PARSEVAL FRAME CONSTRUCTION NATHAN BUSH, MEREDITH CALDWELL, TREY TRAMPEL Abstract. General methods of constructing Parseval frames are desirable in many kinds of signal processing applications, as well as for use by mathematical researchers in frame theory. We begin by examining how the orthonormality of the row space of certain matrices implies the column space is a Parseval frame. We then use these results to study harmonic frames, which are frames arising from generalizations of the discrete Fourier transform. We then discuss a process, analogous to the well-known Gram-Schmidt or- thogonalization process, which converts a frame into a Parseval frame. All our work takes place in finite dimensional spaces. 1. Background Information We begin with a small review of linear algebra and a small introduc- tion to frames. Definition 1.1. We say a complex (real) finite dimensional vector space is a Hilbert space, V , if it is equipped with an inner product hfjgi, that is, a map of V × V ! C which satisfies (1.1) hf + gjhi = hfjhi + hgjhi (1.2) hαfjgi = αhfjgi (1.3) hhjgi = hgjhi hfjfi = 0 then f = 0(1.4) for all scalars α and f; g; h in V . Through this article, all Hilbert spaces will be finite dimensional. Definition 1.2. Recall that operators in a finite dimensional Hilbert space have adjoints, by the Riesz Representation Theorem [1], which we denote T ∗. We say T is a normal operator on V if TT ∗ = T ∗T: That is, T commutes with its adjoint. Definition 1.3. An operator T on V is self-adjoint if T = T ∗. Date: July 6, 2012. 1 2 NATHAN BUSH, MEREDITH CALDWELL, TREY TRAMPEL Definition 1.4. If T is an operator on V , T can be represented by a matrix, denoted [T ] or [aij]. Definition 1.5. A positive operator T is a self-adjoint operator on V such that hT vjvi ≥ 0 for any v 2 V . Theorem 1.6 (Spectral Theorem). [1] Let T be a normal linear oper- ator on a finite-dimensional Hilbert space V with distinct eigenvalues λ1, λ2, ::: , λk, and let Pj be the orthogonal projection of V onto the eigenspace Eλj for 1 ≤ j ≤ k. Then the following are true: (1.5) V = Eλ1 ⊕ Eλ2 ⊕ ::: ⊕ Eλk (1.6) PiPj = 0 for i 6= j (1.7) P1 + P2 + ::: + Pk = I (1.8) T = λ1P1 + λ2P2 + ::: + λkPk Definition 1.7. [1] A frame for a Hilbert space V is a sequence of vectors fxig ⊂ V for which there exist constants 0 < A ≤ B < 1 such that, for every x 2 V , 2 X 2 2 Ajjxjj ≤ jhxjxiij ≤ Bjjxjj : i A frame is a Parseval frame if A = B = 1. k Proposition 1.8. [1] A collection of vectors fxigi=1 is a Parseval frame for a Hilbert space V if and only if the following formula holds for every x in V : k X x = hxjxiixi: i=1 This equation is called the reconstruction formula for a Parseval frame. Theorem 1.9. [1] Suppose that V is a finite-dimensional Hilbert space k and fxigi=0 is a finite collection of vectors from V . Then the following statements are equivalent: k • fxigi=0 is a frame for V k • span fxigi=0 = V . k k Definition 1.10. [2] Let faigi=1 be a set for V . Let Θa : V ! C be 0 hvja1i 1 . defined as Θa(v) = @ . A for v 2 V . We call Θf the analysis hvjaki operator for faig. PARSEVAL FRAME CONSTRUCTION 3 k ∗ k Definition 1.11. [1] Let faigi=1 be a set for V . Let Θa : C ! V be ∗ the adjoint of Θa. We call Θa the reconstruction operator for faig. 0 c1 1 . k ∗ Pk Proposition 1.12. [2] Let w = @ . A 2 C . Then Θa(w) = i=1 ciai c . k k k Proof. Let feigi=1 be the standard orthonormal basis for C and v 2 V . 0 1 0 1 0 hvja1i . *B . C B . C+ B . C B . C ∗ B 1 C B hvja i C Observe that hΘaeijvi = heijΘavi = B C B i C = B . C B . C @ . A @ . A 0 hvjaki ∗ hvjaii = haijvi. Since this is true for any v 2 V , then Θa(ei) = ai. Now note that k k k ∗ ∗ X X ∗ X Θa(w) = Θa( ciei) = ciΘa(ei) = ciai : i=1 i=1 i=1 Definition 1.13. Let S = Θ∗Θ. We call S the frame operator. 2. Orthonormal Vectors to Parseval Frames Theorem 2.1. Let k ≥ n and let A be an n×k matrix in which the rows n form an orthonormal set of vectors in R . Now let F = fv1; : : : ; vkg be the columns of A. Then F is a Parseval frame for Rn. Proof. Let A = [aij]. We know n X hxjvji = xiaij: i=1 Therefore we have k k n !2 X 2 X X jhxjvjij = xiaij j=1 j=1 i=1 k n n X X X = xiaijxlalj j=1 i=1 l=1 n n k ! X X X = xixl aijalj i=1 l=1 j=1 4 NATHAN BUSH, MEREDITH CALDWELL, TREY TRAMPEL Now since the rows of A are orthonormal vectors, then k ( X 1 if i = l a a = ij lj 0 if i 6= l j=1 So, k n n k ! X 2 X X X jhxjvjij = xixl aijalj j=1 i=1 l=1 j=1 n X = xixi i=1 2 = jjxjj Theorem 2.2. Let k ≥ n and let A be an n×k matrix in which the rows k form an orthonormal set of vectors in C . Now let F = fv1; : : : ; vkg be the columns of A. Then F is a Parseval frame for Cn. Proof. Let A = [aij]. We know n X 2 hxjvji = xiaij and also jzj = zz: i=1 Therefore we have k k n ! 2 X 2 X X jhxjvjij = xiaij j=1 j=1 i=1 k n ! n ! X X X = xiaij xlalj j=1 i=1 l=1 k n n ! k n n X X X X X X = xiaij xlalj = xiaij xlalj j=1 i=1 l=1 j=1 i=1 l=1 n n k n X X X X 2 = xixl aljaij = xixi = jjxjj : i=1 l=1 j=1 i=1 3. Harmonic Frames Definition 3.1. Let the vector space be Cn.A harmonic frame is a collection of vectors η0; : : : ; ηm−1 (where m > n) such that for PARSEVAL FRAME CONSTRUCTION 5 0 ≤ k ≤ m − 1 2 k 3 w1 1 . i( h×2π ) ηk = p 4 . 5 where wh = e m m k wn th Noting that wh is an m root of unity. Note that this is a frame for Cn since it is a spanning set in a finite dimensional vector space. In fact it is a Parseval frame. i k−1 Lemma 3.2. For any geometric sequence, fra gi=0 for a; r 2 C; a 6= Pk−1 i r(1−ak) 0; 1; i=0 ra = 1−a : Proof. We first observe that k−1 k X X rai − rai = r + ra + ··· + rak−1 − (ra + ra2 + ··· + rak) i=0 i=0 = r + 0 + ··· + 0 − rak = r − rak : But since k−1 k k−1 k−1 k−1 X X X X X rai − rai = rai − a rai = (1 − a) rai = r − rak; i=0 i=0 i=0 i=0 i=0 we see k−1 X r(1 − ak) rai = ; 1 − a i=0 which concludes our proof. Definition 3.3. [4] The discrete Fourier transform of a function f = N−1 a ··· a ; which we denote f;^ is defined f^ = p1 f P ! ; where 0 n N i=0 i !0; :::; !N−1 denote the column vectors of the N × N Fourier matrix N F Ω. For the entry in the kth row and lth column of this matrix, we N ikl2π kl say F Ωk;l := (!k)l = e N = !N . Corollary 3.4. In the case where m = n, the collection of vectors η0; : : : ; ηm−1 is a Parseval frame. Proof. We first expand the matrix η0 η2 : : : ηm−1 into 6 NATHAN BUSH, MEREDITH CALDWELL, TREY TRAMPEL 21 1 1 ··· 1 3 i 2π 1 i 2π 2 i 2π (m−1) 61 e m e m ··· e m 7 6 i 2π 2 i 2π 4 i 2π 2(m−1) 7 61 e m e m ··· e m 7 6 7 6. .. 7 4. 5 i 2π (m−1) i 2π 2(m−1) i 2π (m−1)(m−1) 1 e m e m ··· e m which is the matrix representation of the inverse discrete Fourier trans- n −2π (k−1)(r−1) form on C : We now observe that e m represents the entry of this matrix in the kth row and rth column. Since i2π (k−1)(r−1) i2π (r−1)(k−1) e m = e m ; we see this matrix is symmetric by the definition of a symmetric matrix. To prove orthogonality of the columns of this matrix we see that for p 6= v, by Lemma 3.1, m−1 X i2π j(p+v) hηpjηvi = e m j=0 i2π m(p+v) 1 − e m = i2π (p+v) 1 − e m 1 − ei2π(p+v) = i2π (p+v) 1 − e m 1 − (cos(2π(p + v)) + i sin(2π(p + v))) = i2π (p+v) 1 − e m 1 − (1 + 0) = i2π (p+v) 1 − e m = 0: Since our matrix is symmetric and we have shown the columns are orthogonal, we know the rows are orthogonal as well.