Notes on Number Theory and Discrete Mathematics ISSN 1310–5132 Vol. 21, 2015, No. 2, 55–58
On right circulant matrices with general number sequence Aldous Cesar F. Bueno Philippine Science High School-Central Luzon Campus Clark Freeport Zone, Angeles City, Pampanga, Philippines e-mail:
[email protected]
Abstract: In this paper, the elements of the general number sequence were used as entries for right circulant matrices. The eigenvalues, the Euclidean norm and the inverse of the resulting matrices were obtained. Keywords: General number sequence, Right circulant matrix. AMS Classification: 15B05.
1
Introduction
The general number sequence as defined in [1] is a sequence whose terms satisfy the recurrence relation wn = pwn−1 − qwn−2 (1) with initial values w0 = a and w1 = b. Here, a, b, p and q ∈ Z. The n-th term of the general number sequence is given by: Aαn + Bβ n (2) wn = α−β where A = b − aβ B = aα − b α+β = p αβ = q p p2 − 4q 6= 0 α−β = The numbers α and β are the roots of the equation x2 −px+q = 0. 55
A right circulant matrix with general number sequence is a matrix of the form w0 w1 w2 ... wn−2 wn−1 w n−1 w0 w1 ... wn−3 wn−2 wn−2 wn−1 w0 ... wn−4 wn−3 RCIRCn (w) ~ = . .. .. . . .. .. , . . . . . . . w2 w3 w4 ... w0 w1 w1 w2 w3 ... wn−1 w0 where wk are the first n terms of the general number sequence.
2
Preliminary result
The following lemma will be used to prove one of the main results. Lemma 2.1. n−1 X (rω −m )k = k=0
1 − rn 1 − rω −m
where ω = e2iπ/n . P −m k ) is a geometric series with first term 1 and common ratio rω −m . Proof: Note that n−1 k=0 (rω Using the formula for the sum of a geometric series, we have n−1 X
(rω −m )k =
k=0
1 − rn ω −mn 1 − rω −m
1 − rn (cos 2π + i sin 2π) 1 − rω −m n 1−r = 1 − rω −m =
This completes the proof.
3
Main results
Theorem 3.1. The eigenvalues of RCIRCn (w) ~ are given by 1 A(1 − αn ) B(1 − β n ) λm = + α − β 1 − αω −m 1 − βω −m where m = 0, 1, ..., n − 1. Proof: From [2], the eigenvalues of a right circulant matrix are given by the Discrete Fourier transform n−1 X λm = ck ω −mk (3) k=0
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where ck are the entries in the first row of the right circulant matrix. Using this formula, the eigenvalues of RCIRCn (w) ~ are λm =
n−1 X Aαk + Bβ k k=0
α−β
ω −mk
n−1 1 X k = Aα + Bβ k ω −mk . α − β k=0
Using Lemma 2.1 we get the desired equation.
Theorem 3.2. The Eucliden norm of RCIRCn (w) ~ is given by s 1 2AB(1 − q n ) A2 (1 − α2n ) B 2 (1 − β 2n ) kRCIRCn (w)k ~ E = n + . + |α − β| 1−q 1 − α2 1 − β2 Proof:
kRCIRCn (w)k ~ E =
=
=
=
v u n−1 u X Aαk + Bβ k 2 tn α−β k=0 v u n−1 u X A2 α2k + B 2 β 2k + 2ABαβ tn (α − β)2 k=0 v u n−1 X 1 u tn [A2 α2k + B 2 β 2k + 2ABαβ] |α − β| k=0 v u n−1 X 1 u tn [A2 α2k + B 2 β 2k + 2ABq]. |α − β| k=0
Note that each term in the summation is from a geometric sequence, so using the formula for sum of geometric sequence, the theorem follows. Theorem 3.3. The inverse of RCIRCn (w) ~ is given by RCIRCn (s0 , s1 , . . . , sn−1 ) where sk
n−1 α−β X (1 − αω −m )((1 − βω −m )ω mk = . n m=0 A(1 − αn )(1 − βω −m ) + B(1 − β n )(1 − αω −m )
Proof: The entries of the inverse of a right circulant matrix can be solved using the Inverse Discrete Fourier transform n−1 1 X −1 mk sk = λ ω (4) n m=0 m 57
where λm are the eigenvalues of the right circulant matrix. Using this equation and Theorem 3.1 we have −1 n−1 1 A(1 − αn ) B(1 − β n ) 1X sk = + ω mk −m −m n m=0 α − β 1 − αω 1 − βω −1 n−1 α − β X A(1 − αn ) B(1 − β n ) + = ω mk n m=0 1 − αω −m 1 − βω −m n−1 (1 − αω −m )(1 − βω −m )ω mk α−β X . = n m=0 A(1 − αn )(1 − βω −m ) + B(1 − β n )(1 − αω −m )
This completes the proof.
References [1] Bozkurt, D. (2012) On the Determinants and Inverses of Circulant Matrices with a General Number Sequence, arXiv: 1202.1068v1 [math.NA] 6 Feb 2012. [2] Karner, H., Schneid, J., Ueberhuber, C. (2003) Spectral decomposition of real circulant matrices, Linear Algebra and Its Applications, 367, 301–311. Available online: http:// www.sciencedirect.com/science/article/pii/S002437950200664X.
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