Linear Algebra and its Applications

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1 Linear Algebra and its Applications 435 (011) Contents lists available at ScienceDirect Linear Algebra and its Applications journal homepage: Boolean invertible matrices identified from two permutations and their corresponding Haar-type matrices Nikolaos Atreas a,, Costas Karanikas b a Aristotle University of Thessaloniki, Department of Mathematics Physics and Computational Sciences, Faculty of Engineering, Thessaloniki, Greece b Aristotle University of Thessaloniki, Department of Informatics, Thessaloniki, Greece ARTICLE INFO Article history: Received 6 July 010 Accepted 4 January 011 Available online 6 January 011 SubmittedbyR.A.Brualdi AMS classification: 15B34 15B10 65T99 Keywords: Boolean matrices Haar-type matrices Permutations ABSTRACT We construct a class R m of m m boolean invertible matrices whose elements satisfy the following property: when we perform the Hadamard product operation R i R j on the set of row vectors R 1,...,R m } of an element R R m we produce either the row R maxi,j} or the zero row. In this paper, we prove that every matrix R R m is uniquely determined by a pair of permutations of the set 1,...,m}. As a by-product of this result we identify Haar-type matrices from a pair of permutations as well, because these matrices emerge from the Gram Schmidt orthonormalization process of the set of row vectors of R matrices belonging in a certain subclass R 0 R m. 011 Elsevier Inc. All rights reserved. 1. Introduction Let g : I m I m I m be an operation on the set I m =1,...,m} (m N) and let A g be the class of all m m matrices A over the field C of complex numbers whose row vectors A 1,...,A m satisfy A i A j = c ij A g(i,j), i, j = 1,...,m (1.1) for some double-indexed sequence of scalars c ij } m i,j=1. Here and hereafter A i A j = ( A i1 A j1,...,a im A jm ), i, j = 1,...,m (1.) Corresponding author. addresses: natreas@gen.auth.gr (N. Atreas), karanika@csd.auth.gr (C. Karanikas) /$ - see front matter 011 Elsevier Inc. All rights reserved. doi: /j.laa

2 96 N. Atreas, C. Karanikas / Linear Algebra and its Applications 435 (011) is the usual Hadamard product operation [11,p.98].ThenotationA = (A 1,...,A m ) is used to denote an m m matrix A with rows A 1,...,A m.wedefinethesupport of the i-row of an m m matrix A by suppa i }= k 1,...,m} : A i,k = 0 } and we use the notation suppa T i } to denote the support of the i-column of A. It is remarkable that well known classes of matrices with numerous applications in signal processing, data manipulation, information and communication } theory satisfy (1.1). Forexampletherowsofthe (k 1)(n 1) m 1 Fourier matrix F m = πi e m [3] satisfy m k,n=1 (F m ) i (F m ) j = m 1/ (F m ) Mod(i+j,m)+1. Also the n n Walsh matrices W n (n = 1,,...)[10] satisfy (1.1) as well. Indeed (W n ) i (W n ) j = n/ (W n ) k, where i 1 = (ε 0,...,ε n 1 ) and j 1 = (e 0,...,e n 1 ) for ε n, e n 0, 1} represent the numbers i 1, j 1 0,..., n 1} in base and k 1 = (ε 0 e 0,...,ε n 1 e n 1 ) under addition mod. Another important class of matrices satisfying (1.1)fori = j is the class of Haar matrices. Recall that the Haar transform described by Haar in 1910 serves as the prototypical wavelet transform providing sparse representations for discrete/piecewise constant signals [6]. The linearity of the Haar transform on finite data of length m = N is expressed via the well known N N Haar matrix H (N) produced recursively by 1 H (1) 1 H (n 1) 1 = 1 1,...,H (n), 1 =, n =,...,N. 1 I n 1, 1 Here the symbol denotes tensor product of matrices [11, p. 43] and I n is the n n identity matrix. From the above definition we derive the following property: H (N) i H (N) j = c ij H (N) maxi,j} : c ij 0} ± 1 } N, i, k/ j = 1,..., N (N > 1), i = j. k= Furthermore in [, p. 198] we made the following observation: Observation 1. The rows of H (N) emerge from the Gram Schmidt orthonormalization process of the setofrowsofa N N boolean invertible matrix U (N) produced recursively by U (1) = 11 U (n 1) (1, 1),...,U (n) =, n =,...,N. (1.3) 10 I n 1 (1, 0) From (1.3) we can see that the row vectors of U (N) satisfy (1.1). More precisely U (N) i U (N) j = c ij U (N) maxi,j} : c ij 0, 1}, i, j = 1,..., N. (1.4)

3 N. Atreas, C. Karanikas / Linear Algebra and its Applications 435 (011) Moreover in [1, Definition 6] we saw that if m = p 1 p N (p 1 p N ) is the prime integer factorization of a natural number m, then(1.3) is a special case of a more general recursive equation which produces m m boolean invertible matrices (called U-matrices) satisfying (1.4) as well. Here we make another important observation. Observation. The Gram Schmidt orthonormalization process of the set of row vectors of U-matrices produces m m Haar-type matrices. Definition 1. We say that an m m orthonormal matrix H m is a Haar-type matrix if the following three conditions are satisfied: (i) for i, j = 1,...,m with i = j,therowsofh m satisfy (H m ) i (H m ) j = c ij (H m ) maxi,j} for some double-indexed sequence of scalars c =c ij }; (ii) the set of all non-zero entries of any row of H m contains only two elements with alternate signs (except for the first row (H m ) 1 = m 1/ 1,...,1}); and (iii) m j=1 (H m) ij = 0, i =,...,m. Motivated from (1.4) we give the following. Definition. For any natural number m we define by R m the class of all m m boolean invertible matrices R =R 1,...,R m } whose row vectors satisfy R i R j = c ij R maxi,j} : c ij 0, 1}, i, j = 1,...,m. Then the following result is straightforward. Lemma 1. LetAbeanm m boolean invertible matrix and let 1 i < j m. Then A R m if and only if suppa j } suppa i } or suppa i } suppa j }=. The aim of this work is to describe all elements R R m in a unique way. Our motivation to do that originates from observations 1 and. Since Haar matrices and their modifications emerge from boolean matrices (1.3) and more generally from U-matrices contained in the class R m, if we could establish an encoding/decoding scheme for any element in R m then we would be able to encode Haar-type matrices as well by the same scheme. Notice that Haar-type matrices and their corresponding transforms [7 9] are used in a variety of applications such as classification [4], identification [5], and compression [7] and so it is important to be able to identify and store these matrices efficiently. The paper is organized in the following sections: In Section we identify all elements R R m from two permutations. In Proposition 1 we define a class A m R m containing all m m unitriangular boolean matrices and we prove that there is an one to one correspondence between the class A m and the symmetric group P m of the set 1,...,m}. Then in Theorem 1 we state our first main result: every element R R m can be uniquely written as R = A r P ρ : (ρ, r) P m P m, where A r is an m m unitriangular boolean matrix in A m associated with a unique permutation r P m and P ρ is an m m permutation matrix associated with another permutation ρ P m.since the symmetric group P m is enumerated, every matrix R R m can be coded by two natural numbers between 1 and m!, see Proposition. InSection 3 we produceorthonormalmatricesfrom R matrices, see (3.) and Proposition 3. InTheorem of this section we determine a subclass R 0 R m whose elements R 0 satisfy the following property: the orthonormalization process of the set of row vectors of any matrix R 0 produces Haar-type matrices. Therefore Haar-type matrices can be identified from two permutations as well.

4 98 N. Atreas, C. Karanikas / Linear Algebra and its Applications 435 (011) The class of R matrices Let R m be the class of m m boolean invertible matrices as in definition. In this section, we determine all elements R R m by two permutations. First we need the following. Definition 3. For any natural number m we denote by A m the class of all m m upper unitriangular boolean matrices A whose row vectors satisfy A i A j = c ij A maxi,j} : c ij 0, 1} i, j = 1,...,m. Then we can prove the following. Proposition 1. The class A m and the set S m = s =s(1),...,s(m)} : s(i) i, s(i) N} (.1) are equivalent, i.e. there exists a bijection f : A m S m. Proof. Clearly A m R m. Therefore for any element A A m and for any 1 i < j m we have suppa j } suppa i } or suppa i } suppa j }= as a result of Lemma 1. LetS m be as in (.1). We define a map f : A m S m : s = f (A) by k, whenever k = max i 1,...,j 1} :supp(a j ) supp(a i ) } s(1) = 1, s(j) =. (.) j, whenever supp(a i ) supp(a j ) = for any i = 1,...,j 1 Using (.) we compute the support of the columns of any element A A m by supp A T j =s k (j) : 0 k μ j }, j = 1,...,m, (.3) where 0 μ j m 1 is the first natural number satisfying s μ j+1 (j) = s μ j (j). Hereweconsider s 0 (j) = j. Alsoweusethenotations k (j) k =,...,m to denote the composition of the sequence s with itself k times. Clearly (.3) ensures that f is injective. Now we shall prove that f issurjective.todothatweconsider an element s S m and we define an m m( boolean ) matrix A by (.3). Then we notice that A is an upper unitriangular matrix because j supp A T j for any j = 1,...,m and s k (j) >s k+1 (j) for any 0 k μ j because of (.1) and (.3). It suffices to prove that A A m and f (A) = s as in (.). To do that we use Lemma 1. First we consider two cases: (i) Let i < j and let i supp A T j.takeanyλ supp(a j ), λ j. Thenj supp A T λ and so from (.3)thereexists1 l λ λ j such that s l λ(λ) = j. Therefore } supp A T λ = s k (λ) : k = 0,...,l λ 1 supp A T j. (.4) Since i supp A T j,by(.4) i supp A T λ, i.e. λ supp(a i ),sosupp(a j ) supp(a i ). (ii) Let i < j and i / supp A T j.takeanyλ supp(a j ). Working as in (i) we obtain (.4). Since s k (λ) > s k+1 (λ) j > i for all k = 0,...,l λ 1 and since i / supp A T j we deduce that i / supp A T λ. Therefore λ/ supp(a i ), i.e. supp(a i ) supp(a j ) =. From (i) and (ii) we deduce that the matrix A satisfies Lemma 1 and so A A m. Furthermore from (i), (.3) and the property s k (j)>s k+1 (j) for any 0 k μ j we obtain s(j)= max i 1,...,j 1} :supp

5 N. Atreas, C. Karanikas / Linear Algebra and its Applications 435 (011) (A j ) supp(a i ) }. Also from (ii) and (.3) we can see that whenever supp(a T j ) =j}, thensupp(a i) supp(a j ) = for any i = 1,...,j 1. Therefore f (A) = s as in (.) and we are done. Now we state our first main Theorem. Theorem 1. Let A m be as in definition 3 and let S m be as in (.1). Consider the bijection f : A m S m : s = f (A) as in (.) and define the elements of the class A m by A s = f 1 (s). Then every element R R m is uniquely written as R = A s P ρ, where P ρ is an m m permutation matrix produced from permuting the columns of the identity matrix I m by a permutationρ P m. Therefore every element R R m is uniquely determined by a pair(ρ, s) P m S m. Proof. Let A s = f 1 (s) be defined above and let P ρ be an m m permutation matrix associated with apermutationρ =ρ k } m k=1 P m as above. Then supp (A s P ρ ) i } = ρ 1 k : k supp (A s ) i } }, where ρ 1 is the inverse permutation of ρ. ThusA s P ρ R m because A s R m. Therefore the set A s P ρ : (ρ, s) P m S m } is a subset of the class R m. Now we assume that there exists an element R R m not written in the form R = A s P ρ.thenwe define a sequence ρ =ρ i } m : ρ i=1 i 1,...,m} by ρ i = max suppr T i }}, i = 1,...,m and we examine two cases: (i) Let ρ P m and let P ρ 1 be the m m permutation matrix produced from permuting the columns of the identity matrix I m by the inverse permutation ρ 1. Then the matrix RP ρ 1 is an upper unitriangular matrix in R m,sorp ρ 1 A m. Therefore there exists a unique sequence s S m such that RP ρ 1 = A s.thusr = A s P ρ, contradiction. (ii) Let ρ / P m. Then there exists at least one pair (i 0, j 0 ), i 0 = j 0 such that ρ i0 = ρ j0 and so i 0, j 0 } suppr ρi0 }. Taking into account Lemma 1 we deduce that suppr T i 0 }=suppr T j 0 }.Thus R is not invertible, contradiction. Remark 1. Let R R m. In order to compute its corresponding pair (ρ, s) P m S m we work in the following steps: (i) We determine a permutation ρ P m by ρ i = maxsuppr T i }}, i = 1,...,m. (ii) Then we compute the matrix RP ρ 1 (P ρ 1 is a permutation matrix as in Theorem 1). (iii) Finally we define a sequence s S m by i, supp(r P ρ 1) T s(i) = }=i} i. max supp(r P ρ 1) T i } i}}, otherwise Now we deal with the set S m in (.1). This set is enumerated because every element s S m is the sequence of digits of a number N 0,...,m! 1} in the factorial number system m s(i) 1 N = m!. (.5) i! i=1

6 100 N. Atreas, C. Karanikas / Linear Algebra and its Applications 435 (011) Clearly the symmetric group P m and the set S m are equivalent. For completeness we demonstrate this equivalence. Let s S m, then we define recursively a unique element ρ P m by ρ (1) =1} ρ () P... ρ = ρ (m) P m : ρ (k) n = k, n = s(k) ρ (k 1) n, n < s(k) n = 1,...,k. ρ (k 1) n 1 n > s(k) (.6) On the other hand for any ρ = ρ (m) P m we find a unique element s S m by reversing (.6). In fact where s(m k) = ( ρ (m k)) 1 ρ (m k 1) n = m k, k = 0,...,m, ρ (m k) n, n < s(m k) ρ (m k) n+1, n s(m k) We present the following example., n = 1,...,m k 1. Example 1. Let s =1, 1,, 1}. We take the trivial permutation ρ (1) =1}. We consider the element s() = 1 and we enlarge ρ (1) by inserting the value as the 1st element of ρ () =, 1}. Then we take the element s(3) = and we enlarge ρ () by inserting the value 3 as the nd element of ρ (3) =, 3, 1}. Finally we take the element s(4) = 1 and we insert the value 4 as the 1st element of ρ (4) = ρ =4,, 3, 1}. Inversely let us assume that ρ =4,, 3, 1}. We delete the biggest element from ρ, store its position to a new sequence s (1) =1} and we keep the residue ρ (1) =, 3, 1}. Thenwedeletethebiggest element of ρ (1), store its position as the 1st element of s () =, 1} and we keep the residue ρ () =, 1}. We proceed in the same way to produce s = s (4) =1, 1,, 1}. Now we can prove the following. Proposition. Every element R R m can be identified from a pair of natural numbers between 1 and m!. Proof. We take any pair of natural numbers (k, n) so that 0 k, n m! 1. For this selection we use (.5) and we produce a unique pair (σ, s) S m S m.thenweuse(.6) to obtain a unique permutation ρ associated with σ. Finally we use Theorem 1 and we construct a unique element R R m. Remark. Proposition imposes some limitations. In order to convert an element of P m (or S m ) to a number N between 1 and m! and vice versa we need m to be small enough so that N can be held in a machine word; for 3-bit words this means m 1 and for 64-bit words this means m 0. Example. Let m = 7. We consider a pair of natural numbers between 0 and 7! 1, for example we select the pair (1637, 935). Thenweuse(.5)toobtain (1637, 935) (σ, s) S m S m : σ =1, 1,, 4, 4, 6, 7}, s =1, 1,, 1, 3,, 5}.

7 N. Atreas, C. Karanikas / Linear Algebra and its Applications 435 (011) Then we use (.6) toobtainσ ρ =, 3, 1, 5, 4, 6, 7}. Finally we use Theorem 1 to produce a matrix R = A s P ρ.indeed R = A s P ρ = = This matrix is uniquely determined by the pair (1637, 935). In the above calculations the matrix A s is produced from (.3) fors =1, 1,, 1, 3,, 5} whereas the permutation matrix P ρ results from permuting the columns of the identity matrix I m by the permutation ρ =, 3, 1, 5, 4, 6, 7}. 3. Haar-type matrices derived from R matrices Let R R m. In this section, we produce orthonormal matrices from R matrices, see (3.) below.also we establish a subclass R 0 R m of matrices whose elements provide Haar-type matrices via (3.). We consider an m m matrix R =R λ : λ = 1,...,m} in R m and we define a map f : Rm S m : s = f (R) by k, whenever k = max i 1,...,j 1} :supp(r j ) supp(r i ) } s(1) = 1, s(j) =. (3.1) j, whenever supp(r i ) supp(r j ) = for any i = 1,...,j 1 Clearly f is not injective but the restriction of f on Am is injective as we showed in Proposition 1, see (.). For s = f (R) we define the row vectors R λ = R λ a λ R λ q λ q λ, (3.) where a λ = 1 δ λ,s(λ) (δ ij is the Knonecker delta symbol), q λ = R s(λ), s(λ) = λ 1 R s(λ) λ 1 μ=s(λ)+1 δ s(μ),s(λ)r μ, s(λ) < λ 1, (3.3) and is the usual norm on the space C m of all complex valued sequences of length m. Lemma. Let R R m, s = f (R) be as in (3.1) and let qλ be a row vector as in (3.3) for some 1 <λ m such that s(λ) λ 1. Then suppr λ } suppq λ } suppr s(λ) }. (3.4) Proof. Under the above assumptions we define Q λ =δ s(μ),s(λ) R μ : δ s(μ),s(λ) = 1, μ= s(λ) + 1,...,λ 1}.

8 10 N. Atreas, C. Karanikas / Linear Algebra and its Applications 435 (011) Taking into account (3.1) we see that all elements in Q λ are those row vectors of R whose support is a proper subset of the support of the row R s(λ). Furthermore, if we consider s(λ)+1 μ 1 <μ λ 1 and R μ1, R μ Q λ,thensuppr μ1 } suppr μ }=,otherwiser μ / Q λ contradiction. From the above observations and (3.3) wededucethat suppr s(λ) } s(λ) = λ 1 suppq λ }=, suppr s(λ) } R μ Q λ suppr μ } s(λ) < λ 1 and so suppq λ } suppr s(λ) }. Also, from (3.1) we see that suppr λ } suppr s(λ) } and suppr λ } suppr μ }= for any μ = s(λ)+1,...,λ 1. Since q λ = R s(λ) λ 1 μ=s(λ)+1 δ s(μ),s(λ)r μ, necessarily suppr λ } suppq λ }.Butq λ cannot coincide with R λ (otherwise the matrix R would not be invertible), so the result follows. Proposition 3. Let R = R1,..., Rm } be an m m matrix whose row vectors Ri are defined in (3.).Then R is an orthogonal matrix and for any λ, rwithλ = rtherowvectorsof R satisfy R λ Rr = c λ,r Rmaxλ,r} for some double indexed sequence of scalars c =c λ,r }. Proof. Let s = f (R) be as in (3.1) and let R = R1,..., Rm } be defined in (3.). Then Rλ, Rr = R λ, R r a λ R λ q λ q λ, R r a r R r q r R λ, q r + a λ a r R λ q λ R r q r q λ, q r, (3.5) where, is the usual inner product on C m. Let λ = r. Without loss of generality we consider 1 λ<r m and a r = 1. Indeed if a r = 0then the two last terms in (3.5) vanish and so Rλ, Rr = R λ, R r a λ R λ q λ q λ, R r. Since a r = 1 δ r,s(r) by definition (3.), the equality a r = 0 implies that s(r) = r and so by (3.1) suppr λ } suppr r }= for any 1 λ r 1. If s(λ) λ 1, then by (3.4) wealsoobtain suppq λ } suppr r }=,becausesuppq λ } suppr s(λ) } suppr r }=. Therefore Rλ Rr = O = 0 Rmaxλ,r}, where O is the m m zero row and Rλ, Rr =0. Similarly, if s(λ) = λ then a λ = 0 and so Rλ, Rr = R λ, R r =0. Also, Rλ Rr = O = 0 Rmaxλ,r}. Now for 1 λ<r m and a r = 1 (i.e. s(r) <r) we examine the following three cases: (a) suppr s(r) } suppr λ }. This inclusion property coupled with (3.4)leadsto suppr r } suppq r } suppr s(r) } suppr λ } suppq λ } suppr s(λ) } (3.6) provided that s(λ) λ 1. Then from (3.6) we derive the estimates R λ, R r = q λ, R r = R r and R λ, q r = q λ, q r = q r. Substituting these estimates in (3.5)weobtain Rλ, Rr =0. If s(λ) = λ then a λ = 0 and (3.6) becomes suppr r } suppq r } suppr s(r) } suppr λ }. From this inclusion property and the fact a λ = 0, (3.5) becomes Rλ, Rr = R λ, R r R r q r R λ, q r = R r R r q r q r = 0.

9 N. Atreas, C. Karanikas / Linear Algebra and its Applications 435 (011) Moreover if we substitute (3.) in(1.) and use (3.6) weobtain ( R λ Rr = 1 a λ R λ ) ( R q λ r = 1 a λ R λ ) q λ R maxλ,r}. (b) suppr s(r) } suppr λ }=. In this case we recall that suppr r } suppq r } suppr s(r) },so R λ, R r = R λ, q r =0 and (3.5) becomes Rλ, Rr = a λ R λ q λ q λ, R r +a λ R λ q λ R r q r q λ, q r. (3.7) Obviously if a λ = 0wehavenothingtoprove.Leta λ = 0 (i.e. s(λ) < λ). It suffices to compute the inner products q λ, R r and q λ, q r, so we need estimates for suppq λ }.Letsuppq λ } suppr s(r) }.ThensuppR λ } suppr s(r) } by (3.4), contradiction. Therefore we examine two sub-cases: (b 1 ) suppq λ } suppr s(r) }=. Then from (3.6), q λ, R r = q λ, q r =0. Thus (3.7) vanishes and Rλ Rr = O. (b ) suppq λ } suppr s(r) }. Then by (3.6)weget q λ, R r = R r and q λ, q r = q r. Substituting these estimates in (3.7) weobtain Rλ, Rr =0. In this case if we substitute (3.) in(1.) and use (3.6) we obtain R λ Rr = R λ q λ R r = R λ R q λ maxλ,r}. (c) suppr s(r) } suppr λ }. In this case r >λ s(λ) s(r), s(r) <r 1 and suppr λ } suppq r }= (see (3.3)), so R λ, R r = R λ, q r =0. Then (3.5) becomes Rλ, Rr = a λ R λ q λ q λ, R r +a λ R λ q λ R r q r q λ, q r. (3.8) Obviously if a λ = 0 we have nothing to prove. Let a λ = 0. Then s(λ) < λ and consequently s(r) s(λ) λ 1 <λ<r, so it suffices to examine the following sub-cases: (c 1 ) s(r) = s(λ) = λ 1. Then from (3.3) wehavesuppq λ }=suppr s(λ) }=suppr s(r) }.Thus q λ, R r = R r and q λ, q r = q r,so(3.8) vanishes. Working as in (b ) we obtain R λ Rr = R λ q λ (c ) s(r) <s(λ) = λ 1. R r = R λ R q λ maxλ,r}. Then suppq λ }=suppr s(λ) } suppr s(r) }. In this case using (3.3) weobtainsuppq λ } suppq r } =,thus q λ, R r = q λ, q r = 0 and all terms in (3.8) vanish. Therefore R λ Rr = O.

10 104 N. Atreas, C. Karanikas / Linear Algebra and its Applications 435 (011) (c 3 ) s(r) s(λ) < λ 1. Then suppq λ } suppr s(λ) } suppr s(r) } and we work as in (c ). Combining the results obtained in (a) (c) the proof is complete. Theorem. Let R 0 be a matrix in R m so that its corresponding sequence s = f (R0 ) in (3.1) satisfies s(i) <i for any i =,...,m. If R0 } is the orthogonal matrix produced from R 0 via (3.), then the matrix ( R0 ) λ 1 ( R0 ) λ : λ = 1,...,m is a Haar-type matrix. Proof. It suffices to check conditions (i) (iii) of definition 1 for the matrix R0 produced from a matrix R 0 R m as in (3.). From Proposition 3 the matrix R0 is orthogonal and satisfies condition (i). Let s = f (R0 ) be as in (3.1). The assumption s(λ) < λ (λ =,...,m) and the definition (3.) combined with the inclusion property (3.4) and the fact that all non-zero entries of R 0 equal 1 show that all non-zero entries of the row ( R0 ) λ belong in the set 1 (R 0) λ, (R 0) λ } q λ q λ which contains only two elements in ( 1, 1) with alternate signs. Therefore condition (ii) is satisfied too. Finally condition (iii) is satisfied because by (3.4) suppr λ } suppq λ } and so 1 (R 0) λ, j supp(r q λ 0 ) λ } ( R0 ) λ,j = (R 0) λ, j suppq q λ λ (R 0 ) λ } 0, otherwise. (3.9) Therefore ( m ( R0 ) λ,j = 1 (R 0) λ q λ j=1 ) (R 0 ) λ (R 0) λ q λ qλ (R 0) λ = 0, λ =,...,m as a result of (3.9). Notice that the assumption s(λ) < λ (λ =,...,m) is essential to obtain ( R0 ) 1 = (R 0 ) 1 =1,...,1} (combine with (3.1) and (3.)). Therefore the orthonormal matrix obtained from R 0 by normalizing its rows is a Haar-type matrix. Remark 3. Let H m be a Haar-type matrix. In order to find its corresponding matrix R 0 R m we replace all negative entries of H m with zeros and all positive entries of H m with ones. Then we work as in Remark 1. On the other hand, for any pair (ρ, s) P m S m (where s satisfies s(i) < i for any i =,...,m) there corresponds a unique matrix R 0 = A s P ρ as in Theorem 1 and a unique orthogonal matrix R0 determined from (3.). The orthonormalization of R0 yields a Haar-type matrix. For example the orthogonal matrix R0 corresponding to the matrix R 0 of Example is obtained from the following process: (i) Given R 0 (as in Example ) we estimate its corresponding sequence s = f (R0 ) from (3.1). It turns out that s =1, 1,, 1, 3,, 5}.

11 N. Atreas, C. Karanikas / Linear Algebra and its Applications 435 (011) (ii) Then we use (3.) to produce an orthogonal matrix R0.Infact /7 /7 5/7 /7 5/7 /7 /7 3/5 /5 0 /5 0 3/5 /5 R 0 = 0 0 1/ 0 1/ /3 0 1/ /3 1/ / / 0 0 1/ Then we normalize all rows( R0 ) i, i = 1,...,7 and we get an orthonormal Haar-type matrix identified from the pair (1637, 945) as in Example. References [1] N. Atreas, C. Karanikas, P. Polychronidou, A class of sparse unimodular matrices generating multiresolution and sampling analysis for data of any length, SIAM J. Matrix Anal. Appl. 30 (1) (008) [] N. Atreas, C. Karanikas, Multiscale Haar unitary matrices with the corresponding Riesz products and a characterization of Cantor-type languages, J. Fourier Anal. Appl. 13 () (007) [3] W.L. Briggs, Henson Van Emden, The DFT. An Owner s Manual for the Discrete Fourier Transform, SIAM, Philadelphia, [4] C. Cattani, Haar wavelet-based technique for sharp jumps classification, Math. Comput. Modelling 39 ( 3) (004) [5] S.L. Chen, H.C. Lai, K.C. Ho, Identification of linear time varying systems by Haar wavelet, Int. J. Systems Sci. 37 (9) (006) [6] Gi-Sang Cheon, B.L. Shader, Sparse orthogonal matrices and the Haar wavelet transform, Discrete Appl. Math. 101 (1 3) (000) [7] F. Dubeau, S. Elmejdani, R. Ksantini, Non-uniform Haar wavelets, Appl. Math. Comput. 159 (3) (004) [8] K. Egiazarian, J. Astola, Tree-structured Haar transforms. Non-linear image processing and pattern recognition, J. Math. Imaging Vision 16 (3) (00) [9] M. Girardi, W. Sweldens, A new class of unbalanced Haar wavelets that form an unconditional basis for L p on general measure spaces, J. Fourier Anal. Appl. 3 (4) (1997) [10] B. Golubov, A. Efimov, V. Skvortsov, Walsh Series and Transforms. Theory and Applications, Kleuwer Academic Publishers, [11] R.A. Horn, C.R. Johnson, Topics in Matrix Analysis, Cambridge University Press, New York, 1991.