Some phenomenon of the powers of certain tridiagonal and asymmetric matrices

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Abstract

In this paper, a new method of calculation of real (and even complex) powers of some asymmetric matrices obeying constance tridiagonals is presented.

Introduction

Let us setTn(a,b,c)acbac0bac0bacban×n,where a,b,cC, nN.

In the course of preparing paper [1] focused on a certain property of tridiagonal matrices and general constant-diagonals matrices, the authors came across a certain phenomenon connected with the real powers of tridiagonal matrix Tn(x,1,1) and then with the complex powers of matrices of a more general form:F3(a,b,c,A)abcbAbcba,where a,b,cC and Aa+c. This phenomenon consists of finding Binet’s formulas of elements of positive integers powers of given matrix and next to extend these ones to complex powers. Our considerations are extended also to matrices of asymmetric forms:G3(a,b,c,A,y)ayby2cbAybcbaandG4(a,b,c,d,e,A,y)ayby2cy3ebAydy2ccdAybecbafor any a,b,c,e,yC, Aa+yc, db+ye.

Section snippets

Positive integers powers of F3

We shall start our deliberations with a simple case, that is with describing the positive integers powers of matrix (2).

Lemma 2.1

Let a1,b1,c1C and A1a1+c1. Then we have:F3(a1,b1,c1,A1)k=F3(ak,bk,ck,Ak)for every kN, where the following formulas hold:Ak=ak+ck,ak+1=a1ak+b1bk+c1ck,bk+1=A1bk+Akb1,ck+1=a1ck+b1bk+c1ak,Ak+1=2b1bk+A1Ak.

The proof of Lemma by induction follows.

Corollary 2.2

We haveak=ck+(a1-c1)k.

Proof

We note that (7), (8), (9)  ak+1  ck+1 = (a1  c1)(ak  ck), k = 1,2,… 

Corollary 2.3

Now let us set (x,yC):za1-c1,A112(x+y),b124(x-y).

Phenomenon of the identity (5)

The identity (5) also holds for all kC! Why is that so? First it should be noted that if in formulas (12), (13), (14), (15) we assume thatxx0m/n,yy0m/n,zz0m/n,mZ, n2N-1{1,3,5,}, x0,y0,z0R, then, by view of (5) we obtain(F3(a1(x,y,z),b1(x,y),c1(x,y,z),A1(x,y)))nr=F3(amr(x0,y0,z0),bmr(x0,y0),cmr(x0,y0,z0),Amr(x0,y0)),which means that(F3(a1(x,y,z),b1(x,y),c1(x,y,z),A1(x,y)))F3(amr(x0,y0,z0),bmr(x0,y0),cmr(x0,y0,z0),Amr(x0,y0))nrfor every rN.

Now, again from the formulas (5) and (12), (13)

An asymmetric generalization of T3(x,1,1)

We shall now designate the form of the powers of the matrix T3(x,1,y). By easily induction argument it can be deduced that the following identities hold:T3(x,1,y)k=akybky(Ak-ak)bkAkybkckbkakfor every kN, wherea1=A1=x,b1=1,c1=0andak+1=xak+ybk,bk+1=xbk+Ak,ck+1=xck+bk,Ak+1=xAk+2ybk,for every kN. We note that alsoAk=ak+yck,kN.From (29) we havebk=ck+1-xck,which by (28) impliesAk=bk+1-xbk=ck+2-2xck+1+x2ck.At last, from (30), (32), (33) we getck+3-2xck+2+x2ck+

Positive integers powers of G4

In this section, we generalize the formulas (39), (40), (41), (42), (43), (44), (45), (46), (47), (48) for G3 to the respective formulas for the matrix G4.

Lemma 5.1

Let a1,b1,c1,e1,y,C andA1a1+yc1andd1b1+ye1.Then we getG4(a1,b1,c1,d1,e1,A1,y)k=G4(ak,bk,ck,dk,ek,Ak,y)for every kN, where the following recurrence formulas hold (again by induction argument):Ak=ak+yck,dk=bk+yek,ak+1=a1ak+yb1bk+y2c1ck+y3e1ek,bk+1=b1ak+A1bk+yd1ck+y2c1ek,ck+1=c1ak+d1bk+A1ck+yb1ek,ek+1=e1ak+c1bk+b1ck+a1ek.

Our next goal is to

Final remarks

  • 1.

    Recall that the definitions of complex powers of matrices F3, G3 and G4 presented in this paper are consequences of extensions of Binet’s formulas for elements of positive powers of matrices F3, G3 and G4. It is the alternative and more effective way than the Jordan decomposition method (see [5]).

  • 2.

    The cases of matrices of the dimensions more than or equal to five will be discussed in the separated papers.

  • 3.

    A nice complement to the problem of diagonalization of a given matrix would be a paper [6].

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