Nov 5, 2008 - tions of results on quaternionic polynomials to the octonionic polynomials ... extending to the octonions some methods valid for the quaternions.
Adv. appl. Clifford alg. 20 (2010), 155–178 c 2008 Birkh¨
auser Verlag Basel/Switzerland 0188-7009/010155-24, published online November 5, 2008 DOI 10.1007/s00006-008-0140-5
Advances in Applied Clifford Algebras
Construction of Octonionic Polynomials Rog´erio Serˆodio Abstract. In a previous paper “[On Octonionic Polynomials”, Advances in Applied Clifford Algebras, 17 (2), (2007), 245–258] we discussed generalizations of results on quaternionic polynomials to the octonionic polynomials. In this paper, we continue this generalization searching for methods to construct octonionic polynomials with a prescribed set of zeros. Two iterative methods, valid for the quaternions, are applied to construct octonionic polynomials with limited results. The non-associativity of the octonion product does not allow the prescribed set of zeros to be the set of zeros of the constructed polynomial. Nevertheless, we will show that one of the methods has some advantage relatively to the other. Finally, a closed form method is given to construct an octonionic polynomial with a prescribed set of zeros. This method requires the inversion of a block Vandermonde matrix. The necessary and sufficient conditions for the existence of the inverse are studied. Mathematics Subject Classification (2000). 11R52, 20G20. Keywords. Division algebra, polynomials, zeros of polynomials, octonions, block Vandermonde matrix.
1. Introduction Although some interest seems to appear on octonionic polynomials, not much has been developed in this theory. In a previous paper [10] we defined the basis of the octonionic polynomials, developed the theory and obtained an algorithm to calculate all the zeros of an octonionic polynomial. It remained an important question to be answered: how can we construct an octonionic polynomial with a prescribed set of zeros? To answer this question we introduce in section 1, a new notation for the octonions. It is a notation based in matrices, which brings some insight to the theory and helps to manipulate the octonions in an easier way. Moreover, this representation is the one we use to manipulate the octonions with the computer. Some results on the octonions and related with this representation are also presented.
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In section 2, we make a brief summary of the theory of octonionic polynomials, so that the reader who hasn’t read the previous work can follow without difficulty. The definition of an octonionic polynomial as well as the most important definition of polynomial product is revised. So we only present here the main results needed for this paper. All the proofs can be found in [10]. In the main section of this paper, section 3, we explore the possibility of extending to the octonions some methods valid for the quaternions. The nonassociativity of the octonion product will limit the obtained results. In order to overcome this limitation, we try a different approach. Instead of having an iterative method, which has its advantages, we search for a closed form method. This means that, given a prescribed set of zeros, we search for the octonion polynomial whose zeros are exactly the prescribed ones. Of course, this set must have some restrictions. It’s impossible to construct, for example, an octonionic polynomial with only two distinct zeros from a conjugacy class. The existence and uniqueness of the octonionic polynomial with a prescribed set of zeros is also studied.
2. The Octonions Let O be the octonion algebra over the real number field R. It is known, by the Cayley-Dickson process [3, 4], that any o ∈ O can be written as o = q1 + q2 k,
(1)
where q1 , q2 ∈ H, (the real quaternion division algebra). The addition and multiplication of any two octonions, o0 = q10 + q20 k, o00 = q100 + q200 k, are defined by o0 + o00
=
(q10 + q100 ) + (q20 + q200 )k,
(2)
and o0 o00
=
(q10 q100 − q200 q20 ) + (q200 q10 + q20 q100 )k,
(3)
where q100 , q200 denote the conjugates of the quaternions q100 , q200 , respectively. Thus, O is an eight-dimensional non-associative division algebra over the real numbers R. A natural basis of this algebra as a space over R is formed by the elements
1, i, j, ij, k, ik, jk, ijk. The multiplication table for the basis of O is
(4)
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∗
1
i
j
ij
k
ik
jk
ijk
1
1
i
j
ij
k
ik
jk
ijk
i
i
−1
ij
−j
ik
−k
ijk
−jk
j
j
−ij
−1
i
jk
−ijk
−k
ik
ij
ij
j
−i
−1
−ijk
−jk
ik
k
k
k
−ik
−jk
ijk
−1
i
j
−ij
ik
ik
k
ijk
jk
−i
−1
−ij
−j
jk
jk
−ijk
k
−ik
−j
ij
−1
i
ijk
ijk
jk
−ik
−k
ij
j
−i
−1
For the sake of convenience, the elements of the basis can also be written as e0 = 1,
e1 = i,
e2 = j,
e3 = ij,
e4 = k, e5 = ik, e6 = jk, e7 = ijk.
(5)
Under this notation, all octonions take the form o=
7 X
o` e` ,
(6)
`=0
where the coefficients o` are real. Also, every o ∈ O can be simply written as o = Re (o) + Im (o), where Re (o) and Im (o) are called the real and imaginary part, respectively. P7 P7 Given o = `=0 o` e` ∈ O, the conjugate of o is defined as o = o0 − `=1 o` e` . For convenience we will define the norm of o, denoted by no , as no = oo = oo = P 7 2 `=0 o` and the trace of o, denoted by to , as two times the real part of o, i.e., to = o + o. It is straightforward to see that the norm and the trace of an octonion are real numbers. The inverse of a non-zero octonion o is o−1 = o/no . For theoretical and computational reasons, we will introduce a matrix representation of an octonion. Due to the non-associativity of the octonion algebra, we have to use a different matrix product. The cross product concept in R7 is useful for the construction of this matrix representation of the octonions. b be two vectors in R7 . We define cross product of the b and b Definition 1. Let a b b as b and b, and represent by a b × b, vectors a b=a b + ab, b×b b·b a (7)
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b and · denotes the scalar vector product. b, b = 0 + b where a = 0 + a have
b we b and b = b0 + b, From Definition 1 and given two octonions a = a0 + a b + a0 b b + b0 a b b·b b+a b × b. ab = a0 b0 − a b= To the imaginary part of an octonion a
Sba
=
0 a3 −a2 a5 −a4 a7 −a6
−a3 0 a1 a6 −a7 −a4 a5
a2 −a1 0 −a7 −a6 a5 a4
7 P
(8)
a` e` we can associate the matrix
`=1
−a5 −a6 a7 0 a1 a2 −a3
a4 a7 a6 −a1 0 −a3 −a2
−a7 a4 −a5 −a2 a3 0 a1
a6 −a5 −a4 a3 a2 −a1 0
.
(9)
This matrix Sba will perform, as we will see, an important role in the matrix representation of an octonion. b of R7 , then b and b Theorem 1 (Leite and Vitoria [8]). Given two vectors a b=a b b × b. Sba b
(10)
Definition 2. The matrix representation φ(a) of the octonion a =
7 P
ai ei , is de-
i=0
fined by φ(a) = b = [a1 a2 · · · where a In full, a0 a1 a2 a3 φ(a) = a4 a5 a6 a7
a0 b a
−b aT Sba + Ia0
−a4 −a5 −a6 a7 a0 a1 a2 −a3
−a5 a4 a7 a6 −a1 a0 −a3 −a2
,
(11)
a7 ]T . −a1 a0 a3 −a2 a5 −a4 a7 −a6
−a2 −a3 a0 a1 a6 −a7 −a4 a5
−a3 a2 −a1 a0 −a7 −a6 a5 a4
−a6 −a7 a4 −a5 −a2 a3 a0 a1
−a7 a6 −a5 −a4 a3 a2 −a1 a0
.
(12)
As the octonionic algebra is non-associative, it can not be isomorphic to the matrix algebra with the usual multiplication. Aiming at the introduction of a convenient matrix multiplication, we will show two other ways of representing the octonions by matrices which are closely related to the representation (12).
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Definition 3. Let b = b0 + b1 e1 + · · · + b7 e7 ∈ O. We will call the column representation, or vectorial, or also, ket representation1 of an octonion b, to the column b T. matrix |bi = [b0 b1 · · · b7 ]T , or |bi = [b0 b] Definition 4. Let b = b0 +b1 e1 +· · ·+b7 e7 ∈ O. We will call the row representation, or bra representation, of an octonion b, to the row matrix hb| = [b0 − b1 · · · − b7 ],
(13)
b or, hb| = [b0 − b]. The following result connects both concepts, bra and ket representations, of an octonion. Proposition 1. For all a ∈ O, we have |aiT = ha|. Proof. By direct verification.
(14)
Now we will present results aiming the use of the matrix (12) in the octonion product. Theorem 2 (Tian [11]). Let a, b ∈ O and λ ∈ R. Then i) a = b if, and only if, φ(a) = φ(b), ii) φ(a + b) = φ(a) + φ(b), iii) φ(λa) = λφ(a), iv) φ(1) = I8 , v) φ(a) = φ(a)T . Corollary 2.1. Let a, b ∈ O and λ ∈ R. Then i) a = b if, and only if, |ai = |bi, ii) |a + bi = |ai + |bi, iii) |λai = λ|ai, iv) |1i = [1 0 0 0 0 0 0 0]T . Corollary 2.2. Let a, b ∈ O and λ ∈ R. Then i) a = b if, and only if, ha| = hb|, ii) ha + b| = ha| + hb|, iii) hλa| = λha|, iv) h1| = [1 0 0 0 0 0 0 0]. In the following result we establish the octonion multiplication using matrix representations, of which we give a proof. 1 This
terminology is inspired on the one used in Quantum Mechanics. Its origin comes from the word “bracket”. That’s why the left side is called “bra” and the right side “ket”.
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Theorem 3 (Tian [11]). Let a, b ∈ O. Then |abi = φ(a)|bi.
(15)
b We already know, b and b = b0 + b. Proof. Let a, b ∈ O, written as a = a0 + a b b b Or rather, b · b + a0 b + b0 a b+a b × b. from the expression (8), that ab = a0 b0 − a b c b b b · b and Im(ab) ≡ ab = a0 b + b0 a b+a b × b. Re(ab) = a0 b0 − a Thus, from ket of ab we obtain " # b b·b a0 b0 − a |abi = (16) b + b0 a b b+a b×b a0 b b0 a0 −b aT = (17) b b Sba + Ia0 a b = φ(a)|bi.
(18)
Hence |abi = φ(a)|bi.
Corollary 3.1. Let a, b ∈ O. Then hab| = ha|φ(b).
(19)
Proof. From Theorem 3, we have |cdi = φ(c)|di.
(20)
Transposing each member of this equality, we have T
|cdiT = (φ(c)|di) .
(21)
Using Proposition 1 and item v) of Theorem 2, we obtain hcd| = |diT φ(c)T
(22)
hdc| = hd|φ(c).
(23)
Writing a = d and b = c, we conclude that hab| = ha|φ(b).
Although the octonionic algebra is not isomorphic to the matrix algebra of the type (12), in the following result we establish that the algebra generated by a single octonion is isomorphic to the algebra of these matrices, where the matrix multiplication is the usual one. Theorem 4 (Tian [11]). Let a ∈ O. Then φ(a2 ) = φ(a)2 . Next we present, without proof, a corollary of this theorem. Corollary 4.1. Let a ∈ O. Then φ(an ) = φ(a)n , for all n ∈ N. Observation 1.1. The importance of these matrix representations is linked with the possibility of “forgetting” the octonion non-associativity when they are chained. For example, when we have products of type a(b(cd))
or
((ab)c)d.
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In the following result this observation is emphasized. We also present an original proof, using the ket notation of an octonion. Theorem 5 (Eganova and Shirokov [5]). Given a, b, c ∈ O, to compute a(bc) is sufficient to obtain the first column of the usual matrix product φ(a)φ(b)φ(c). Formally, |a(bc)i = φ(a)φ(b)|ci. Proof. Applying successively the Theorem 3, we have |a(bc)i = φ(a)|bci = φ(a)φ(b)|ci.
(24)
When a chained multiplication is in the reverse order, i.e., from left to right, we have a similar result. We also show an original proof for this result. Corollary 5.1 (Eganova and Shirokov [5]). Given a, b, c ∈ O, to compute (ab)c is sufficient to calculate the first row of the usual matrix product φ(a)φ(b)φ(c). Formally, h(ab)c)| = ha|φ(b)φ(c). Proof. Let a, b, c ∈ O. Transposing the bra of (ab)c using the Proposition 1, we have h(ab)c|T
= |(ab)ci
(25)
= |c abi
(26)
= |c(ba)i.
(27)
Applying Theorem 5 to the last expression of the second member, using item v) of Theorem 2 and again Proposition 1, we obtain h(ab)c|T
= φ(c)φ(b)|ai T
(28) T
= φ(c) φ(b) ha|
(29)
T
(ha|φ(b)φ(c)) .
(30)
=
T
Hence h(ab)c| = ha|φ(b)φ(c).
(31)
It is well known that any two octonions generate an associative subalgebra isomorphic to the quaternion algebra [3, 12]. Octonion multiplication satisfies three important identities related to the associativity of multiplication, which we will present without proofs. They are the so called Moufang identities (see, for example, [6] for proofs): (xyx)z = x(y(xz)), z(xyx) = ((zx)y)x, (xy)(zx) = x(yz)x.
(32) (33) (34)
An equivalence relation ∼ over O is defined as: for any two octonions o and o0 , o ∼ o0 , if there exist σ ∈ O, σ 6= 0, such that o0 = σoσ −1 . In this case o
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and o0 are said to be similar. The conjugacy class of o, denoted by [o], is the set {x ∈ O : x ∼ o}. Finally, we present a theorem and a corollary that may be found in [10], that will be useful in the last section where we present some methods to construct octonionic polynomials. Theorem 6 (Serˆ odio [10]). Given any three octonions α, β, γ, we have (αβ)γ + (αγ)β = αtγβ .
(35)
where tγβ is the trace of the octonion γβ. Corollary 6.1 (Serˆ odio [10]). Given any three octonions α, β, γ, it follows that (αβ)γ + (αγ)β + β(γα) + γ(βα) = tα tγβ .
(36)
3. Octonionic Polynomials Before we start to describe some methods for the construction of octonionic polynomials, let us first agree on some definitions. In this sense, in this section we will give a summary of the main results on octonionic polynomials. For more detail and for proofs see [10]. Let O[X] denote the polynomial ring in the variable x over O. Every polynomial p ∈ O[X] can be written as p(x) = am xm + am−1 xm−1 + · · · + a1 x + a0 for some nonnegative integer m, ar ∈ O, r = 0, 1, . . . , m. If am 6= 0, m is called the degree of p. If am = 1, we say that the polynomial is monic; otherwise it is said to be non-monic. For the particular case where the coefficients of the polynomial are real, the corresponding polynomial ring will be denoted by R[X]. The equality and the addition of two polynomials are defined as the usual way. Contrarily, the multiplication of two octonionic polynomials is defined in an unexpected way. Definition 5. The product of two octonionic polynomials q(x) =
m X
al xl ,
s (x) =
` X
bj xj ,
(37)
j=0
l=0
is denoted q ? s and is given by (q ? s )(x) =
m+` X
ck xk ,
(38)
k=0
where ck =
k X
ai bk−i ,
k = 0, . . . , m + `,
i=0
regarding to ai = 0 if i > m and bi = 0 if i > `.
(39)
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Thus, the multiplication of octonionic polynomials is defined as if the variable x was real, commuting with the octonion coefficients. As a particular case of polynomial multiplication Pm is the multiplication of an octonion c and an octonionic polynomial q(x) = i=0 ai xi ∈ O[X]. From the above definition, this product is given by p(x)
=
(c ? q)(x)
=
m
(40) m−1
(cam )x + (cam−1 )x
+ · · · + (ca1 )x + ca0 .
(41)
The evaluation of an octonionic polynomial p at an octonion o can only be done after expressing p so that all the coefficients are on the left and the variable on the right. Then, the evaluation of p at o is simply to substitute o for x. The evaluation at o is not a ring homomorphism from O[X] to O. This is, if p(x) = (q ? s)(x) ∈ O[X], then, in general, p(o) 6= q(o)s(o). An octonion o is said to be a zero of p if p(o) = 0. We denote the set of all zeros of p by Zero(p). Definition 6. The classity of an octonionic polynomial p, denoted Ω(p), is given by the number of distinct conjugacy classes that the zeros of p belong to. For any o ∈ O, let ∆o (x) = (x − o) ? (x − o) = x2 − to x + no be the characteristic polynomial of the octonion o. Since to , no ∈ R, for all o ∈ O, this means that ∆0 ∈ R[X], for all o ∈ O. Note also that ∆o is an irreducible real quadratic polynomial if o ∈ O\R. Lemma 1. Given o ∈ O, ∆o (o) = 0. From the definition of conjugacy class and characteristic polynomial of an octonion, it follows that two octonions are similar if and only if they have the same characteristic polynomial. Pm an octonionic polynomial p(x) = r=1 ar xr ∈ O[X], we define p(x) = Pm Given r r=1 ar x ∈ O[X] and the normal polynomial np , given by X np (x) = (p ? p)(x) = (p ? p)(x) = ar as xr+s . (42) 0≤r,s≤m
For all octonionic polynomial p, we have that np ∈ R[X]. The main results in the theory of octonionic polynomial are the following ones. Theorem 7 (Serˆ odio [10]). Let p ∈ O[X] and ∆o be the characteristic polynomial of an octonion o. If ∆o divides p, then o is a zero of p. Theorem 8 (Serˆ odio [10]). Given p ∈ O[X] and ∆o , if ∆o does not divide p and exists o0 , o00 ∈ [o] such that p(o0 ) = p(o00 ) = 0, then o0 = o00 . As a consequence of this theorem, we have the following important result. Corollary 8.1 (Serˆ odio [10]). Given p ∈ O[X], if o0 ∼ o00 such that o0 6= o00 and 0 00 p(o ) = p(o ) = 0, then [o0 ] ⊂ Zero(p).
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Theorem 9 (Serˆ odio [10]). Given p ∈ O[X] and o ∈ O, ∆o divides np if, and only if, there exists at least an o0 ∼ o such that p(o0 ) = 0. This last theorem gave rise to the algorithm to compute the zeros of an octonionic polynomial (see [10]). The following corollaries are some consequences from the previous theorem and will be needed later on. Corollary 9.1 (Serˆ odio [10]). Given p ∈ O[X], if the degree of p is equal to m, then Ω(p) ≤ m. Corollary 9.2 (Serˆ odio [10]). Let p ∈ O[X]. If Zero(p) = {oi : i = 1, . . . , m}, then the degree of p is equal to m. To conclude this section we will present the result where the relation between the normal polynomial of a non-monic octonionic polynomial and its corresponding monic polynomial is established. Theorem 10. Let p ∈ O[X] be a monic polynomial. Define the non-monic polynomial q ≡ c ? p for a non-zero c ∈ O. Then nq (x) = nc np (x). Proof. Let p(x) = a0 + a1 x + · · · and q(x) = c ? p(x) = ca0 + ca1 x + · · · . It can be seen that the monomials of nq = q ? q involve products of the form (car ) (as c) + (cas ) (ar c) if r 6= s and (car ) (ar c), depending on the considered monomial. The second one involves octonions in an associative subalgebra so it can be seen that (car ) (ar c) = nc nar . The first one involves more than two octonions which in general do not associate. To overcome this difficulty, we will apply the third Moufang Identity. It is known that c = tc − c. Hence, (car ) (as c)
= (car ) [as (tc − c)] = tc (car ) as − (car ) (as c) .
(43) (44)
But (car ) (as c) = c (ar as ) c by the third Moufang Identity. Hence, (car ) (as c) = tc (car ) as − c (ar as ) c.
(45)
(cas ) (ar c) = tc (cas ) ar − c (as ar ) c.
(46)
Similarly,
Adding the two last expressions we get (car ) (as c) + (cas ) (ar c) = tc [(car ) as + (cas ) ar ] − c (ar as + as ar ) c.
(47)
By Theorem 6, it follows that (car ) (as c) + (cas ) (ar c)
= tc ctar as − c2 tar as = ctar as (tc − c) = nc tar as .
Thus, all terms have nc in common. Hence it follows that nq = nc np .
(48) (49)
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4. Methods for Constructing Octonionic Polynomials In this section, we propose some methods to construct an octonionic polynomial from a prescribed set of zeros. For the quaternionic case, this problem is already solved [1, 2, 7]. Naturally, we tried to extend to the octonions some algorithms valid to the quaternions. We verified that, using these algorithms, we were able to get a solution although weaker and not completely satisfactory. This means that, starting from a prescribed set of zeros and using the referred algorithms we obtain an octonionic polynomial whose zeros may not be the prescribed ones. Nevertheless, they belong to the conjugacy classes of the prescribed zeros. These are the results presented in Propositions 2 and 3. However, making use of other techniques it was possible to determine a polynomial with a prescribed set of zeros. We have already seen that any of two octonions give origin to an associative subalgebra and hence isomorphic to the quaternions. This implies that if we want to construct an octonionic polynomial whose zeros belong to an associative subalgebra of the octonions, the algorithms known for the quaternions continue valid. In this case, the set of zeros of the constructed polynomial will be exactly the prescribed set, as expected. We start with a generalization to the octonions of Beck’s algorithm [1] for the quaternionic case. In general, the zeros of the octonionic polynomial constructed with this algorithm are not the prescribed ones. We can only prescribe the classes to which they will belong. Proposition 2. Let a1 , a2 , . . . , am ∈ O, be not pairwisely similar octonions. The iteration p0 (x) = 1, h i −1 pi (x) = pi−1 (x)x − pi−1 (ai )ai (pi−1 (ai )) ? pi−1 (x), (50) i = 1, . . . , m, is defined and gives an octonionic polynomial of degree m, whose zeros are oi ∈ [ai ], i = 1, . . . , m. Proof. Consider a1 , a2 , . . . , am ∈ O such that [ai ] 6= [aj ], if i 6= j. Beginning with the polynomial p0 (x) = 1, we construct p1 according to the relation (50). That is p1 (x)
= x − a1 .
From this polynomial we construct p2 in the same way. Thus h i −1 ? p1 (x). p2 (x) = p1 (x)x − p1 (a2 )a2 (p1 (a2 )) It is easy to verify that p2 is defined because p1 (a2 ) 6= 0.
(51)
(52)
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Writing α2 = p1 (a2 )a2 (p1 (a2 )) np2 (x)
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and q1 (x) = (α2 ? p1 )(x), we have
(p2 ? p2 ) (x) = [p1 x − q1 ] ? [p1 x − q1 ] (x)
(53)
([p1 x − q1 ] ? [p1 x − q 1 ]) (x),
(55)
=
=
(54)
and from the definition of the product of octonionic polynomials, it follows that np2 (x)
=
(p1 ? p1 )(x)x2 − [(p1 ? q 1 )(x) + (q1 ? p1 )(x)] x +(q1 ? q 1 )(x)
(56)
But q1 (x) = (α2 ? p1 )(x), hence q 1 (x) = (p1 ? α2 )(x). Now, the expression p1 ? q 1 + q1 ? p1 in (56) has terms whose factors involve only two octonions and products of the form presented in Theorem 6. Thus, everything goes as if the factors associate. Therefore (p1 ? (p1 ? α2 )) (x) + ((α2 ? p1 ) ? p1 ) (x) = (p1 ? p1 ? α2 )(x) + (α2 ? p1 ? p1 )(x),
(57)
and from the definition of normal polynomial we conclude that (p1 ? (p1 ? α2 )) (x) + ((α2 ? p1 ) ? p1 ) (x)
= np1 (x)α2 + α2 np1 (x).
(58)
Since the normal polynomial has real coefficients, we obtain (p1 ? (p1 ? α2 )) (x) + ((α2 ? p1 ) ? p1 ) (x)
= np1 (x) (α2 + α2 ) = np1 (x)tα2 .
(59) (60)
From Theorem 10, we have that nq1 = nα2 np1 . Thus, from (56), (58) and (60), we obtain np2 (x)
= np1 (x)x2 − np1 (x)tα2 x + np1 (x)nα2 ,
(61)
and since all terms have in common the factor np1 (x), np2 (x)
= np1 (x) x2 − tα2 x + nα2 ,
(62)
or np2 (x)
= np1 (x)∆α2 (x).
(63)
Since np1 = ∆α1 , we have np2 = ∆α1 ∆α2 .
(64)
Hence, by Theorem 9 and having in account that αi ∼ ai , i = 1, 2, the zeros of p2 belong to the conjugacy classes [a1 ] and [a2 ]. Therefore, since by hypothesis a3 is not similar to neither a1 nor a2 , the iteration h i −1 p3 (x) = p2 (x)x − p2 (a3 )a3 (p2 (a3 )) ? p2 (x), (65) is defined.
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Thus, proceeding in a similar way, we obtain np3 = ∆α1 ∆α2 ∆α3 ,
(66)
−1
where α2 = p1 (a2 )a2 (p1 (a2 )) . So, proceeding successively up to index m, we see that the polynomials are defined and we obtain npm = ∆α1 ∆α2 . . . ∆αm .
(67)
As αi ∼ ai , i = 1, . . . , m, we conclude that ∆ai divides npm . Hence, by Theorem 9, there exist oi ∈ [ai ], i = 1, . . . , m, such that pm (oi ) = 0. Example 4.1. Let us consider the following set of octonions {i , 1 + j , −1 + ijk} . We verify that the three octonions belong to three distinct conjugacy classes. Using the iteration given in Proposition 2, we obtain the following polynomial: p(x) = x3 + ax2 + bx + c where a =
8i − 37j + 29ij + 33k − 4ik + 29jk − 37ijk 69
b =
−18 − 210j + 75ij − 50k − 50ik − 25jk + 210ijk 207
249i − 186j − 123ij − 41k − 152ik + 17jk − 186ijk . 207 Calculating the zeros of p we obtain Zero(p) = {o1 , o2 , o3 }, where c
=
o1
= −1 −
o2
=
1105i + 3382j + 1381ij + 87k − 1060ik + 431jk − 14719ijk 15249
1559i − 442j − 442ij − 228k + 10ik + 1042jk − 212ijk 2001
1105i − 320j + 1105ij + 3123k − 2026ik − 2881jk + 2807ijk . 15249 We see that, although the zeros were not the prescribed ones, they belong to the prescribed class of octonions. More specifically, the zeros o1 , o2 and o3 belong to the conjugacy classes of the octonions [−1 + ijk], [i] and [1 + j], respectively. o3
=
1−
In Proposition 2, the construction of polynomials is done by means of left multiplications of first degree polynomials. We observe that we can rewrite the iterative process as follows −1 ? pi−1 (x). (68) pi (x) = x − pi−1 (ai )ai (pi−1 (ai ))
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In the following result we show that there are advantages if the multiplication is done on the right instead of left. In this case, at least one of the zeros will be one of the prescribed. Lemma 2. Let q ∈ O[X] and u ∈ O. The octonionic polynomial p(x) = (q ? (x − u)) (x) has u as a zero. Proof. Let us consider the octonionic polynomial, q(x) = am xm + am−1 xm−1 + · · · + a1 x + a0 ,
(69)
and an octonion u. Thus, the octonionic polynomial p(x) = (q ? (x − u)) (x),
(70)
p(x) = am xm+1 + (am−1 − am u) xm + · · · + (a0 − a1 u) x − (a0 u) .
(71)
has the following form Evaluating p at u, we have p(u) = am um+1 + (am−1 − am u) um + · · · + (a0 − a1 u) u − (a0 u) ,
(72)
and, since each term involves only two octonions, the non-associativity is not posed. We conclude that p(u) = 0. Next we will present an extension to the octonions of Gordon-Motzkin’s Theorem, valid to the quaternions [7]. In the following result we present an algorithm to construct octonionic polynomials. This algorithm seems to be better than the proposed one in Proposition 2, since the set of zeros of the constructed octonionic polynomial has at least one of the prescribed zeros. Proposition 3. If pm (x) = [[[(x − a1 ) ? (x − a2 )] ? (x − a3 )] ? . . .] ? (x − am ),
(73)
where a1 , a2 , . . . , am ∈ O, then every zero of pm is similar to one of the octonions ai , i = 1, . . . , m. In particular, pm (am ) = 0. Proof. The proof is very similar to the one given for Proposition 2. Writing pm (x) = (pm−1 ? (x − am ))(x), we have npm (x) = (pm ? pm )(x) = [pm−1 (x) ? (x − am )] ? [pm−1 (x) ? (x − am )], (74) = [pm−1 (x) ? (x − am )] ? (x − am ) ? pm−1 (x) .(75) But, the products in this last relation are of the same kind as those in the proof of Proposition 2. Thus, we conclude that npm (x)
= npm−1 (x)∆am (x).
(76)
Now, in a similar way, since pm−1 (x) = (pm−2 ? (x − am−1 ))(x), we conclude that npm−1 = npm−2 ∆am−1 , and proceeding successively, we obtain npm = ∆a1 . . . ∆am−1 ∆am .
(77)
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Hence, from Theorem 9, we conclude that there exists a zero belonging to each conjugacy class [ai ], i = 1, . . . , m. In particular, by Lemma 2, we see that one of the zeros is precisely am . Example 4.2. Consider the following set of octonions {i, 1 + j, −1 + ijk}, whose elements belong to distinct conjugacy classes. 2 Using the algorithm, we obtain the following octonionic polynomial p(x) = ((x − i) ? (x − 1 − j)) ? (x + 1 − ijk) = x3 − (i + j + ijk)x2 − (1 + j − ij − ik + jk − ijk)x + i + ij − k + jk. Calculating the zeros of this polynomial, we find the following set Zero(p) = 6j + 4k − 2ik + 4jk + 3ijk i + 2j + 2ij + 6jk + 6ijk = −1 + ijk, 1 + , . 9 9 We see that octonion in the last factor, −1 + ijk, belongs to set of zeros and that the other two are similar to 1 + j and i, respectively. Up to now, we saw two algorithms to construct octonionic polynomials. As showed, these algorithms do not give, in general, octonionic polynomials with the prescribed set of zeros. What we will show next is how to construct an octonionic polynomial with a prescribed set of zeros. This set must verify some conditions as we will see. Consider the monic octonionic polynomial of degree m, pm (x) = xm + am−1 xm−1 + · · · + a1 x + a0 ,
(78)
and λ1 , . . . , λm ∈ O. We want the octonions λi , i = 1, . . . , m, to be the zeros of the polynomial pm . Thus, we have to determine the coefficients ai , such that they satisfy the following m equations . m−1 pm (λ1 ) = λm + .. + a1 λ1 + a0 = 0, 1 + am−1 λ1 . m−1 pm (λ2 ) = λm + .. + a1 λ2 + a0 = 0, 2 + am−1 λ2 (79) .. . . m−1 pm (λm ) = λm + .. + a1 λm + a0 = 0. m + am−1 λm 2 Observe
that Proposition 3 does not impose this restriction. We will see later the case where some of the elements of the set belong to the same conjugacy class.
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In other words, the problem consists in solving the following system of linear equations with octonions, . am−1 λm−1 + .. + a1 λ1 + a0 = −λm 1 , 1 .. m−1 am−1 λ2 + . + a1 λ2 + a0 = −λm 2 , . .. . am−1 λm−1 + .. + a1 λm + a0 = −λm m m.
(80)
If we rewrite this system of equations in a matrix form [am−1 . . . a1 a0 ]
λm−1 1
λm−1 2
···
m−1 λm
.. .
.. .
.. .
λ1
λ2
···
λm
1
1
···
1
m = − [ λm 1 . . . λ2
λm m ] , (81)
we see that the problem of determining the coefficient passes through the inversion of an octonionic Vandermonde matrix. Since the inversion of octonionic matrices is much more complicated than the quaternionic case, we will use the matrix representations for the octonions and appeal to the algebra of block matrices. In this way, using the bra representation, we can rewrite the system of equations (80) in the following way: . ham−1 λm−1 + .. + a1 λ1 + a0 | = −hλm 1 |, 1 .. m−1 ham−1 λ2 + . + a1 λ2 + a0 | = −hλm 2 |, .. . . + .. + a1 λm + a0 | = −hλm ham−1 λm−1 m |. m
(82)
By Corollaries 2.1 and 3.1, we have . ham−1 |φ(λm−1 ) + .. + ha1 |φ(λ1 ) + ha0 | = −hλm 1 |, 1 . ham−1 |φ(λm−1 ) + .. + ha1 |φ(λ2 ) + ha0 | = −hλm 2 |, 2 . . . . ham−1 |φ(λm−1 ) + .. + ha1 |φ(λm ) + ha0 | = −hλm m m |,
(83)
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or, in matrix notation, [ham−1 | . . . ha1 | ha0 | ]
φ(λm−1 ) 1
φ(λm−1 ) 2
.. .
.. .
φ(λ1 )
φ(λ2 )
···
φ(λm )
φ(1)
φ(1)
···
φ(1)
···
φ(λm−1 ) m
.. .
m m = − [hλm 1 | . . . hλ2 | hλm | ] ,
(84)
or [ham−1 | . . . ha1 | ha0 | ]
Λ1m−1
Λm−1 2
.. .
.. .
Λ1
Λ2
···
Λm
I
I
···
I
···
Λm−1 m
.. .
m m = − [hλm 1 | . . . hλ2 | hλm | ] ,
(85)
where Λj = φ(λj ), j = 1, . . . , m, and having in account Corollary 4.1. Thus, the determination of an octonionic polynomial with a prescribed set of zeros, passes through the inversion of a block Vandermonde matrix. In this sense, we enunciate the following result. Proposition 4. Let λ1 , λ2 , . . . , λm ∈ O be distinct, and V (Λ1 , Λ2 , . . . , Λm ) be the block Vandermonde matrix m−1 Λ1 Λm−1 · · · Λm−1 m 2 .. .. .. . . . , V (Λ1 , Λ2 , . . . , Λm ) = Λ1 Λ2 ··· Λm I
I
···
I
where Λi = φ(λi ), i = 1, . . . , m, m ≥ 3. The Vandermonde matrix V (Λ1 , Λ2 , . . . , Λm ) is singular if, and only if, there exist at least three octonions λi , i = 1, . . . , m, belonging to a same conjugacy class. Proof. Only if: Let λ1 , λ2 , . . . , λm ∈ O, m ≥ 3, be distinct and Λi = φ(λi ), for i = 1, . . . , m. Let us suppose that the block Vandermonde matrix V = V (Λ1 , Λ2 , . . . , Λm ) is singular.
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Consider a row matrix with 8m columns of real elements, A = [ a(m−1)1 . . . a(m−1)8 · · · a01 . . . a08 ].
(86)
Multiplying A by the Vandermonde matrix V and equaling to zero, we obtain an homogeneous system of equations, AV = 0. Since V by hypothesis is singular, the solution set of this system contains a non-trivial solution. This means that there exists at least a solution for which the 8m real elements aij , with i = 0, . . . , m − 1 and j = 1, . . . , 8, are not simultaneously null. Partitioning the matrix A in m matrices 1 × 8, we see that there exist m octonions, not simultaneously null, such that A = [ ham−1 | · · · ha0 | ],
(87)
where hai | = [ ai1 . . . ai8 ], with i = 0, . . . , m − 1. The Vandermonde matrix structure suggests that, in the study of the homogeneous system, AV = 0, we consider the m equations of the form ham−1 |Λm−1 + ham−2 |Λm−2 + · · · + ha1 |Λi + ha0 | = h0|, i i
(88)
for i = 1, . . . , m. Having into account that Λi = φ(λi ), by Corollaries 2.2 and 3.1, we can write ham−1 λm−1 + am−2 λm−2 + · · · + a1 λi + a0 | = h0|, i i
(89)
am−1 λm−1 + am−2 λim−2 + · · · + a1 λi + a0 = 0, i
(90)
or for i = 1, . . . , m. As ai are not simultaneously null, we conclude that exists the polynomial p(x) = am−1 xm−1 + am−2 xm−2 + · · · + a1 x + a0 ,
(91)
of degree, at most, m − 1, and that the λi , i = 1, . . . , m, are zeros of this polynomial. From Corollary 9.1, the octonions λi , i = 1, . . . , m, belong, at most, to m−1 distinct conjugacy classes. Let us suppose that in each of these classes exist, at most, two zeros. Let s1 and s2 be the number of conjugacy classes with one and two zeros, respectively. Hence, we have that s1 + s2 ≤ m − 1 and s1 + 2s2 = m. From Corollary 9.2, for each single zero of a class corresponds a degree of the polynomial. Besides this, by Corollary 8.1, if two distinct zeros are similar, then all elements of this conjugacy class are also zeros, being the characteristic polynomial of this class a divisor of the polynomial. Thus, to each pair of distinct similar zeros corresponds two degrees to the polynomial. Hence, we have that deg(p) = s1 + 2s2 = m > m − 1, which contradicts (91).
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We conclude that, if the Vandermonde matrix V (Λ1 , Λ2 , . . . , Λm ) is singular, then at least three λi , i = 1, . . . , m, belong to the same conjugacy class. If: Let λ1 , λ2 , . . . , λm ∈ O, m ≥ 3, be distinct. Let us suppose that, among these octonions, there exists at least three belonging to a same conjugacy class. Without loss of generality, let us suppose that these octonions are λ1 , λ2 and λ3 . As, by hypothesis, these three octonions are similar, they have the same characteristic polynomial, and so they verify the following relations ∆λ1 (λi ) = λ2i − tλi + n = 0,
i = 1, 2, 3,
(93)
where t = 2Re(λ1 ) and n = |λ1 |2 . In matrix notation, we have the following relations Λ2i − tΛi + nI = 0,
i = 1, 2, 3.
(94)
Furthermore, multiplying these relations by matrix Λji , we obtain the following relations Λj+2 − tΛj+1 + nΛji = 0, i i
j ≥ 0, i = 1, 2, 3.
(95)
Consider the Vandermonde matrix V =
Λm−1 1 .. . Λ31 Λ21 Λ1 I
Λ2m−1 .. . Λ32 Λ22 Λ2 I
Λm−1 3 .. . Λ33 Λ23 Λ3 I
Λm−1 4 .. . Λ34 Λ24 Λ4 I
···
··· ··· ··· ···
Λm−1 m .. . Λ3m Λ2m Λm I
,
(96)
and denote each row by Li , i = 1, . . . , m. Using operations, over rows, of the form L0i = Li − tLi+1 + nLi+2 ,
i = 1, . . . , m − 2,
(97)
where t = 2Re(λ1 ) and n = |λ1 |2 , and having into account the relations (95), by gaussian elimination on the matrix V , we obtain the matrix
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Let us partition the matrix V 0 in two blocks: J composed by the first 3 × 8 columns and K by the remaining (m − 3) × 8 columns. By hypothesis, m ≥ 3. As the first 8 × (m − 2) rows of J are null, we have that rank (J ) ≤ 16 rank (K) ≤ 8(m − 3).
(98) (99)
Using for the matrix [J K] a known inequality (see [9], pp. 158), which relates the rank of the sum of two matrices with the sum of rank of each matrix, we obtain rank([A B])
= rank([J K]) = rank([J 0] + [0 K]) ≤ ≤ ≤ ≤
