STABILITY AND LINEAR INDEPENDENCE ASSOCIATED WITH SCALING VECTORS. JIANZHONG WANG y

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1 STABILITY AND LINEAR INDEPENDENCE ASSOCIATED WITH SCALING VECTORS JIANZHONG WANG y Abstract. In this paper, we discuss stability and linear independence of the integer translates of a scaling vector = (1; ; r) T, which satises a matrix renement equation nx (x) = P k (2x? k); k=0 P where (P k ) is a nite matrix sequence. We call P (z) = 1 2 P k z k the symbol of. Stable scaling vectors often serve as generators of multiresolution analyses (MRA) and therefore play an important role in the study of multiwavelets. Most useful MRA generators are also linearly independent. The purpose of this paper is to characterize stability and linear independence of the integer translates of a scaling vector via its symbol. A polynomial matrix P (z) is said to be two-scale similar to a polynomial matrix Q(z) if there is a polynomial matrix T (z) such that P (z) = T (z 2 )Q(z)T?1 (z): This kind of factorization of P (z) is called two-scale factorization. We give a necessary and sucient condition, in terms of two-scale factorization of the symbol, for stability and linear independence of the integer translates of a scaling vector. Key words. stability, linear independence, scaling vectors, multiwavelets, multiresolution analysis, two-scale similarity AMS subject classications. Primary 41A63, 42C05. Secondary 42A05, 1. Introduction. In this paper, we discuss stability and linear independence of the integer translates of a scaling vector. A distribution vector (x) = ( 1 (x); 2 (x); ; r (x)) T ; x 2 R; is said to be a scaling vector if it is compactly supported and satises a matrix renement equation (1.1) (x) = X k2z P k (2x? k); where the matrix sequence (P k ) k2z is called a mask of : Taking the Fourier transform of both sides of equation (1.1), we obtain (1.2) ^(!) = P (z)^(!=2); z = e?i!=2 ; where P (z) := 1 2 P Pk z k is a symbol of : We call ^(0) the moment (of order 0) of since ^(0) = R R (x)dx: When ^(0) = 0; we call a zero-moment scaling vector. We will characterize stability and linear independence of the integer translates of a scaling vector via a special factorization of its symbol. Our study of scaling vectors is based on shift-invariant spaces. Hence, we rst introduce some notions and results in the theory of shift-invariant spaces. Let S be a linear space of distributions on R. We say that S is shift-invariant if f 2 S =) f(? j) 2 S 8j 2 Z: This paper is supported by NSF Grant DMS y Departmentof Mathematics and Information Sciences, Sam Houston State University, Huntsville, TX ( mth jxw@shsu.edu Phone: (409) Fax: (409) ) 1

2 2 JIANZHONG WANG We are interested in shift-invariant spaces generated by entries of a scaling vector. In this paper, since only scaling vectors are considered, any distribution vector is assumed to be compactly supported. Let l denote the space containing all sequences of complex numbers and l 0 denote the space of all compactly supported sequences in l. For a distribution vector ; the semi-convolution of with a vector sequence a := (a k ) 2 (l) r ; denoted by a; is dened by We dene a = rx X j=1 k2z a j;k j (x? k): S 0 () := f a; a 2 (l 0 ) r g ; and, if 2 (L p ) r ; 1 p 1, S() := f a; a 2 (l) r g ; S p () := Clos L p(s 0 ()): It is obvious that these sets are shift-invariant spaces of functions and distributions. The vector is called the generator of S() (S 0 (); S p ()). P For 2 (L p ) r ; 1 p 1, its L p r norm is dened by kk p = ( i=1 k ik p p )1=p. Similarly, for a vector sequence a P 2 (l p ) r, its l p r P norm is kak p = ( j=1 k2z ja j;kj p ) 1=p. The integer translates of a distribution vector 2 (L p ) r are said to be l p -stable, (1 p 1), if there exist two positive constants C 1 and C 2 such that, for any a 2 (l p ) r ; (1.3) C 1 kak p k ak p C 2 kak p : It is known that (1.3) holds if and only if the following r sequences (1.4) (^ l (! + 2k)) k2z ; l = 1; 2; ; r; are linearly independent for all! 2 R (see [12]), where ^f denotes the Fourier transform of function f: ^f(!) := Z x2r f(x) exp(?ix!) dx: Note that the linear independence of sequences (1.4) does not depend on p and that it does make sense even for the distribution vectors not in (L p ) r. (Recall that distribution vectors in this paper are assumed to be compactly supported. Hence their Fourier transforms are entire function vectors.) Therefore, we give the following denition. A distribution vector is said to have stable integer translates if the sequences in (1.4) are linearly independent for all! 2 R ([12] ). Another important notion for distribution vectors is linear independence. A distribution vector is said to have linearly independent integer translates if, for any a 2 (l) r, a = 0 =) a = 0: Jia and Micchelli in [12] proved the following result. The integer translates of are linearly independent if and only if the sequences in (1.4) are linearly independent

3 LINEAR INDEPENDENCE OF SCALING VECTORS 3 for all! 2 C. Hence the linear independence of the integer translates of implies the stability of the integer translates of. For a distribution vector, we also introduce the notion of nitely linear independence. A distribution vector is said to have nitely linearly independent integer translates if, for any a 2 (l 0 ) r, a = 0 =) a = 0: It is clear that if the integer translates of are stable (or linearly independent), then they are also nitely linearly independent. But the converse is not true. For convenience, in the rest of the paper, we will simply say that is stable (linearly independent, nitely linearly independent) instead of that has stable (linearly independent, nitely linearly independent) integer translates. Stability and linear independence are important properties of distribution vectors. Bases of shift-invariant spaces are often required to be stable so that the duality principle can be applied eciently (see [2]). In wavelet theory, generators of multiresolution analyses (MRA) are stable. Moreover, most useful generators of MRA are also linearly independent. Hence it is desirable to give a criterion for stability and linear independence of scaling vectors in terms of their symbols. Scaling vectors are discussed in several papers for dierent purposes. Goodman and Lee [6] and Goodman, Lee, and Tang [7] discussed scaling vectors in generality and constructed multiwavelets using MRA generated by scaling vectors. In [8], Heil and Colella discussed the existence and uniqueness of the solution of a matrix renement equation, and the regularity of the solution as well. Discussions of the supports of scaling vectors can be found in Massopust, Ruch, and Van Fleet [15], Ruch, So, and Wang [19], and So and Wang [20]. Approximation order provided by scaling vectors was studied by Heil, Strang, and Strela [9], Jia, Riemenschneider, and Zhou [13], and Plonka [16]. In [16], Plonka introduced two-scale similarity for the symbol of a scaling vector to characterize its approximation order. This method was also successfully applied in the construction of the scaling vectors with required regularity (see [3]) and with symmetry (see [17] ). In this paper, we use two-scale similarity to characterize stability, linear independence, and nitely linear independence of scaling vectors. In section 2, we discuss stability and linear independence of a distribution vector. A result of [11] shows that any generator of a shift-invariant space can be obtained as a nitely transformed linearly independent generator in the same space. Using this result, we obtain necessary and sucient conditions for stability (linear independence) of a scaling vector in terms of properties of the corresponding transform. In Section 3, we discuss twoscale similarity of the symbol of a scaling vector. Based on the results in Section 2, we derive necessary and sucient conditions for stability (linear independence) of a scaling vector in terms of two-scale similarity. Some examples are given in Section Transform of Generators of A Shift-Invariant Space. In this section, we discuss stability and linear independence of a distribution vector. We denote by dim the dimension (i.e. the number of the components) of a distribution vector : We start with the following result in [11]. Theorem 2.1 (Jia). Let be a distribution vector. Then there exists a linearly independent distribution vector 2 S() such that dim dim; S 0 ( ); and S( ) = S(): We will call dim the index of and denote it by # : It is obvious that # dim; and # < dim if and only if is nitely linearly dependent. Let and be two generators of a shift-invariant space S: Then there is a

4 4 JIANZHONG WANG rational function matrix T (z) such that ^(!) = T (e?i! ) ^(!); where T (z) is called the transform (matrix) from to : By Theorem 2.1, If 2 S() is linearly independent, then the transform T (z) is a Laurent polynomial matrix. We now introduce some notations for transform matrices. In the rest of this paper, a polynomial always means a Laurent polynomial. We denote the polynomial ring over C by P and denote the eld of rational functions over C by R. We shall simply call a polynomial matrix a P- matrix and call a matrix with rational function entries an R-matrix. The set of all rs P-matrices is denoted by P rs ; and the set of all r s R-matrices is denoted by R rs : When r = s; we use P r (R r ) instead of P rr (R rr ). A matrix P (z) 2 P r is said to be invertible if there is an R-matrix R(z)(:= P?1 (z)) 2 R r such that R(z)P (z) = I; where I is the identity matrix, and P (z) is said to be invertible everywhere if P (z) is invertible for any z 2 C n f0g : Hence an invertible polynomial matrix may be not invertible everywhere. If P (z) is invertible everywhere, then det P (z) = cz k with c 6= 0; which implies that its inverse P?1 (z) is also a polynomial matrix. For a nonsquare P-matrix, its rank and its everywhere rank can be dened in the same way. For an arbitrary P-matrix, the following three types of operations are called elementary row (column) operations. 1. Multiplying the i-th row (column) by cz k ; where c is a nonzero constant and k is an integer. 2. Interchanging the i-th and the j-th row (column). 3. Adding the product of p(z) 2 P and the j-th row(column) to the i-th row (column), where i 6= j: A matrix that performs an elementary operation is said to be an elementary P-matrix and a nite product of elementary matrices is said to be a fundamental P-matrix. It is clear that a P-matrix is fundamental if and only if it is invertible everywhere. Recall that a (non-square) everywhere full rank P-matrix can always be extended to a (square) fundamental P-matrix. Hence, for convenience, we also call a (non-square) everywhere full rank matrix a fundamental P-matrix. Fundamental P-matrices play an important role in the study of transforms between the generators of a shift-invariant space. The rst observation is that if dim = dim and the transform from to is a fundamental matrix, then S() = S( ) and is linearly independent (nitely linearly independent, stable) if and only if is linearly independent (nitely linearly independent, stable). Generally, let and be two generators of a shift-invariant space S. The transform T from to is not necessarily fundamental. However, since both and are compactly supported, T is at least an R-matrix, and T reduces to a P-matrix if 2 S 0 ( ): Now we begin to discuss nitely linear dependence with the following lemma. Lemma 2.2. If is nitely linearly dependent, then there is a nitely linearly independent distribution vector S 0 (); such that dim = # ; S 0 ( ) = S 0 (); and therefore the transform matrix from to is fundamental. Proof. Set r = dim and s = # : By Theorem 2.1, there is a linearly independent generator I S() such that dim I = s and S 0 ( I ): Therefore, there is a full rank matrix T (z) 2 P rs such that ^(!) = T (e?i! )^ I (!): Let D(z) be the Ds Canonical Smith's form of T (z): Since T (z) has full rank, we have D(z) = ; O where D s (z) = diagfd 1 (z); d 2 (z); ; d s (z)g with d i (z) 6= 0; 1 i s, and O is the (r? s) s zero matrix. Furthermore, there are two fundamental matrices L(z) 2 P r

5 and R(z) 2 P s such that LINEAR INDEPENDENCE OF SCALING VECTORS 5 (2.1) T (z) = L(z)D(z)R(z): Dening I by ^ I(!) = D(e?i! )R(e?i! )^ I (!); we obtain I = ( 1 ; 2 ; ; s ; 0; ; 0) T : Let = ( 1 ; 2 ; ; s ) T and L s (z) be the r s matrix containing the rst s columns of L(z): Then is nitely linearly independent and ^(!) = L s (z) ^ (!); which implies that S 0 ( ): On the other hand, we also have ^(z) = L^s(e?i! )^(!); where L^s(z) is the sr matrix containing the rst s rows of L?1 (z): Since L(z) is fundamental, so are L?1 (z); L s (z); and L^s(z). Hence S 0 (): The proof is completed. The following theorem characterizes nitely linear dependence of distribution vectors. Theorem 2.3. A distribution vector (dim = r) is nitely linearly dependent if and only if there exists a non-identity matrix T (z) 2 P r such that ^(!) = T (e?i! )^(!): Proof. \ =) 00 : Set # = s. Since is nitely linearly dependent, s < r: Let I ; ; L(z); L s (z) and L^s(z) be dened as in the proof of the above lemma. Then we have ^(!) = L s (e?i! )L s^(e?i! )^(!) and L^s(e?i! )L s (e?i! ) 6= I: \ (= ": Suppose that there is a P-matrix T (z) 6= I; such that Then we have ^(!) = T (e?i! )^(!): 0 = (I? T (e?i! ))^(!); which implies that is nitely linearly dependent. Now we characterize linear independence of a distribution vector. Theorem 2.4. is linearly dependent if and only if there is a distribution vector S() and a non-fundamental P-matrix P (z) such that (2.2) ^(!) = P (e?i! ) ^ (!): Proof. \ =) 00 : If is linearly dependent, P then, by [12], there is a non-zero vector r (a i ) such that the distribution = i=1 a i i is linearly dependent. Without loss of generality, we may assume a 1 = 1: By [18], there is a polynomial p(z) with at least one zero in C n f0g and a distribution g in S() such that ^(!) = p(e?i! )^g(!): Therefore, we have 0 ^ 1 (!) ^ 2 (!). ^ r (!) 1 C A = 0 p(e?i! )?a 2?a r C A ^g(!) ^ 2 (!). ^ r (!) where p(z) has at least one zero in C n f0g : Hence the transform matrix is not fundamental. 1 C A ;

6 6 JIANZHONG WANG \ (= ": Assume there exists a distribution vector S() and a non fundamental P-matrix P (z) such that (2.2) holds. Write dim = r and dim = s: If r > s; then is nitely linearly dependent. We now assume r s: Choose two square fundamental matrices L(z) and R(z) so that P (z) = L(z)D(z)R(z); where D(z) is the Canonical Smith's form of P (z). Setting ^ l (!) = L?1 (e?i! )^(!) and ^ r(!) = R(e?i! ) ^(!) respectively, we have ^ l (!) = D(e?i! ) ^ r(!): Let D =? D r O ; where D r (e?i! ) = diagfd 1 (e?i! ); ; d r (e?i! )g and O is the r (s? r) zero matrix. Since P (z) is non-fundamental, at least one of d 1 (e?i! ); ; d r (e?i! ) vanishes at an! 0 2 C: Hence, one of the components of l is linearly dependent. It follows that is also linearly dependent. Corollary 2.5. The transform between two linearly independent generators (of a shift invariant space) must be a fundamental matrix. We have similar results for stability of a distribution vector. Theorem 2.6. Let? = fz 2 C; jzj = 1g : is unstable if and only if there is a distribution vector S() such that the transform from to is a P-matrix which is singular at a value z 2?: Corollary 2.7. The transform between two stable generators (of a shift-invariant space) is an R-matrix invertible over?: 3. Stability and Linear Independence of A Scaling Vector. We now characterize the linear independence (nitely linear independence, stability) of a scaling vector in terms of its symbol. We are interested in the scaling vector with a P- matrix symbol. It is clear that if a scaling vector has symbol P (z); then its moment ^(0) satises ^(0) = P (1)^(0); which implies that ^(0) is either a zero vector or a 1-eigenvector of P (1): If P (1) has more than one 1-eigenvectors, then dierent scaling vectors may share one symbol. We now assume that is a solution of equation (1.1) with a non vanished moment v; which is a 1-eigenvector of P (1): If equation (1.1) also has a non-trivial zero-moment solution 0, then all functions + c 0 ; c 2 C; have the same symbol P (z) and the same moment v: On the other hand, if two scaling vectors 1 and 2 have the same symbol and the same moment, then 1? 2 is a zero-moment scaling vector. Hence, we have the following conclusion. Let v 6= 0 be a 1-eigenvector of P (1): The solution of (1.1) with a given moment v is unique if and only if equation (1.1) has no non-trivial zero-moment solution. As [8] mentioned, if equation (1.1) has nontrivial zero-moment solutions, then P (1) must have eigenvalue 2, ( 1; 2 Z). We now denote by E(M) the set of all eigenvalues of matrix M; and let and OP r = A 2 P r ; 2 62 E(A(1)); 2 Z; 1 ; SP r = fa 2 OP r ; 1 2 E(A(1))g : Therefore, if P 2 SP r and v is a 1-eigenvector of P (1); then equation (1.1) has a unique solution with the moment v: In contrast with the fact that several scaling vectors may share a symbol, a scaling vector may have more than one symbols. However, the following lemma conrms the uniqueness of the symbol of a nitely linearly independent scaling vector. Lemma 3.1. The symbol of a scaling vector (with dim = r ) is unique in P r if and only if is nitely linearly independent.

7 LINEAR INDEPENDENCE OF SCALING VECTORS 7 Proof. Assume is nitely linearly independent. Then dim = #. If satises the following two scaling equations (x) = X k2z C k (2x? k) and (x) = X k2z D k (2x? k) ; then (3.1) 0 = X k2z(c k? D k ) (2x? k) : Since () is nitely linearly independent, so is (2). By (C k? D k ) 2 (l 0 ) rr ; we have C k? D k = 0: The converse is trivial. Since the relation between a scaling function and its symbol(s) has been cleared, we are ready to characterize scaling vectors via their symbols. The notion of two-scale similarity plays an important role in characterizing scaling vectors ([17]). A matrix P 2 P r is said to be two-scale similar to Q 2 P r if there exists an invertible matrix T 2 P r such that (3.2) P (z) = T (z 2 )Q(z)T?1 (z): The matrix T (z) in (3.2) is called two-scale similar transform. Since the matrix T?1 (z) need not necessarily to be a polynomial matrix, T (z) may be singular at some points in C n f0g. Hence, P being two-scale similar to Q does not imply Q being two-scale similar to P: However, if P is two-scale similar to Q with a fundamental transform, then Q is also two-scale similar to P. In this case, we say that P and Q are fundamentally two-scale similar. As shown in Section 2, and with S() = S( ) have same linear independence (nitely linear independence, stability) if and only if their symbols P and Q are fundamentally two-scale similar. The following fact is often used in the rest. If P is fundamentally two-scale similar to Q with the two-scale transform T; and satises the two-scaling equation then the scaling vector I determined by ^(!) = P (e?i!=2 )^(!=2); (3.3) ^ I(!) = T?1 (e?i! )^(!) satises (3.4) ^ I(!) = Q(e?i!=2 ) ^ I(!=2): The following theorem characterizes a nitely linearly dependent scaling vector. Theorem 3.2. Let P 2 SP r be a symbol of (dim = r). Then is nitely linearly dependent if and only if the following two conditions are satised. (1) P (z) is fundamentally two-scale similar to a matrix (3.5) Qs (z) Q(z) = Y (z) 0 X(z)

8 8 JIANZHONG WANG where Q s 2 SP s ; X 2 OP r?s ; and s < r: (2) Let T be the two-scale similar transform: P (z) = T (z 2 )Q(z)T?1 (z): Then the last r? s components of T?1 (1)^(0) vanish, that is (3.6) T?1 (1)^(0) = (u 1 ; ; u s ; 0; ; 0) T : Furthermore, if P (1) has only a single 1-eigenvector, then the solution of (1.1) with ^(0) 6= 0 is unique, and therefore is nitely linearly dependent if and only if (1) holds. Proof. \ =) ": Assume is nitely linearly dependent with a symbol P 2 SP r. By Lemma 2.2, there is a nitely linearly independent scaling vector such that ^(!) = T (e?i! ) ^ (!); where T (z) is an r s fundamental matrix. Let I = ( 1; ; s; 0; ; 0) T ; and T (z) be an r r fundamental matrix extended from T (z): Then ^(!) = T (e?i! ) ^ I(!): Let Q(z) = T?1 (z 2 )P (z)t (z): Then, setting z = e?i!=2 ; we obtain ^ (!) = T?1 (z 2 )^(!) = T?1 (z 2 )P (z)^(!=2) = T?1 (z 2 )P (z)t (z) ^ (!=2) = Q(z) ^ (!=2); which implies that Q(z) is a symbol of I : Writing Qs (z) Y (z) Q(z) = W (z) X(z) ; we have (3.7) ^ (!) 0 Qs (z) = W (z) Y (z) X(z) ^ (!=2) 0 ; z = e?i!=2 : From (3.7), we obtain W (z) ^ (!=2) = 0: Since is nitely linearly independent, by Lemma 3.1, W (z) = 0: We now prove Q s 2 SP s : From (3.7), we have (3.8) ^(!) = Q s (e?i!=2 ) ^ (!=2); which derives ^(0) = Q s (1) ^(0) and therefore 1 is an eigenvalue of Q s (1): Note that matrix P (1) is similar to matrix Q(1): Hence they have the same set of eigenvalues. However, all eigenvalues of both Q s (1) and X(1) are also eigenvalues of Q(1): Therefore, we have the following. P 2 SP r () Q 2 SP r () Q s 2 SP s and X 2 OP r?s ; which implies (1). Choosing! = 0 in ^(!) = T (e?i! ) ^ I(!); we obtain T?1 (1)^(0) = ( ^1(0); ; ^s(0); 0 ; 0) T ; which is (3.6). \ (= ": Let be a scaling vector whose symbol P is fundamentally two-scale similar to a Q(z) : P (z) = T (z 2 )Q(z)T?1 (z);

9 LINEAR INDEPENDENCE OF SCALING VECTORS 9 where Qs (z) Y (z) Q(z) = 0 X(z) ; s < r; with Q s 2 SP s and X 2 OP r?s ; and T?1 (1)^(0) = (u 1 ; ; u s ; 0; ; 0) T := u I : We prove that is nitely linearly dependent. Note that Besides, Q s 2 SP s and X 2 OP r?s =) Q 2 SP r : Q(1)u I = T?1 (1)P (1)T (1)u I = u I : Hence there is an unique distribution I = ( I;1 ; ; I;r ) T satisfying equation (3.4) with the moment u I : Write u = (u 1 ; ; u s ) T : By (3.2) and (3.5), we have Q s (1)u = u: Let = ( 1 ; ; s ) T be the unique solution of (3.8) with the moment u: Then the scaling vector ( 1 ; ; s ; 0; ; 0) T is a solution of (3.4) with the moment u I : Since the solution of (3.4) with the moment u I is unique, we have I = ( 1 ; ; s ; 0; ; 0) T : By ^(!) = T (e?i! ) ^ I(!); is nitely linearly dependent. Finally, we prove that if P (1) has only a single 1-eigenvector, then (2) will be unconditionally satised. In fact, if P (1) has only a single 1-eigenvector, so do Q and Q s (z): Let u = (u 1 ; ; u s ) T be the 1-eigenvector of Q s. Then u I = (u 1 ; ; u s ; 0; ; 0) T is the only 1-eigenvector (without counting the scalar multiples) of Q(1): Recalling that ^(0) is the only 1-eigenvector (without counting the scalar multiples) of P (1); we have T?1 (1)^(0) = cu I = (cu 1 ; ; cu s ; 0; ; 0) T ; which implies that (2) is true. Remark 1. If P (1) has more than one (linearly independent) 1-eigenvectors, then the solutions of equation (1.1) corresponding to dierent moments (without counting the scalar multiples) may have linearly independent properties. We illustrate this phenomenon in the following example. Example 1. Let ( 1+z P (z) = 2 )2 0 0 ( 1+z : 2 )4 It is clear that P (z) 2 SP 2 ; and vector (0; 1) and (1; 1) are two linearly independent 1-eigenvectors of P (1): Obviously, P (z) is two-scale fundamentally similar to itself with transform T (z) = I: Denote by N m (x) the m-th order cardinal B-spline ([2]). The scaling vectors 1 = (N 2 (x); 0) T and 2 = (N 2 (x); N 4 (x)) T are solutions of (1.1) with the moments ^ 1 (0) = (1; 0) T and ^ 2 (0) = (1; 1) T respectively. Note that 1 is nitely linearly dependent but 2 is not. Note also that the vector (1; 0) T satises (3.6) while (1; 1) T does not. Remark 2. If, in Theorem 3.2, P (z) is merely assumed to be in P r ; then the solution of equation (1.1) with a certain mean value is not necessarily unique. In this case, if a solution of (1.1) with ^(0) 6= 0 is nitely linearly dependent, then its symbol P is still fundamentally two-scale similar to a matrix Q in the form (3.5), but Q s is no longer necessarily in SP r and X is no longer necessarily in OP r?s : On the

10 10 JIANZHONG WANG other hand, if one of 's symbols is fundamentally two-scale similar to Q in the form (3.5) with a matrix X such that the renement equation ^F(!) = X(e?i!=2 ) ^F(!=2) has only the trivial solution F = 0 with ^F(0) = 0; and ^(0) satises the condition (2) in 3.2, then is nitely linearly dependent. The proof of this remark is similar to that for Theorem 3.2. According to Theorem 3.2, from the set of symbols of a nitely linearly dependent scaling vector, we can select a relatively simple symbol for it. Corollary 3.3. If ( dim = r) is nitely linearly dependent and # = s and one of its symbols is in SP r ; then has a symbol fundamentally two-scale similar to a matrix Qs (z) 0 (3.9) : Proof. Let P be a symbol of : By Theorem 3.2, P is fundamentally two-scale similar to a matrix Q in the form (3.5) with a two-scale transform matrix T, and T?1 (1)^(0) = (u 1 ; ; u s ; 0 ; 0) T : Let I be dened by ^(!) = T (e?i! ) ^ I(!): Since X 2 OP r and we have, ^ I(0) = (u 1 ; ; u s ; 0; ; 0) T ; I = ( 1 ; ; s ; 0; ; 0) T which satises Setting we have Qs (e?i!=2 ) 0 P(z) = T 2 (z) Qs (z) 0 ^ (!=2) = ^(!): T?1 (z); ^(!) = P (e?i!=2 )^(!=2); where P(z) is a symbol of. We now characterize linear independence of a scaling vector. First, we prove the following lemma. Lemma 3.4. The symbol of a linearly independent scaling vector is a P-matrix. Proof. Assume the scaling vector is linearly independent, so is (2). By [22], a compactly supported function in S((2)) is also in S 0 ((2)). Hence 2 S 0 ((2)) and therefore its symbol is a P-matrix. The following theorem characterizes linear independence of a scaling vector. Theorem 3.5. Assume the scaling vector with non-zero moment is nitely linearly independent and its symbol is a P-matrix P 2 P r (SP r ): Then is linearly dependent if and only if P is two-scale similar to a matrix Q 2 P r (SP r ) with a non-fundamental two-scale similar transform T such that (3.10) rank T (1) = rank T (1); ^(0) :

11 LINEAR INDEPENDENCE OF SCALING VECTORS 11 Remark 3. If the matrix T (1) is non-singular, then the condition (3.10) is always true. Hence the condition (3.10) is eective whenever T (1) is singular. Note that the condition (3.10) is equivalent to \there is a u 2 C r such that T (1)u = ^(0)". To prove this theorem, we need some notations. Let m > 1 be an odd integer, and let h m be the smallest positive integer such that 2 hm 1(mod m). For a primitive m-th root of unit z 0, we call p m (z) = (z 0? z)(z 2 0? z) (z2hm?1 0? z) an m-cycle polynomial. Since z0 2hm = z 0 ; it can be veried that p m (z0 2l ) = 0 for any integer l 0 and therefore p m (z 2 ) = p m (z)p m (?z): We say that p(z) has m- cycle zeros if p m (?z) (NOT p m (z)!) is a factor of p(z): When we will not stress the index m; m-cycle zeros and m-cycle polynomial are simply called cycle zeros and cycle polynomial respectively. A polynomial p(z) is said to have symmetric zeros if there is an 2 C; 6= 0 such that p() = p(?) = 0: Note that any polynomial p(z) has an unique factorization (3.11) p(z) = cz k ly j=1 p mj (z) ty l=1 (z? z i ); where p mj ; j = 1; ; l; are all cycle polynomials and z i 6= 0; i = 1; ; t: Lemma 3.6. Assume the scaling vector 's symbol P 2 P r is two-scale similar to a P-matrix Q; (3.12) with the two-scale transform matrix (3.13) P (z) = T c (z 2 )Q(z)T c?1 (z); T c (z) = diag(1; 1; ; c(z); ; 1); {z } j?1 where c(z) is either (z? c), c 6= 0; or an m-cycle polynomial p m (z): Also assume ^ j (0) = 0 whenever c(z) = z? 1: Then there is a compactly supported function 2 S() such that ^ j (!) = c(e?i! ) ^(!); and therefore is linearly dependent. Furthermore, if P 2 SP r then Q 2 SP r : Proof. Without loss of generality, we assume j = 1. By (3.12) and (3.13), we have (3.14) ^ 1 (!) = p 11 (e?i!=2 )^ 1 (!=2) + rx i=2 c(e?i! )q i1 (e?i!=2 )^ i (!=2); where p 11 (z) = q 11 (z) c(z2 ) c(z) : If c(z) = z? c; c 6= 0 and 1; then c(z) can not be a factor of c(z 2 ): It follows that c(z 2 ) is a factor of p 11 (z): Let! 0 satisfy e?i!0? c = 0: We have, for an integer k; ^ 1 (! 0 + 2k) = p 11 ((?1) k e?i! 0 2 )^1 (! k) + rx i=2 c(e?i!0 )q i1 ((?1) k e?i! 0 2 )^i (! ) = q 11 ((?1) k e?i!0=2 c(e?i!0 ) ) c((?1) k e?i!0=2 ) ^ 1 (! k) = 0:

12 12 JIANZHONG WANG In the case c(z) = z? 1; by the assumption, ^ 1 (0) = 0: For k 6= 0; writing k = 2 l j with odd j; we have, by (3.14), ^ 1 (2k) = p 11 (e?ik )^ 1 (k) = (e?i2l j + 1)q 11 (e?i2l j )^ 1 (2 l j) = (2q 11 (1)) l (e?ij + 1)q 11 (?1)^ 1 (j) = 0: We now prove ^ 1 (! 0 +2k) = 0; k 2 Z; for some! 0 2 R in the case c(z) = p m (z): The proof is similar to that one of Theorem 1 in [14]. For reader's convenience, we include it here. Let z 0 be a primitive mth root. Then z 0 has the form e?i2n=m, where n is an integer relatively prime to m; and p m (e?i2d+1 n=m ) = 0; for all integers d 0: We claim that for all (3.15) ^ 1 (2n=m + 2k) = 0: To prove (3.15), we write n + km in the form 2 l j; where l is a nonnegative integer and j is an odd integer. Recalling that p m (?z) = p m (z 2 )=p m (z); we have ^ 1 ( 2n m + 2k) = ^ 1 (2 l+1 j=m) = p 11 (e?i2l j=m )^ 1 ( 2l j m ) + rx i=2 = q 11 (e?i2lj=m ) p m(e?i2 l+1 j=m ) p m( e?i2l j=m ) ^ 1 ( 2l j m ) = p m(? e?i2lj=m )q 11 (e?i2lj=m )^ 1 ( 2l j m ) = ly t=0 p m (?e?i2t j=m ) ly t=0 p m (e?i2l+1 j=m )q i1 (e?i2l j=m )^ i ( 2l j m ) q 11 (?e?i2t j=m )! ^ 1 ( j m ): Hence in order to prove (3.15) it suces to show that p m (?e?ij=m ) = 0: For this purpose, we invoke Euler's theorem to nd an integer s > l such that 2 s 1(modm): It follows that (3.16) j 2 s j 2 s?l (2 l j) 2 s?l n (mod m): Since j is odd, by (3.16), j? 2 s?l n = (2t + 1)m for some integer t: From this we see that p m (?e?ij=m ) = p m (e?i2s?l n=m ) = 0: In all cases, 1 is linearly dependent, so is : It is also clear that the function dened by ^ 1 (!) = c(e?i! ) ^(!) is compactly supported and in the space S(): Finally, we prove P 2 SP r =) Q 2 SP r : When c(z) 6= z? 1; Q(1) is similar to P (1) and therefore P 2 SP r () Q 2 SP r. When c(z) = z? 1; equation(3.12) implies P (1) = p11 (1) 0 P r?1 (1) ; Q(1) = p11 (1)=2 0 P r?1 (1) Therefore Q(1) preserves all eigenvalues of P (1) except p 11 (1); which changes to p 11 (1)=2: Note that P 2 SP r : Hence, there is no positive integer such that p 11 (1) = :

13 LINEAR INDEPENDENCE OF SCALING VECTORS 13 2 : It follows that p 11 (1)=2 =2 2 ; 0 and therefore Q 2 OP r We now prove 1 2 E(Q(1)) so that Q 2 SP r : Let 1 = ( 2; ; r ) T. Then P (1)^(0) = ^(0) and ^ 1 (0) = 0 imply ^ 1 (0) 6= 0 and P r?1 (1)^ 1 (0) = ^ 1 (0): It follows that 1 2 E(P r?1 (1)) and 1 2 E(Q(1)): The following is the proof of Theorem 3.5. Proof. \=)". If is linearly dependent, then, by Theorem 2.1, there exists a linearly independent scaling vector 2 S() such that S 0 ( ) and S( ) = S(): Since is nitely linearly independent, dim = dim : By Lemma 3.4, 's symbol Q is a P-matrix. Let T be the polynomial matrix determined by ^(!) = T (e?i! ) ^(!): Then P (z) is two-scale similar to Q(z) with the transform T (z): By Corollary 2.5, T is non-fundamental. Besides, we have T (1) ^ (0) = ^(0): Thus (3.10) is true. \(= ": We assume P (z) is the symbol of ; and P 2 P r (SP r ) is two-scale similar to Q 2 P r (SP r ) with a non-fundamental two-scale transform matrix T 2 P r : We now factor T into the form T (z) = L(z)D(z)R(z); where L and R both are fundamental and D is the Canonical Smith's form of T. It is clear that D is a nonfundamental diagonal matrix. We dene E by ^ E (!) = L?1 (e?i! )^(!). Then is linearly independent if and only if E is linearly independent. The symbol of E is P(z) ~ = L?1 (z 2 )P (z)l(z): It is obvious that rank T (1) = rank T (1); ^(0) if and only if rank D(1) = rank D(1); ^ E (0) : Let Q(z) ~ = R(z2 )Q(z)R?1 (z). Then ~P(z) = D(z 2 ) Q(z)D ~ Q Q?1 (z). We now factor D(z) into D(z) = s T s(z) j U j(z); where T s is the diagonal matrix of the form (3.13) with det(t s (1)) 6= 0; while U j (z) is the diagonal matrix of the form (3.13) with det(u j (1)) = 0: If the product Q s T s(z) has at least one factor, by Lemma 3.6, E and hence are linearly dependent. Now if Q s T s(z) = I, then the product Q j U j(z) has at least one factor. Without loss of generality, we can assume U 1 (z) = diag (z? 1; 1; ; 1). It follows that the rst row of D(1) is zero. Since there is a v 2 C r ; such that D(1)v = ^ E (0), ^ E;1 (0) = 0: By Lemma 3.6, E and hence are linearly dependent. The proof of P 2 PS r =) Q 2 SP r is similar to that one in Lemma 3.6. From Theorem 3.5, we derive a sucient condition for linear independence of a scaling vector. Corollary 3.7. Assume the scaling vector is nitely linearly independent and its symbol is a P-matrix P 2 P r : Then is linearly independent if (i) the matrix (P (?1); ^(0)) has the full rank, and (ii) det P (z) has neither symmetric zeros nor cycle zeros. Proof. Assume the conditions (i) and (ii) hold and is linearly dependent. By Theorem 3.5, there is a P-matrix Q and a non-fundamental P-matrix T such that P (z) = T (z 2 )Q(z)T?1 (z) and Note that rank T (1) = rank T (1); ^(0) : (3.17) det P (z) = det Q(z) det T (z 2 )= det T (z): Write t(z) = det T (z), p(z) = det P (z); and q(z) = det Q(z). Then (3.17) becomes p(z) = q(z)t(z 2 )=t(z); where t; p; and q are polynomials with deg(t) 1: We factor the polynomial t(z) into t(z) = (1? z) s d(z) with d(1) 6= 0: If deg(d) 1; then p(z) has either symmetric zeros or cycle zeros. This contradicts the condition (ii). Now if deg(d) = 0; then t(z) = c(1? z) s ; s 1: In this case, T (1) is singular, and therefore

14 14 JIANZHONG WANG (T (1); ^(0)) does not have full rank. Since P (?1) = T (1)Q(?1)T?1 (?1); the matrix (P (?1); ^(0)) does not have full rank. This contradicts condition (i). Similarly, we have the following result for the stability of a scaling vector. Theorem 3.8. Assume the scaling vector is nitely linearly independent and its symbol is a P-matrix P 2 P r (SP r ): Then is unstable if and only if P is two-scale similar to a P-matrix Q 2 P r (SP r ) with a non-fundamental two-scaletransform matrix T such that det T (z) has zero(s) on jzj = 1 and rank T (1) = rank T (1); ^(0) : Corollary 3.9. Assume the scaling vector is nitely linearly independent and its symbol is a P-matrix P 2 P r : Then is stable if (i) the matrix (P (?1); ^(0)) has full rank and (ii) det P (z) has neither symmetric zeros on jzj = 1 nor cycle zeros. Remark 3. When r = 1; the scaling vector reduces to a single function. A sucient and necessary condition for linear independence (stability) of a scaling function has been obtained by Jia and Wang [14]. Besides, T. A. Hogan [10] also obtained the same results as Corollary 3.7 and 3.9 in a dierent way. 4. Examples. In this section we give more examples. Example 2. Consider the following scaling equation: 1?1 (4.1) (x) = (2x) + 0?1 Its symbol is where matrix P (1) = is v 0 =? 1 z 2? (2x? 2): (2x? 1) P (z) = z + z 2?z 2 z + z 2?1 + z 2 ; 3=2?1=2 has eigenvalues 1 and 1=2: A right 1-eigenvector T 1 : P (1)v0 = v 0. It is easy to check that z + z 2?z 2 z + z 2?1 + z 2 = (1 + z)=2 z=2 0 (?1 + z 2 + z 3 )=2 z?1 and the equation x() =?x(2) + x(2?2) + x (2?3) has only the trivial compactly supported solution x() = 0. Hence, by Theorem 3.2, the solution of (4.1) with the mean value? ^1 (0) ^2 (0) T =? 1 1 T is nitely linearly dependent. In fact,? this solution is (x) = (0; 1] (1 + z)=2 0 z 2 = (1 + z)=2 0 z 2?1 We can verify that (x) satises the following equation. (x) = (2x) + + (2x? 2) + T (1; 2]. By Corollary 3.3, one of 's symbols is (1 + z)=2 0 : (2x? 1) z?1 (2x? 3):

15 LINEAR INDEPENDENCE OF SCALING VECTORS 15 By Theorem 3.2, (x) will satisfy any equation with a symbol having the form of (1 + z)=2 Y (z) z 2 ;? X(z) z?1 where Y (z) and X(z) are arbitrary polynomials such that X(1) 6= 2 ( 1; 2 Z): For example, if we set Y (z) =?1=2; X(z) = z=2; then, by Theorem 3.2, the matrix 1=2 1=2 (1 + z)=2?1=2 0 (z + z 2 = )=2 z 2? z=2 z?1 ought to be a new symbol of : It can be veried that (x) satises the following equation. 1 1 (x) = (2x? 2): (2x) (2x? 1) Example 3. Now we analyze the scaling vector in Example 4.6 in [15] (for the case of L = r = 2). The symbol of the scaling vector = ( 1 ; 2 ) T is P (z) = ( 1 + z 1=2 1=2 2 )2 : z=2 1=2 Using the formula of Theorem 2.3 in [20], we obtain supp 1 [0; 2 1] and supp 3 2 [0; 2 2]: The graphs of ( 3 1; 2 ) T ; which can be found in Figure 4.1 of [15], show that supp 1 = [0; 2 1] and supp 3 2 = [0; 2 2 ]: Since the right end-point of the support of 3 any nitely linear combination of the integer translates of 1 is k +1=3 while the right end-point of the support of any nitely linear combination of the integer translates of 2 isk + 2=3; scaling vector ( 1 ; 2 ) T is nitely linearly independent. Matrix 1=2 1=2? T? T P (1) = has eigenvalues 1 and 0; and ^1 (0) ^2 (0) = 1 1 1=2 1=2 is an 1-eigenvector. P (z) has the following two-scale similarity: P (z) = T (z 2 )Q(z)T?1 (z); where T (z) = 1 z? 1 ; Q(z) = ( 1+z 2 )2 ( z?1 z+1 )( 2 2 )2 1+z 0 8 : Note that T (z) is non-fundamental and equation T (1)u? T = 1 1 has a solution u P = (1; 0) T. By Theorem 3.5, ( 1; 2 ) T is linearly dependent. Indeed, we have k2z ( 1(x? k)? 2 (x? k)) = 0: Example 4. Finally, we consider the orthonormal fractal scaling vector in [5]. Its symbol is p 2 6 p 2(1 + z) 16 P (z) = 20?(1 + z)(1? 10z + z 2 )? p 2(3? 10z + 3z 2 ; ) and the initial vector is ^(0)? = p T 2 1 : Using the formula of Theorem 2.3 in [20], we obtain that supp 1 [0; 1] and supp 2 [0; 2]: We rst point out that is

16 16 JIANZHONG WANG nitely linearly independent. Indeed, since supp 1 [0; 1]; the integer translates of 1 are linearly independent. It follows that the entire function ^ 1 has no 2-periodic zero in C. Assume is nitely linearly dependent, then there are two polynomial a(z) and b(z) such that a(z)^ 1 (!)? b (z) ^ 2 (!) = 0; z = e?i! : Hence ^ 2 (!) = a(z) b(z) ^ 1 (!): Since ^ 2 is also an entire function and ^ 1 has no 2-periodic zero in C, s(z) := a(z) b(z) must be a polynomial. Furthermore, since supp 1 [0; 1] and supp 2 [0; 2]; s(z) must be a linear polynomial. Assuming that s(z) = (c + dz); we have ^ 2 (!) = (c + de?i! )^ 1 (!): Therefore, T (e?i! )^(!) = (^ 1 (!); 0) T ; where T (z) = By Theorem 3.2, we also have T?1 (z 2 )P (z)t (z) = ; thus 0 (c + dz 2 ) 6 p 2(1 + z) + 16(c + dz) + (1 + z)(1? 10z + z 2 ) + p 2(c + dz)(3? 10z + 3z 2 ) = 0: c + dz?1 : However, this equation has no solution for c and d: Hence, is nitely linearly independent. Finally we conrm the linearindependence p pof from the fact that det P (z) =? 1 (z )3 and (P (?1)^(0)) = has full rank. 0?8=5 1 Acknowledgment. The author would like to thank the referees for their helpful comments. REFERENCES [1] C. K. Chui and Jianzhong Wang, A general Framework of Compactly Supported Spline and Wavelets, J. Approx. Theory, 71(1993), pp. 263{304. [2] C. K. Chui and Jianzhong Wang, On compactly supported spline wavelets and a duality principle, Trans. Amer. Math. Soc., 330(1992), pp. 903{916. [3] A. Cohen, I. Daubechies, and G. Plonka, Regularity of renable function vectors, preprint. [4] C. de Boor, R. DeVore, and A. Ron, Approximation from shift-invariant subspaces of L 2 (R d ), J. Functional Analysis, 119(1994), pp. 37{78. [5] G. Donovan, J. S. Geronimo, D. P. Hardin, and P. R. Massopust, Construction of orthogonal wavelets using fractal interpolation functions, SIAM J. Math. Anal., 27(1996), pp. 1158{1192. [6] T. N. T. Goodman and S. L. Lee, Wavelet of multiplicity r, Trans. Amer. Math. Soc., 342(1994), pp. 307{324. [7] T. N. T. Goodman, S. L. Lee, and W. S. Tang, Wavelets in wandering subspaces, Trans. Amer. Math. Soc., 338(1993), pp. 639{654. [8] C. Heil and D. Colella, Matrix renement equations: Existence and uniqueness, J. Fourier Anal. Appl., 2(1996), pp. 363{377. [9] C. Heil, G. Strang, and V. Strela, Approximation by translates of renable functions, Numer. Math., 73(1996), pp. 75{94. [10] T. A. Hogan, Stability and independence of the shifts of nitely many renable functions, preprint. [11] Rong-Qing Jia, Shift-invariant spaces on the real line, Proc. Amer. Math. Soc., 125(1997), pp. 785{793. [12] Rong-Qing Jia and C. A. Micchelli, On linear independence for integer translates of a nite number of functions, Proc. Edinburgh Math. Soc., 36(1992), pp. 69{85. [13] Rong-Qing Jia, S. D. Riemenschneider, and Ding-Xuan Zhou, Approximation by multiple renable functions, Canadian J. Math., to appear. [14] Rong-Qing Jia and Jianzhong Wang, Stability and linear independence associated with wavelet decompositions, Proc. Amer. Math. Soc., 117(1993), pp. 1115{1124.

17 LINEAR INDEPENDENCE OF SCALING VECTORS 17 [15] P. Massopust, D. Ruch, and P. Van Fleet, On the support properties of scaling vectors, Applied and Computational Harmonic Analysis, 3(1996), pp. 229{238. [16] G. Plonka, Approximation order provided by renable function vectors, Constr. Approx., to appear. [17] G. Plonka and V. Strela, Construction of multi-scaling functions with approximation and symmetry, preprint. [18] A. Ron, A necessary and sucient condition for the linear independence of the integer translates of a compactly supported distribution, Constructive Approx., 5(1989), pp. 297{308. [19] D. Ruch, W. So, and Jianzhong Wang, Global support of a scaling vector, preprint. [20] W. So and Jianzhong Wang, Estimating the support of a scaling vector, SIAM J. Matrix Anal. Appl., 18(1997), pp. 66{73. [21] V. Strela, Multiwavelets: Regularity, orthogonality and symmetry via two-scale similarity transform, preprint. [22] Kang Zhao, Global linear independence and nitely supported dual basis, SIAM J. Math. Anal., 23(1992), pp. 1352{1355.