Matrices and Polynomials

Transcription

1 APPENDIX 9 Matrices and Polynomials he Multiplication of Polynomials Let α(z) =α 0 +α 1 z+α 2 z 2 + α p z p and y(z) =y 0 +y 1 z+y 2 z 2 + y n z n be two polynomials of degrees p and n respectively. hen, their product γ(z) = α(z)y(z) is a polynomial of degree p + n of which the coefficients comprise combinations of the coefficient of α(z) and y(z). A simple way of performing the multiplication is via a table of which the margins contain the elements of the two polynomials and in which the cells contain their products. An example of such a table is given below: α 0 α 1 z α 2 (1) y 0 α 0 y 0 α 1 y 0 z α 2 y 0 z 2 y 1 z α 0 y 1 z α 1 y 1 z 2 α 2 y 0 z 3 y 2 z 2 α 0 y 1 z 2 α 1 y 2 z 3 α 2 y 2 z 4 he product is formed by adding all of the elements of the cells. However, if the elements on the SW NE diagonal are gathered together, then a power of the argument z can be factored from their sum and then the associated coefficient is a coefficient of the product polynomial. he following is an example from the table above: γ 0 + γ 1 z + γ 2 z 2 α 0 y 0 +(α 0 y 1 + α 1 y 0 )z +(α 0 y 2 + α 1 y 1 + α 2 y 0 +)z 2 (3) + γ 3 z 3 = +(α 1 y 2 + α 2 y 1 )z 3 + γ 4 z 4 + α 2 y 4 z 4. he coefficients of the product polynomial can also be seen as the products of the convolutions of the sequences {α 0,α 1,α 2...α p } and {y 0,y 1,y 2...y n }. 1

2 D.S.G. POLLOCK: ECONOMERICS he coefficients of the product polynomials can also be generated by a simple multiplication of a matrix by a vector. hus, from the example, we should have (3) γ 0 y γ 1 y 1 y 0 0 γ 2 = y 2 y 1 y 0 α α 0 α 1 α 0 0 α 1 = α 2 α 1 α 0 y 0 y 1 γ 2 0 y 2 y 1 α 2 0 α 2 α 1 y 2 γ v 0 0 y α 2 o form the elements of the product polynomial γ(z), powers of z may be associated with elements of the matrices and the vectors of values indicated by the subscripts. he argument z is usually described as an algebraic indeterminate. Its place can be taken by any of a wide variety of operators. Examples are provided by the difference operator and the lag operator that are defined in respect of doubly-infinite sequences. It is also possible to replace z by matrices. However, the fundamental theorem of algebra indicates that all polynomial equations must have solutions that lie in the complex plane. herefore, it is customary, albeit unnecessary, to regard z as a complex number. oeplitz Matrices Polynomials with Matrix Arguments here are two matrix arguments of polynomials that are of particular interest in time series analysis. he first is the matrix lag operator. he operator of order denoted by (4) L =[e 1,e 2,...,e 1, 0] is formed from the identity matrix I =[e 0,e 1,...,e 1 ] by deleting the leading vector e 0 and by appending a column of zeros to the end of the array. In effect, L is the matrix with units on the first subdiagonal band and with zeros elsewhere. Likewise, L 2 has units on the second subdiagonal band and with zeros elsewhere, whereas L 1 has a single unit in the bottom left (i.e. S W ) corner, and L +r = 0 for all r 0. In addition, L 0 = I is identified with the 2

3 POLYNOMIALS AND MARICES identity matrix of order. he example of L 4 is given below: L 0, L, (5) L 2, L 3 Putting L in place of the argument z of a polynomial in non-negative powers creates a so-called oeplitz banded lower-triangular matrix of order. An example is provided by the quadratic polynomials α(z) =α 0 + α 1 z + α 2 z 2 and y(z) =y 0 + y 1 z + y 2 z 2. hen, there is α(l 3 )y(l 3 )=y(l 3 )α(l 3 ), which is written explicitly as (6) α α 1 α 0 0 y y 1 y 0 0 = y y 1 y 0 0 α α 1 α 0 0 α 2 α 1 α 0 y 2 y 1 y 0 y 2 y 1 y 0 α 2 α 1 α 0 he commutativity of the two matrices in multiplication reflects their polynomial nature. Such commutativity is available both for lower-triangular oeplitz and for upper-triangular oeplitz matrices, which correspond to polynomials in negative powers of z. he commutativity is not available for mixtures of upper and lower triangular matrices; and, in this respect, the matrix algebra differs from the corresponding polynomial algebra. An example is provided by the matrix version of the following polynomial identity: (7) (1 z)(1 z 1 )=2z 0 (z + z 1 )=(1 z 1 )(1 z) Putting L in place of z in each of these expressions creates three different matrices. his can be illustrated with the case of L 3. hen, (1 z)(1 z 1 ) gives rise to (8) = , whereas (1 z 1 )(1 z) gives rise to (9) =

4 D.S.G. POLLOCK: ECONOMERICS A oeplitz matrix, in which each band contains only repetitions of the same element, is obtained from the remaining expression by replacing z by L 3 in 2z 0 (z + z 1 ), wherein z 0 is replaced by the identity matrix: (10) = Example. It is straightforward to derive the dispersion matrices that are found within the formulae for the finite-sample estimators from the corresponding autocovariance generating functions. Let γ(z) ={γ 0 +γ 1 (z+z 1 )+γ 2 (z 2 +z 2 )+ } denote the autocovariance generating function of a stationary stochastic process. hen, the corresponding dispersion matrix for a sample of consecutive elements drawn from the process is (11) Γ = γ 0 I + 1 γ τ (L τ + F τ ), where F = L is in place of z 1. Since L and F are nilpotent of degree, such that L q,fq = 0 when q, the index of summation has an upper limit of 1. Circulant Matrices In the second of the matrix representations, which is appropriate to a frequency-domain interpretation of filtering, the argument z is replaced by the full-rank circulant matrix (12) K =[e 1,e 2,...,e 1,e 0 ], which is obtained from the identity matrix I =[e 0,e 1,...,e 1 ] by displacing the leading column to the end of the array. his is an orthonormal matrix of which the transpose is the inverse, such that K K = K K = I. he powers of the matrix form a -periodic sequence such that K +q = K q for all q. he periodicity of these powers is analogous to the periodicity of the powers of the argument z = exp{ i2π/}, which is to be found in the Fourier transform of a sequence of order. 4

5 (13) POLYNOMIALS AND MARICES he example of K 4 is given below K4 0 =, K, K4 2 =, K 3 he matrices K 0 = I,K,...,K 1 form a basis for the set of all circulant matrices of order a circulant matrix X =[x ij ] of order being defined as a matrix in which the value of the generic element x ij is determined by the index {(i j) mod }. his implies that each column of X is equal to the previous column rotated downwards by one element. It follows that there exists a one-to-one correspondence between the set of all polynomials of degree less than and the set of all circulant matrices of order. herefore, if α(z) is a polynomial of degree less that, then there exits a corresponding circulant matrix (14) A = α(k )=α 0 I + α 1 K + + α 1 K 1. A convergent sequence of an indefinite length can also be mapped into a circulant matrix. hus, if {γ i } is an absolutely summable sequence obeying the condition that γ i <, then the z-transform of the sequence, which is defined by γ(z) = γ j z j, is an analytic function on the unit circle. In that case, replacing z by K gives rise to a circulant matrix Γ = γ(k ) with finite-valued elements. In consequence of the periodicity of the powers of K, it follows that (15) { { { } Γ= γ j }I + γ (j+1) }K + + γ (j+ 1) K 1 j=0 j=0 = ϕ 0 I + ϕ 1 K + + ϕ 1 K 1. Given that {γ i } is a convergent sequence, it follows that the sequence of the matrix coefficients {ϕ 0,ϕ 1,...,ϕ 1 } converges to {γ 0,γ 1,...,γ 1 } as increases. Notice that the matrix ϕ(k) =ϕ 0 I + ϕ 1 K + + ϕ 1 K 1, which is derived from a polynomial ϕ(z) of degree 1, is a synonym for the matrix γ(k ), which is derived from the z-transform of an infinite convergent sequence. 5 j=0

6 D.S.G. POLLOCK: ECONOMERICS he polynomial representation is enough to establish that circulant matrices commute in multiplication and that their product is also a polynomial in K. hat is to say (16) If X = x(k ) and Y = y(k ) are circulant matrices, then XY = YX is also a circulant matrix. he matrix operator K has a spectral factorisation that is particularly useful in analysing the properties of the discrete Fourier transform. o demonstrate this factorisation, we must first define the so-called Fourier matrix. his is a symmetric matrix (17) U = 1/2 [W jt ; t, j =0,..., 1], of which the generic element in the jth row and tth column is (18) W jt = exp( i2πtj/) = cos(ω jt) i sin(ω j t), where ω j =2πj/. he matrix U is a unitary, which is to say that it fulfils the condition (19) Ū U = U Ū = I, where Ū = 1/2 [W jt ; t, j =0,..., 1] denotes the conjugate matrix. he operator can be factorised as (20) K = Ū D U = U D Ū, where (21) D = diag{1,w,w 2,...,W 1 } is a diagonal matrix whose elements are the roots of unity, which are found on the circumference of the unit circle in the complex plane. Observe also that D is -periodic, such that D q+ = D q, and that Kq = Ū D q U = U Dq Ū for any integer q. Since the powers of K form the basis for the set of circulant matrices, it follows that such matrices are amenable to a spectral factorisation based on (13). Example. Consider, in particular, the circulant autocovariances matrix that is obtained by replacing the argument z in the autocovariance generating function γ(z) by the matrix K. Imagine that the autocovariances form a doubly infinite 6

7 POLYNOMIALS AND MARICES sequence, as is the case for an autoregressive or an autoregressive movingaverage process: (22) Ω = γ(k )=γ 0 I + = ϕ 0 I + 1 γ τ (K τ + K τ ) ϕ τ (K τ + K τ ). Here, ϕ τ ; τ =0,..., 1 are the wrapped coefficients that are obtained from the original coefficients of the autocovariance generating function in the manner indicated by (15). he spectral factorisation gives (23) Ω = γ(k )=Ūγ(D)U. he jth element of the diagonal matrix γ(d) is (24) γ(exp{iω j })=γ 0 +2 γ τ cos(ω j τ). his represents the cosine Fourier transform of the sequence of the ordinary autocovariances; and it corresponds to an ordinate (scaled by 2π) sampled at the point ω j from the spectral density function of the linear (i.e. non-circular) stationary stochastic process. 7