHΦx = 0 =⇒ N (Φ) ⊆ N (ΦHΦ).

Let x ∈ N (ΦHΦ). Then,

ΦHΦx = 0 =⇒ xHΦHΦx = 0 =⇒ ‖Φx‖22 = 0 =⇒ x ∈ N (Φ).

Thus N (Φ) = N (ΦHΦ). Since rank of Φ is M , hence rank(G) = M . Thus exactlyM of eigen values of G are positive.

Definition 6 The redundancy of a frame is defined as ρ = NM .

An orthonormal basis (ONB) for FM has exactly M vectors. The redundancy of aframe tells us about relative number of extra vectors in a frame viz. an ONB.

Definition 7 When the frame vectors φkNk=1 are linearly independent, thenthe 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 dependentand 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 vectorsatisfies ‖φk‖2 = 1.

A unit norm frame is also known as a dictionary.

ETF 5

The diagonal entries of the Grammian for a unit norm frame are 1. Naturallyfor 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) =∑Mk=1 λi is bounded

by

αM = λminM ≤M∑k=1

λi = tr(ΦΦH) ≤ λmaxM = βM.

But

tr(ΦΦH) = tr(ΦHΦ) = 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 = ΦHΦis 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

λ2k ≥

N2

M. (1.14)

The lower bound is achieved only if λk = NM 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 − NM . Then

M∑k=1

ek =

M∑k=1

(λk −

N

M

)= N −M N

M= 0.

6 SHAILESH KUMAR

Thus,M∑k=1

λ2k =

M∑k=1

(N

M+ ek

)2

=

M∑k=1

N2

M2+

2N

M

M∑k=1

ek +

M∑k=1

e2k

=N2

M+

M∑k=1

e2k ≥

N2

M.

The lower bound is achieved only if ek = 0, thus λk = NM for k = 1, . . . ,M .

1.2. Dual frame and pseudo inverse. The reconstruction of x from its framecoefficients ΦHx is calculated with a pseudo-inverse of ΦH .

Theorem 11 If φkNk=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 φkNk=1 is a frame but not a Riesz basis,then φkNk=1 is linearly dependent. Thus, there exists an a ∈ N (Φ) with a 6= 0.Since the restriction of ΦH to R(ΦH) is bijective thus invertible, hence ΦH admitsa 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 onR(ΦH)⊥.

Theorem 12 If φkNk=1 is a frame, then S = ΦΦH is invertible and thepseudo inverse satisfies

ΦH†

= (ΦΦH)−1Φ. (1.15)

Alternatively, the right pseudo-inverse for the frame synthesis operator Φ isgiven by

Φ† = ΦH(ΦΦH)−1. (1.16)

Proof. Since S is positive definite, it is invertible. Next, we see that

ΦH†ΦHx = (ΦΦH)−1ΦΦHx = x. (1.17)

Thus, ΦH†

is a left inverse. For any a ∈ R(ΦH)⊥ = N (Φ), we have Φa = 0. Thus,

ΦH†a = (ΦΦH)−1Φa = 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.

ETF 7

Definition 9 Let φkNk=1 be a frame. The vectors of a dual frame aredefined by

φk = (ΦΦH)−1φk, 1 ≤ k ≤ N. (1.19)

The dual frame consists of φkNk=1. The dual frame synthesis operatoris defined by

Φ ,[φ1 . . . φN

]= (ΦΦH)−1Φ = S−1Φ. (1.20)

The dual frame analysis operator is given by ΦH .

Theorem 13 The dual frame operator S = S−1.

Proof.

ΦΦH = (ΦΦH)−1Φ((ΦΦH)−1Φ

)H= (ΦΦH)−1ΦΦH(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 1β and 1

α . Specifically,

1

β‖x‖22 ≤

N∑k=1

|〈x, φk〉|2 ≤1

α‖x‖22, ∀ x ∈ FM (1.21)

Proof. Note that

N∑k=1

|〈x, φk〉|2 = ‖ΦHx‖22 = xH Sx = xHS−1x.

Since S is positive definite, hence S−1 is positive definite too. The eigen values ofS−1 are the inverse of eigen values of S. Thus if α is the minimum eigen value of Sthen 1

α is the maximum eigen value of S−1. Similarly for β. The result follows.

1.3. Analysis and Synthesis.

Theorem 15 We havex = ΦΦHx = ΦΦHx. (1.22)

This can also be expressed as

x =

N∑k=1

〈x, φk〉φk =

N∑k=1

〈x, φk〉φk. (1.23)

Proof.

ΦΦHx = (ΦΦH)−1ΦΦHx = x.

8 SHAILESH KUMAR

Similarly

ΦΦHx = ΦΦH(ΦΦH)−1x = x.

1.4. Tight frames.

Definition 10 When the frame bounds α and β are equal, a frame is said tobe tight.

Definition 11 A tight frame is called unit norm tight frame (UNTF) ifeach 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‖22 = ‖ΦHx‖22 ∀ 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 onlyeigen value of S.

Proof. Let S = V ΛV H be the eigen value decomposition of S. Since α = β, henceall eigen values of S are equal. Thus Λ = αI. Thus S = αV IV H = αI.

It is obvious that Φ = S−1Φ = 1αΦ. 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. Exceptfor an extra factor of 1

α which is the inverse of the frame redundancy factor, thereconstruction formula is identical to ONB.

Theorem 18 Rows of the frame synthesis operator Φ of a tight frame ΦNM areorthogonal. The squared norm of each row is α. Specifically, the norm of the

rows of a UNTF is√

NM .

ETF 9

Proof. We have S = ΦΦH = αI. Clearly the inner product of distinct rows is 0and 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 φkNk=1 be a unit norm frame with the frame synthesisoperator Φ. The coherence or the maximum correlation of the frame isdefined as

µ(Φ) , maxk 6=l|〈φk, φl〉|. (1.27)

Whenever the frame is obvious from the context, we may denote the coherencesimply by µ.

Definition 13 Fix N ≥M . Let Ω denote the set of all unit norm frames overFM consisting of N vectors. A frame Φ ∈ Ω is called a Grassmannian frameif

µ(Φ) = infX∈Ωµ(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 areorthogonal to each other. This notion is generalized in frames through the idea ofequiangular tight frames.

Definition 14 A unit norm tight frame φkNk=1 is called equiangular if

|〈φk, φl〉| = p whenever k 6= 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 matrixis also the same. We note that 0 ≤ p < 1 where 0 is achieved only when Φ is anorthonormal 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 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 6= 0. (2.1)

Definition 16 ~ Let φkNk=1 be a unit norm frame with the frame synthesisoperator Φ. The minimum correlation of the frame is defined as

µmin(Φ) , mink 6=l|〈φk, φl〉|. (2.2)

Theorem 19 The coherence of a frame ΦNM is invariant to unitary transfor-mation.

Proof. Let Ψ be an orthonormal basis for FM . Let ΨΦ be the frame constructedby the orthonormal transformation. Then

〈Ψφk,Ψφl〉 = 〈φk,ΨHΨφl〉 = 〈φk, φl〉since the inner products are invariant to unitary transformation. It is obvious that

maxk 6=l|〈Ψφk,Ψφl〉| = max

k 6=l|〈φk, φl〉|.

Theorem 20 The Grammian of a frame ΦNM is invariant to unitary transfor-mation.

Proof. Let Ψ be an orthonormal basis for FM . Let ΨΦ be the frame constructedby the orthonormal transformation. Then

(ΨΦ)H(ΨΦ) = ΦHΨHΨΦ = ΦHΦ.

This result is simply due to the invariance of inner products over orthonormaltransformation.

Theorem 21 ~ Let ΦNM be a frame in FM . Let Γ be a diagonal matrix whosediagonal entries are chosen from 1,−1 when F = R or eiθ : 0 ≤ θ < 2πwhen F = C. Let α, β be frame bounds of Φ. Then α, β are also frame boundsof 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.

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)Hx‖22 =

N∑k=1

|〈x, γkφk〉|2 =

N∑k=1

|γk|2|〈x, φk〉|2 =

N∑k=1

|〈x, φk〉|2 = ‖ΦHx‖22.

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 . . . 1γ1 γ2 . . . γN...

.... . .

...

γM−11 γM−1

2 . . . γM−1N

(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 thesubmatrices are non-singular as long as the bases γkNk=1 are distinct.

Theorem 22 The only Vandermonde matrices that are equal norm and tighthave 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 unitcircle, provided N ≥ 2M .

Proof. Since the matrix is a tight frame, hence it is full rank. Its rows have equalnorms.

4. Grassmannian frames

4.1. Existence of Grassmannian frames. Let SM−1 = x ∈ FM : ‖x‖2 =1denote the unit sphere in FM . A unit norm frame is constructed by picking upvectors from SM−1.

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(xkNk=1).

Define a norm over X = FM × · · · × FM (N times) given by∥∥xkNk=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 xkNk=1 in X. Weneed to show that for any ε > 0 there exists a δ > 0 such that |µ(xkNk=1) −µ(ykNk=1)| < ε whenever ‖xkNk=1 − ykNk=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 isbounded over S and a minimum as well maximum value exists. Thus, unit normsequences xkNk=1 exist in S with minimum coherence. What is left to show thatthey also span FM . If xkNk=1 doesn’t span FM , then we can easily replace one ofthe vectors in the sequence with a unit vector from the orthogonal complement ofxkNk=1 in FM . This process doesn’t increase coherence. We can keep doing it tillthe sequence spans FM . Thus there are indeed some sequences which satisfy all thecriteria for a frame with minimum coherence. This completes the proof.

Theorem 25 Let ΦNM be a Grassmannian frame and let Ψ be an orthonormalbasis (unitary matrix) for FM . Then, ΨΦ is also a Grassmannian frame.

Proof. Since (due to theorem 19) coherence is invariant to unitary transformation,hence ΦNM and ΨΦNM have identical coherence. Thus, ΨΦNM is also a Grassmannianframe. 4.2. Optimal Grassmannian frames.

ETF 13

Theorem 26 Let ΦNM with N ≥M be a unit norm frame. Then

µ(Φ) ≥

√N −MM(N − 1)

. (4.1)

The equality holds if and only if Φ is equiangular tight frame with frame boundsα = 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 framewith the specified frame bound. Such a Grassmannian frame is known as optimalGrassmannian frame.

Proof. Let Φ be a Grassmannian frame and G = ΦHΦ ∈ 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∑Mk=1 λk = N and

∑Mk=1 λ

2k ≥ N2

M . The lower bound is

achieved only if λk = NM for all M positive eigen values. Let G =

[g1 . . . gN

]in terms of its column vectors. Consider the matrix G2. Its eigen values areλ2

1 ≥ · · · ≥ λ2N . Since G2 = GHG. Thus,

M∑k=1

λ2k = tr(G2) =

N∑k=1

gHk gk.

But

gHk gk = ‖gk‖22N∑l=1

|〈φk, φl〉|2.

We get

N2

M≤

M∑k=1

λ2k =

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 + 2N(N − 1)

2maxk 6=l|〈φk, φl〉|2.

Rearranging this, we obtain

µ2(Φ) ≥ N −MM(N − 1)

.

14 SHAILESH KUMAR

We now show that equality holds only if Φ is an equiangular tight frame. We notethat µ2(Φ) = N−M

M(N−1) forces that

N2

M=

M∑k=1

λ2k =

N∑k=1

N∑l=1

|〈φk, φl〉|2.

This can happen only if λk = NM for k = 1, . . . ,M . Immediately, since the largest

and smallest eigen values of the frame operator ΦΦH are equal, hence Φ is a tightframe. To see that Φ is also equiangular, we proceed as follows. Recall from abovethat

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 maxk 6=l|〈φk, φl〉|2 = N−M

M(N−1) , hence for any k 6= l,

|〈φk, φl〉|2 =N −MM(N − 1)

− εk,l

where εk,l ≥ 0. Summing over k < l, we get

N(N −M)

2M=∑k<l

(N −MM(N − 1)

− εk,l)

=

(N(N − 1)

2

)(N −MM(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 −MM(N − 1)

.

We have shown that if the lower bound for coherence is satisfied, then Φ has to bean ETF. We now show the converse.

Let Φ be an equiangular tight frame. Then the frame bounds are α = β = NM .

Further, there is µ ∈ [0, 1), such that |〈φk, φl〉|2 = µ2 for k 6= l. Since Φ is tight,hence λk = N

M for k = 1, . . . ,M and zero otherwise. We have

N2

M=

M∑k=1

λ2k =

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.

ETF 15

Proof. Let ΦNM be an equiangular frame. Let Pk : RM → RM be the projection ofx on to the line spanned by φk, i.e., Pkx = 〈x, φk〉φk (or Pk = φkφ

Tk ). 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 6= 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 = lµ2 if k 6= l

Then, the Grammian G of the set P1, . . . , PN ⊂ V is

[G]k,l ==

1 if k = lµ2 if k 6= l

Determinant of G is given by

det(G) = (1 + (N − 1)µ2)(1− µ2)N−1 6= 0.

Therefore G is invertible and full rank. We now have

N = rank(G) = dim(spanP1, . . . , 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 letV 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 = φkφ

Hk . The

inner product definition remains the same. Rest of the argument is identical.

Theorem 29 For optimal Grassmannian frame, the minimum and maximumcorrelations are identical.

Proof. The reason is that all off-diagonal entries in G = ΦHΦ have same magnitude.

16 SHAILESH KUMAR

Theorem 30 [4] Let M,N ∈ N with N ≥ M . Assume R is a HermitianN ×N matrix with entries Rk,k = 1 and

Rk,l =

±√

N−MM(N−1) if F = R

±i√

N−MM(N−1) if F = C

, (4.3)

for k, l = 1, . . . , N ; k 6= l. If the eigenvalues λ1, . . . , λN of R are such thatλ1 = · · · = λM = N

M and other eigen values are 0, then there exists a frame

ΦNM that achieves the bound (4.1).

Proof. SinceR is Hermitian, it has a spectral factorization of the formR = 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 NM IWH . Alternatively

R =

(√N

MW

)(√N

MW

)H.

Let φk =√

NM Wk,lMl=1. In other words, φk is the k-th row (transposed) of(√

NM W

). Clearly,

〈φk, φl〉 = φHl φk = φTk φl = Rl,k.

Hence, φkNk=1 is equiangular. φkNk=1 is also tight since the non-zero eigenvalues of R are identical. Hence by theorem 26, φkNk=1 achieves the optimalGrassmannian frame coherence bound.

Theorem 31 ~ Let ΦNM be an optimal Grassmannian frame with coherenceµ. Then for every vector unit norm vector x ∈ FM , there exists at least onevector φ in ΦNM such that

|〈x, φ〉| ≥ µ. (4.4)

Proof. Since ΦNM spans FM , hence x is a linear combination of M or less vectorsfrom ΦNM . If x is a scalar multiple of one of the vectors in ΦNM (say φk) , then|〈x, φk〉| = 1 > µ.

We now consider the case where x is a linear combination of more than onevectors in ΦNM . Thus, it is different from all vectors in ΦNM .

For the sake of contradiction, assume that |〈x, φ〉| < µ for every vector in ΦNM .Now construct a new frame XNM by replacing one of the vectors (say φ1) in ΦNMwith x. Clearly coherence of XNM is also µ (the absolute inner product of any pairof vectors in XNM excluding x). Thus, XNM achieves the lower bound in (4.1). Butthis is possible only if XNM is equiangular, i.e. |〈x, φk〉| = µ for k = 2, . . . , N . Wereach a contradiction. Hence, there exists at least one vector φ in ΦNM such that|〈x, φ〉| ≥ µ.

ETF 17

5. Full Spark Frames

This section considers the design of full spark frames. i.e. unit norm frames ΦMNsuch that every set of M frame vectors in the frame is linearly independent andspans FM .

A frame with larger spark can provide unique sparse representations for signalswith larger sparsity levels.

Definition 17 A frame ΦMN 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 aredistinct.

6. Subspace angles

Consider an arbitrary vector x ∈ FM . The angle between x and standard basisvectors is given by

cos θi =|xi|‖x‖2

where we ignore the sign / phase of xi. We consider the angle between x and thevector closest to it in the standard basis. The minimum angle is given by

cos θmin = max1≤i≤M

|xi|‖x‖2

.

For a general orthonormal basis Ψ, we have

cos θmin = max1≤i≤M

|ψHi x|‖x‖2

.

If x is one of the basis vectors in Ψ, then θmin is zero degree. How large canθmin be? Since, the angle calculation is independent of the norm of x, hence let usassume x to be a unit norm vector.

We will call this angle as the principal angle and the corresponding vector asthe 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. Alsov = ΨHx, thus vi = ψHi x. The principal angle is given by

cos θmin = max1≤i≤M

|ψHi x| = max1≤i≤M

|vi|.

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,

max1≤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 = max1≤k≤N

|φHk x|‖x‖2

.

Let Φ be an optimal Grassmannian frame (i.e. it is also a unit norm equiangulartight frame). Let x be an arbitrary unit norm vector. What is the minimum angleof 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 isdefined by

cos θ1 = supu∈U,v∈V

uHv = uH1 v1, ‖u‖2 = 1, ‖v‖2 = 1. (6.1)

For k = 2, . . . , q, the k-th principal angle θk is defined by

cos θk = supu∈Uk−1,v∈Vk−1

uHv = uHk vk, ‖u‖2 = 1, ‖v‖2 = 1. (6.2)

where Uk(Vk) is orthogonal complement of spanu1, . . . , uk (spanv1, . . . , vk)in U (V) respectively. The vectors u1, . . . , uk and v1, . . . , vk are calledprincipal vectors.

If we have U and V as matrices whose ranges are U and V respectively, thenin order to find the principal angles, we construct orthonormal bases QU and QV .We then compute the inner product matrix G = QHUQV . The SVD of G gives theprincipal 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 QVbe orthonormal bases for U and V respectively. Put G = QHUQV . Let the SVDof this p× q matrix be

G = Y ΣZH , Σ = diag(σ1, . . . , σq).

ETF 19

We assume that σ1 ≥ · · · ≥ σq. Then the principal angles and principal vectorsassociated with this pair of subspaces are given by

cos θk = σk, U = QUY, V = UV Z. (6.3)

Proof. Note that QHUQU = Ip and QHV QV = Iq. Also, Y HY = Y Y H = Ip andZHZ = ZZH = Iq.

The singular values and singular vectors of a matrix G are characterized by

σk = max‖y‖2=1,‖z‖2=1

(yHGz) = yHk Gzk,

subject toyHyj = zHzj = 0, j = 1, . . . , k − 1.

Let u = QUy ∈ U and v = QV z ∈ V, then it follows that ‖u‖2 = ‖y‖2, ‖v‖2 = ‖z‖2and

yHyi = uHui, zHzi = vHvi.

Since yHGz = yHQHUQV z = uHv, hence we can rewrite

σk = max‖u‖2=1,‖v‖2=1

(uHv) = uHk vk,

subject touHuj = vHvj = 0, j = 1, . . . , k − 1.

This is exactly the definition of the principal angles and vectors which concludesthe proof.

6.2. One dimensional subspaces. The principal angle between the subspacesspanned 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 = ΦHx1 and y2 = ΦHx2, theangle 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‖22 = ‖ΦHx‖22.

Thus

‖ΦHx‖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〉 = 〈ΦHx1,ΦHx2〉 = x2HΦΦHx1 = x2HSx1 = x2H N

Mx2Hx1.

Thus

〈y1, y2〉 =N

M〈x1, x2〉.

20 SHAILESH KUMAR

This gives us

cos θy =|〈y1, y2〉|‖y1‖2‖y2‖2

. =NM |〈x

1, x2〉|NM ‖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 anglebetween the one dimensional subspaces spanned by two vectors x1 and x2 in FM

is same as the principal angle between the one dimensional subspaces spannedby the vectors y1 = ΦHx1 and y2 = ΦHx2 in FN .

6.3. Optimal Grassmannian Frame. Recall that an optimal Grassmannian frameis an equiangular tight frame (with unit norm columns). The correlation between

any two vectors of the frame is given by µ =√

N−MM(N−1) .

Let ΦNM be an optimal Grassmannian frame. It is obvious that the angle betweenany two one dimensional subspaces spanned the vectors in ΦNM is given by cos θ1 =√

N−MM(N−1) .

Following discussion focuses on subspace angles between independent subspacescreated by choosing atoms from ΦNM .~ Let us consider two subspaces as U = spanφk, φl and V = spanφm 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‖22 = ‖qk‖22 + ‖ql‖22 ± 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 + µµ.

On the other hand if 〈ql, qm〉 = −µ, then 〈qa, qm〉 = 0 and 〈qb, qm〉 = 2µ. We have

〈qb, qm〉 =

√2

1− µµ.

In both cases, we have found a vector in U whose inner product with qm is largerthan µ. If µ 1, then in both cases, the inner product is approximated by

√2µ.

This result is summarized in following theorem.

ETF 21

Theorem 36 ~ Let ΦNM be a real optimal Grassmannian frame with coherenceµ where µ 1. Let us consider two subspaces as U = spanφk, φl andV = spanφm where k, l,m are different. Further, assume that U and V V Vare independent. Then the smallest principal angle between U and V V V isapproximated by

cos θ1 ≈√

2µ.

The question now is, can we generalize this result for more vectors in the subspacesU 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 pelsewhere. Let Cn be a matrix identical to Hn except that its (1, 1)-th elementequals p. Then

det(Hn) = (1 + (n− 1)p)(1− p)n−1 (A.1)

anddet(Cn) = p(1− p)n−1. (A.2)

References

[1] John J Benedetto and Joseph D Kolesar. Geometric properties of grassmannianframes for and. EURASIP Journal on Advances in Signal Processing, 2006(1):049850, 2006.

[2] ke Bjorck and Gene H Golub. Numerical methods for computing angles betweenlinear subspaces. Mathematics of computation, 27(123):579–594, 1973.

[3] Stephane Mallat. A wavelet tour of signal processing: the sparse way. AccessOnline via Elsevier, 2008.

[4] Thomas Strohmer and Robert W Heath. Grassmannian frames with appli-cations to coding and communication. Applied and computational harmonicanalysis, 14(3):257–275, 2003.

Index

Coherence, 9

Dictionary, 4Dual frame, 7

Equiangular tight frame, 9

Frame, 2

Frame analysis operator, 2

Frame coefficients, 2Frame operator, 2

Frame redundancy, 4

Frame synthesis operator, 2Full spark frame, 17

Gram matrix, 3Grammian, 3

Grassmannian frame, 9

Maximum correlation, 9

Minimum correlation, 10

Principal angle, 18Principal vector, 18

Redundancy, 4Riesz basis, 4

Smallest principal angle, 18Spark, 10

Tight frame, 8

Unit norm frame, 4

Unit norm tight frame, 8

22