# Polynomials and vandermonde matrices over the field of quaternions.

1. Introduction. Let B be a compact topological Hausdorff space and X := C(B) the normed vector space of all real valued, continuous functions defined on B with norm || f || := [max.sub.x [member of] B] [absolute value of f (x)]. Consider the set of all n-dimensional subspaces of X with n [greater than or equal to] 2 (let B contain sufficiently many points). We investigate whether there is a Haar space in this set. This is a space V with the following property: Given arbitrary, but pairwise distinct points [t.sub.j] [member of] B and arbitrary real numbers [u.sub.j], there is a unique v [member of] V, such that v([t.sub.j]) = [u.sub.j], j = 1,2, ..., n. Thus, in Haar spaces of dimension n all interpolation problems in the above sense can be solved uniquely, regardless of the choice of [t.sub.j] and [u.sub.j], j = 1,2, ..., n. The only restriction on the [t.sub.j] is that they be pairwise distinct. This type of space is also called unisolvent. The prototype of a Haar space is [[PI].sub.n-1], the space of all real polynomials of degree at most n - 1 on a compact interval of positive length. A counter-example is the span (x, [x.sup.2], ..., [x.sup.n]) on a compact interval containing the origin. The fact that Haar spaces do not exist if B is a subset of [R.sup.k] with k [greater than or equal to] 2 is known for a long time, Haar [6, p. 311]. For a proof we refer to the original paper. The essential ingredients of the proof are properties of the determinant of the matrix which describes the interpolation problem and the intermediate value theorem for real valued, continuous functions.

Since there is no intermediate value theorem for complex valued functions, the proof does not carry over to the case B [subset] C, though B may be regarded as two dimensional in this case. However, as is also well known, C(B) contains Haar spaces if B [subset] C and if C(B) is now the space of complex valued functions on B. The set of complex polynomials, also denoted by [[PI].sub.n-1] is again a prototype. A more precise information on what subsets of C allow the definition of Haar spaces is given by Mairhuber . In the quaternionic case all essential ingredients of the proof are missing. There is no determinant, Fan , and no intermediate value theorem. The quaternionic case is the topic of the next sections.

A comprehensive bibliography on quaternions ordered with respect to subjects has been published by Gsponer .

2. Quaternionic polynomials. By H we denote the (skew) field of quaternions. A polynomial on H is already a very complicated item. A monomial of degree j [greater than or equal to] 0 is a mapping [m.sub.j] : H [right arrow] H defined by

(2.1) [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII.]

A polynomial of degree n is any finite sum of monomials of degree < n. Therefore, the space of all polynomials of degree [less than or equal to] n has no finite dimension. According to Eilenberg and Niven  a polynomial p of degree [greater than or equal to] 1 with the property that the monomial with the highest degree in p occurs exactly once, has at least one zero. This is called The Fundamental Theorem of Algebra for Quaternions by the two mentioned authors. And there is no hope that the restriction on the monomial with the highest degree can be weakened, since p(x) := a[x.sup.n] - [x.sup.n] a - 1 has no zero. This follows by application of the real part R which is linear and commutative, hence, Rp(x) = -1, implying that p cannot have a zero. This example is taken from Pumpliin and Walcher , which also contains a review and expansion of some results on the number of zeros of polynomials on quaternions H. The Fundamental Theorem can be applied to the polynomial p(x) := [(x - a).sup.2] = [x.sup.2] - xa - ax + [a.sup.2], a [member of] H\{R} and shows that, in general, we cannot expect more than one zero. This applies even to polynomials without repetition of monomials of the same degree. Examples are p(x) := [x.sup.2] - x(i + j) + k and q(x) := [x.sup.2] - (i + j) x + k where i := (0,1,0,0), j := (0,0,1,0), k := (0,0,0,1). We have p(i) = q(j) = 0 and there are no other roots. 

3. Quaternionic simple polynomials. We will turn our attention to polynomials of one of the following types:

(3.1) pl(x) := [a.sub.0] + [a.sub.1]x + [a.sub.2][x.sup.2] + xxx + [a.sub.n][x.sup.n],

(3.2) pr (x) := [a.sub.0] + x[a.sub.1] + [x.sup.2][a.sub.2] + xxx + [x.sup.n][a.sub.n], [a.sub.0], [a.sub.1], ..., [a.sub.n]; x [member of] H.

We will call these polynomials simple. If the coefficients are real, the two types coincide. They also coincide with the polynomials of general type if all coefficients are real. Thus, a real polynomial (i. e. having real coefficients) is always simple. There is a tight connection between pl and pr, which is explained in Janovska and Opfer .

Theorem 3.1. Let p be a real polynomial. If p has (as a polynomial over C) only real zeros, then p as a polynomial over H has no other zeros. If p has (as a polynomial over C) also complex zeros, then, p as a polynomial over H has infinitely many zeros. More precisely: Let z be one of the complex zeros of p, then, [h.sup.-1] zh is also a zero for all h [member of] H\{0}. There are no other zeros.

Proof. It is easy to see that p([h.sup.-1] zh) = [h.sup.-1] p(z)h for all real polynomials and all h [member of] H\{0}; see also Janovska and Opfer  and Pumplun and Walcher .

The mapping

x [right arrow] [h.sup.-1] xh, h [member of] H\{0}

defines an equivalence relation in H with equivalence classes

(3.3) [x] := {u := [h.sup.-1] xh : h [member of] H\{0}}.

The following lemma makes it easy to recognize equivalent elements.

Lemma 3.2. Two quaternions a, b are equivalent (in the sense a = [h.sup.-1] bh for some h [not equal to] 0) if and only if

(3.4) Ra = Rb and [absolute value of a] = [absolute value of b]

Proof. Janovska and Opfer .

By using (3.3), (3.4) and x := ([x.sub.1], [x.sub.2], [x.sub.3], [x.sub.4]) we can also write

[x] = {([x.sub.1], [u.sub.2], [u.sub.3], [u.sub.4]) [member of] H : [u.sup.2.sub.2] + [u.sup.2.sub.3] + [u.sup.2.sub.4] = [r.sup.2] := [x.sup.2.sub.2] + [x.sup.2.sub.3] + [x.sup.2.sub.4]}.

This is apparently a sphere in H where the first component, [x.sub.1], is fixed. If x [member of] R then we have [x] = {x} which means that [x] contains exactly the element x. Let z [member of] C. Then, the complex conjugate [bar.z] is also belonging to [z], and if z is nonreal, then [z] contains infinitely many elements. We can put Theorem 3.1 into a simpler form.

COROLLARY 3.3. Let p be a real polynomial over H of degree n. Then the set of all zeros can be partitioned into at most n equivalence classes.

For simple polynomials (with quaternionic coefficients) the zeros fall in two classes. Let z be a nonreal zero. Then, either all elements of the equivalence class [z] consist of zeros, or apart from z there is no zero in [z]. In the first case the zero z (and all zeros in [z] as well) is called a spherical zero and in the second case the zero is called an isolated zero. If a zero is real, then it will also be called isolated. See Pogorui and Shapiro  and Janovska and Opfer  for details. The next theorem contains a statement on the number of zeros of a simple, quaternionic polynomial. For a proof see also [10, 13].

Theorem 3.4. Let p be a simple polynomial over H of degree n. Then, p has [n.sub.1] [greater than or equal to] 0 isolated zeros and [n.sub.2] [greater than or equal to] 0 equivalence classes of zeros with 1 [less than or equal to] [n.sub.1] + [n.sub.2] [less than or equal to] n.

That means, that Corollary 3.3 is also valid for simple polynomials. We called the polynomials defined in (3.1) and in (3.2) simple. In the literature one finds also other words for simple. There is an Italian group, Gentile et al. [3, 4], who refers to these polynomials as regular, and there are two other groups, a Portuguese one, Serodio et al. ), and a Brazilian one, de Leo et al. , who refer to these polynomials as unilateral. It should be remarked, that both the Italian and the Brazilian groups did not take notice of the mentioned paper by Pogorui and Shapiro .

4. Location of zeros of quaternionic polynomials. For complex polynomials there are some theorems saying that all roots are outside a certain disk centered at the origin, and that all roots are inside some other disk also centered at the origin. We will show that analogous results hold for simple and even for almost all types of quaternionic polynomials. Without loss of generality we may assume that [a.sub.n] = 1 and [a.sub.0] [not equal to] 0 in the simple polynomials (3.1), (3.2) if we are interested in their zeros.

Theorem 4.1. Let p be a simple polynomial over H of degree n with an = 0, and [a.sub.0] [not equal to] 0. Then, the open ball {z [member of] H : [absolute value of z] < r} does not contain any zero of p, where r is the only positive root of the real polynomial

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII.]

Proof. We have [??] (0) = -[absolute value of [a.sub.0]] < 0 and [??](x) > 0 for sufficiently large x > 0. In addition, [??]'(x) > 0 for all x > 0, implying that p is strictly increasing for all x > 0. Thus, there is exactly one positive zero, denoted by r. Let p := pl and p(x) = 0, hence

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII.]

The proof for p := [p.sub.r] is the same.

Theorem 4.2. Let p be a simple polynomial over H of degree n with [a.sub.n] = 1 and [a.sub.0] [not equal to] 0. Then, all zeros of p are contained in the ball

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII.]

Proof. Let p := pl and x be a zero of p and assume the contrary, hence [absolute value of x] > R, in particular, [absolute value of x] > 1 . Then,

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII.]

a contradiction. Thus, [absolute value of x] [less than or equal to] R. The proof for p = [p.sub.r] is the same. Example 4.3. The two simple, quadratic polynomials

pl (x) := [x.sup.2] + jx + i, pr(x) := [x.sup.2] + xj + i

have the following roots:

Roots of pl : [x.sub.1] := 0.5 (-1,1, -1, -1), [x.sub.2] := 0.5 (1, -1, -1, -1).

Roots of pr : [x.sub.1] := 0.5 (-1,1, -1, 1), [x.sub.2] := 0.5 (1, -1, -1, 1).

All roots have absolute value one. Applying Theorem 4.1 yields r := [square root of (5-1)]/2 [approximately equal to] 0.618, and Theorem 4.2 yields R := 2.

Both bounds, r, R, are sharp for p(x) := [x.sup.n] + 1. Interestingly, Theorem 4.1 can be carried over to general polynomials.

COROLLARY 4.4. Let p(x) := [[SIGMA].sup.n.sub.j = 0 [mu].sub.j](x) be a polynomial of degree n over H, where each [[mu].sub.j] is a finite sum of monomials of degree j defined in 2.1:

(4.1) [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII.]

with [absolute value of ([A.sub.n])] [not equal to] 0, [absolute value of ([A.sub.n])] [not equal to] 0. Then, there is no zero of p located in the open ball

{z [member of] H : [absolute value of (z)] < [??]},

where [??] is the only positive zero of the real polynomial

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII.]

Proof. Repeat the proof of Theorem 4.1 replacing [a.sup.j][x.sup.j] with [[mu].sub.j] (x).

For the proof it is apparently sufficient to assume that [[absolute value of ([A.sub.j])] [not equal to] for one 1 [less than or equal to] j [less than or equal to] n.

In order to generalize Theorem 4.2 we need to make the assumption that the highest degree monomial occurs only once.

Corollary 4.5. Let p(x) := [[SIGMA].sup.n.sub.j=0[[mu].sub.j] (x) be a polynomial of degree n over H, where each [[mu].sub.j] is a finite sum of monomials of degree j defined in (2.1), with the exception that [[mu].sub.n] consists only of a single monomial of degree n. Define [[absolute value of ([A.sub.j])] as in (4.1) and assume that [absolute value of ([A.sub.n])] = 1, [absolute value of ([A.sub.0])] [not equal to] 0. Then, all zeros of p are contained in the ball

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII.]

Proof. Repeat the proof of Theorem 4.2 and replace [a.sub.j[x.sup.j] with [[mu].sub.j].

Example 4.6. Let p(x) := [x.sub.2] - ax - xa + [a.sup.2] with a [member of] H\{R}. The polynomial p has the single zero x = a. Corollary 4.4 yields [??](x) := [x.sup.2] + 2[absolute value of a] - [[absolute value of x].sup.2]. The only positive root is [??] := ((square root of (2)]) - 1)[absolute value of a]. Corollary 4.5 yields [??] := max{1, [([absolute value of a] + 1).sup.2] - 1}. The smallest [??] = 1 is obtained for [absolute value of a] := [square roof of (2)] - 1 [approximately equal to] 0.41 which yields [??] := 3 - 2 [square root of (2)] [approximately equal to] 0.17.

5. The interpolation problem and the Vandermonde matrix. Let B [subset] H be a compact set and X := C(B) the space of all quaternion valued, continuous functions defined on B. The general question is whether there are Haar spaces V [subset of] C( B) . We shall show, that the polynomial space composed of simple polynomials is not a Haar space. For this purpose let us study two interpolation problems: Given arbitrary, but pairwise distinct points [t.sub.j] [member of] H and arbitrary numbers [u.sub.j] [member of] H, j = 0,1,..., n, we are interested in whether the interpolation problems

(5.1) [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII.]

(5.2) [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII.]

have a solution, where [p.sub.l]; [p.sub.r] are simple polynomials of degree n, defined by (3.1) and (3.2), respectively. The following matrix V will be called Vandermonde matrix:

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII.]

The two problems (5.1), (5.2) are equivalent to the following two matrix problems, respectively:

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII.]

where

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII.]

The main point here is, that these problems are different in general. Since quaternionic matrices have no determinants, Fan , the singularity of a matrix must be defined in a more elementary way.

6. Short excursion to linear systems over H. We will give some elementary properties of matrices with quaternions as entries and mappings defined by matrices; see Zhang  for an overview of such mappings and for possible decompositions for such matrices. For more general types of linear mappings; cf. Janovska and Opfer .

Definition 6.1. Let A [member of] [H.sup.pxq]. The maximal number of right independent columns of A will be called right column rank of A. Let p = q. The square matrix A will be called nonsingular, if the right column rank is maximal, i. e. rank(A) = p, where rank is to be understood as right column rank. The mapping f : [H.sup.p] [right arrow]] [H.sup.p] defined by

(6.1) f(x) := Ax

will be called nonsingular if A is nonsingular.

In the same fashion left column rank and left, right row rank of A are defined. A standard theorem in non commutative linear algebra is: The right column rank coincides with the left row rank and the left column rank coincides with the right row rank. Thus, a quaternionic matrix has two ranks, the right and the left column rank. The above definition already suggests that the right column rank will be more important than the left column rank, since in Ax the components of x are always on the right of the matrix elements of A.

Theorem 6.2. Let A [member of] [H.sup.pxp] be a square matrix. The mapping f as defined in (6.1) is singular (i. e. not nonsingular) if and only if the homogeneous system f (x) = 0 has non trivial solutions. The system f (x) = c has a unique solution for all c [member of] [H.sup.p] if and only if f is nonsingular. The mapping g defined by g(x) := [x.sup.T]A is singular if and only if f is singular, where [x.sup.T] denotes the transpose of x.

Proof. The mapping f defined in (6.1) may be regarded as a right linear combination of the columns of A. The mapping g may be regarded as a left linear combination of the rows of A, and the right column rank and the left row rank coincide. The remaining part is easy.

Definition 6.3. Let A [member of] [H.sup.pxp]. The right column rank of A will be called rank of A. That f (x) := Ax and f[(x).sup.T] := [x.sup.T][A.sup.T] do not define the same mapping (apart from transposition) will be shown by the following example; see Zhang . Example 6.4. Let

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII.]

This matrix has right column rank 2 and left column rank 1. The latter statement can be easily verified by multiplying the first column of A from the left by i. The result is the second column. The transpose [A.sup.T] has right column rank 1 and left column rank 2. Thus, A is nonsingular, whereas [A.sup.T] is singular.

7. Vandermonde matrices, continued. We return now to the Vandermonde matrix and the corresponding interpolation problems (5.1), (5.2). It is clear that the Vandermonde matrix and its transpose are nonsingular for n [less than or equal to] 1. In order to show that it is possible to find singular Vandermonde matrices for general n [greater than or equal to] 3 the idea is the following: Try to find pairwise distinct points [t.sub.0], [t.sub.1], ..., [t.sub.n] such that the sum of the first and penultimate row equals the sum of the second and last row. If this is possible, the left and right row rank are not maximal, which implies that the rank is not maximal. Hence, V and [V.sub.T] are singular.

Theorem 7.1. Let n = 2. Define [t.sub.0] := i := (0,1,0,0), [t.sub.1] := j := (0,0,1,0), [t.sub.2] := k := (0, 0, 0, 1) , and

(7.1) [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII.]

Let n [greater than or equal to] 3 and h [member of] H\{C} be arbitrary. Define the following Vandermonde matrix V: Put to = 1 and set

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII.]

where [e.sub.1], [e.sub.2], ..., [e.sub.n-1] are the real or complex roots of

(7.2) [t.sup.n-1] + 1 = 0.

If -1 is one of the roots, then let [e.sub.1] = -1 , otherwise, choose any enumeration of the roots. Then V and its transpose VT are singular.

Proof. For n = 2 it is obvious that [t.sub.0], [t.sub.1], [t.sub.2] are pairwise distinct and that V defined in (7.1) and [V.sub.T] are singular. Let n [greater than or equal to] 3. It is clear that the second row of V contains only pairwise distinct entries, in particular, [t.sub.n] := [h.sup.-1][e.sub.n-1] h [??] C. The formula (7.2) implies that the first and penultimate row sum to (2,0,..., 0). Formula (7.2) implies [t.sup.n] + t = 0 for all roots, which implies that row 2 and the last row also sum to (2,0,..., 0). Thus, the (right and left) row rank are not maximal. It follows that the rank is not maximal and in all cases n [less than or equal to] 2 the Vandermonde matrix V and its transpose [V.sup.T] are singular.

We observe that for all selected points (second row of Vandermonde's matrix) the absolute value is one. That means, we can restrict our considerations to the unit ball B := {z [member of] H : [absolute value of x] [less than or equal to] 1} or to the unit sphere [partial derivative]B := {z [member of] H : |z| = 1}.

8. Unisolvency and the number of zeros. In the theory of real or complex valued continuous functions the existence of Haar spaces of dimension n is equivalent to the fact that the elements in the Haar space do not have more than n - 1 zeros with the only exception of the zero function. We will see that even in quaternionic spaces the situation is analogue.

Theorem 8.1. Let V [subset of] C(B) be a vector space with (left or right) dimension n, where the set C( B) is the set of quaternion valued, continuous functions on B, and B [subset] H a compact set. The space V is a Haar space if and only if all functions in V\{0} have at most n - 1 zeros.

Proof. (a) Assume V is a Haar space. Then, v([t.sub.j]) = 0, j = 1,2,..., n for pairwise distinct [t.sub.j] [member of] implies v = 0. Thus, any v [not equal to] 0 can have at most n - 1 zeros in B. (b) Assume V is not a Haar space. Then there are pairwise distinct points [t.sub.j] and values [u.sub.j], j = 1, 2..., n, such that the interpolation problem v ([t.sub.j]) = [u.sub.j], j = 1, 2 ..., n, has no or two different solutions [v.sub.1], [v.sub.2]. In the latter case, v := [v.sub.1] - [v.sub.2] is not the zero function but has n zeros at the given n points tj. If the interpolation problem has no solution, then the homogeneous problem v ([t.sub.j]) = 0, j = 1, 2 ..., n must have a non trivial solution. In all cases there is a non zero function v with at least n zeros.

The fact that polynomials have too many zeros is, thus, responsible for the fact that polynomials do not form a Haar space. The question, whether there are Haar spaces in the quaternionic C(B) remains open.

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GERHARD OPFER ([dagger])

* Received November 11, 2008. Accepted for publication December 1, 2008. Published online June 12, 2009. Recommended by L. Reichel.

([dagger]) University of Hamburg, Faculty for Mathematics, Informatics, and Natural Sciences [MIN], BundestraBe 55, 20146 Hamburg, Germany, opfer@math.uni-hamburg.de

(1) The first example was communicated by Fabio Vlacci, Firenze, Italy.
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