Download Equations in Quaternions

Equations in Quaternions
Author(s): Ivan Niven
Source: The American Mathematical Monthly, Vol. 48, No. 10 (Dec., 1941), pp. 654-661
Published by: Mathematical Association of America
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654
EQUATIONS
[December,
IN QUATERNIONS
lines Ax = By and Bx = Cy. A constructionfor the hyperbola b2x2
-a2y2 = a2b2
y=2bt/(1-t2). The
series for logeN was exhibited as the result of certain rearrangementsof the
series 1-1/2+1/3-1/4+
. Unusual weighted sums were considered as
approximations for a definiteintegral and contrasted with the usual "rules."
Finally, xrwas computed by the use of the inversesine.
7. ProfessorPawley described extensionsof Newton's method forapproximating real roots in which the desired root is approximated by x interceptsof
curves of higherorderof contact than the tangent.He derived an upper bound
to the errorinvolved in these approximations. In particular,he simplifiedthe
well known parabolic approximationby expanding an x interceptof the parabola into a convergentalternating series. An upper limit to the errorinvolved
in this approximationwas derived and illustratedby an example.
A. J. LEWIS, Secretary
was relatedto the parametricformx=a(l+t2)/(l-t2),
EQUATIONS
IN QUATERNIONS
of Illinois
IVAN NIVEN, University
1. Introduction.We prove the existenceof a quaternion root of the equation
(1)
a(x) = xm+ alxml + a2xm-2+ . .. + am=O,
am 7
,
with coefficientsfromthe algebra of real quaternions. The writerhad proved
this result when m is odd, but the proof was renderedobsolete when Nathan
Jacobson pointed out that the result(without restrictionon m) can be obtained
as a simple consequence of some work of Ore [I]. This is given in detail in ?2.
In ?3 we give a method for obtaining the roots of (1), which is not very
practical in the sense that it involves the simultaneoussolving of two real equations of degree 2m-1. The method used is a generalizationof Sylvester's treatment [2 ] of the quadratic equation correspondingto (1). Sylvester's conclusion
that a quadratic equation has six roots is incorrectbecause he neglectsto show
that they exist,and also overlooksthe possibilityof an infinitenumberof roots;
a complete analysis is given in ?4 (Theorem 2). The number of roots of (1) is
discussed in ?5 (Theorem 3), necessaryand sufficientconditionsbeing given for
an infinitenumberof roots.
The proofgiven here of the existenceof a root of (1) is stated forthe general
case where the coefficients
of the equation are quaternions over any real-closed
fieldR (i.e., no sum of squares in R is equal to -1, and no algebraic extensionof
R has this property).
Reinhold Baer, on hearing of this existence proof,proved the converse, so
that we have the followingstrongresult:
THEOREM 1. Let D be a non-commutative
divisionalgebrawithcentrumC. Then
everyequation (1) withcoefficients
fromD has a solutionin D if and onlyif C is
a real-closedfield,and D is thealgebraof real quaternionsoverC.
1941]
EQUATIONS
655
IN QUATERNIONS
The necessityof these conditionsis shown in ?6; the proofgives a slightly
strongerresultthan stated in the theoremabove, since only those equations with
coefficientsfrom C are used. The writeris indebted to Jacobson and Baer for
permissionto give theirproofshere.
Note that (1) is a special equation. The most general quadratic term, for
example, would have the formbxcxd,involvingthree coefficients.However, the
resultsare valid forequations similar to (1) having all coefficientsto the right
of the powers of the unknown.
2. The existence of a root. Let the coefficientsof (1) be quaternions over a
real-closed fieldR. By replacing the quaternions a, by their conjugates Jr, we
obtain a polynomialj(x). We multiplythis on the rightby a(x), and allow x to
be commutativewith the coefficients.Thus we obtain a polynomial a(x) with
coefficientsin R, which is, by the fundamentaltheorem of algebra, factorable
into linear factorsin R(i, x), and hence a fortioriin R(i, j, x). Theorem 1 on p.
494 of Ore's paper [1 ] states that any other factorizationof a(x) in R(i, j, x)
must also have linear factors.Now a(x) can be factoredinto irreduciblefactors,
and these are factorsof a(x). Hence by Ore's theoremthey are linear. Taking
x-c as the rightlinear factor,we can write
a(x) =
(xm-l
+
blxm-2 +
. - -
+ bm1)(x - c).
That x = c is a root of a(x) = 0 is not apparent fromthis equation, since we have
assumed that x commutes with the coefficients.However, upon rewr;tingthe
above equation in a formanalogous to (1),
a(x) =
xm +
(b1 -
c)xm-l
+ (b2-
bic)Xm-2 +
+
(bm_- -
bm-2C)X -
b.-ic,
we verifyimmediatelythat c is a root.
3. A right-divisionalgorithm.The normn of any quaternion x is definedas
the product of x and its conjugate x; and the addition of x and x gives t, the
trace of x. It is well known that x satisfiesthe equation
X2-tx + n = O.
(2)
We now divide a(x) on the rightby the expressionin this equation, and obtain
the algorithm
(3)
a(x) =
q(x2
-
tx + n)
+ f(ar, t, n)x + g(a,, t, n),
the remainderbeing comprisedof the last two terms,polynomialsin t, n and the
of a(x). The nature of the quotient q does not interestus. Note that
coefficients
the remaindervanishes forany root of equation (1), and conversely.
When fzF0, the vanishing of this remaindercan be expressed in the form
1
(4)
--i
(fg)
f
ff
wheref,the conjugate off, is obtained by replacingeach a,. in f by its conjugate
656
EQUATIONS
[December,
IN QUATERNIONS
J,. Since the conjugate of a product equals the product of the conjugates in reverse order,we have
ff
By multiplicationand addition of the last two equations we get the norm and
trace of x; thus
(5)
n
f(kg),
fi
t=
y(fg
ff
+ gf),
since ff and gg have real coefficients
and are commutativewith the other polynomials. These equations may be writtenin the forms
(6)
N(t, n)=nff-gg
= 0,
T(t, n) = tff+ fg + gf=0,
where N(t, n) and T(t, n) are polynomialsin t and n with real coefficients.
First we note that any rootxo of (1) has a trace toand a normnowhichsatisfy
equations (6). This is apparent except whenf(ar, to,no) = 0, in which case equation (4) is meaningless. But in this case equation (3) implies that g(r, to,no)
vanishes, and equations (6) are satisfied.
Conversely,any simultaneous real solution (to,no) of (6) gives one or more
roots of (1). First suppose thatf(ar, to,no)#0. The values toand no can be substituted in (4) to give a quaternion xo, and since these quantities satisfy (2),
equation (3) indicates that xo is a root of a(x) =0. It is importantto note that
in this case one solution of equations (6) gives-exactlyone solution of (1).
On the other hand, iff(ar, to,no) = 0, then the firstequation (6) gives
g(ar, to, no)g(ar,
to, no) =
0.
But the product of a quaternion and its conjugate is zero only if the quaternion
is zero, and hence g(ar, to,no) =0. Returningto (3), we see that any solution of
(7)
x2-tox+no=0
is also a solutionof (1). The above analysis enables us to inquire into the number
of roots of (1), but firstwe must know the numberof solutions in quaternions
of equation (7).
4. Quadratic equations. Consider the equation
(8)
x2 + bx + c =
0,
cG
0,
b and c being real quaternions. We assume that t(b), the trace of b, is zero, for
otherwisethe substitutionx =y -4t(b) gives a quadratic equation with the required property.For example, the substitutionx=y+-to in equation (7) gives
(9)
2
d 0
d=
2to-nO
1941]
657
EQUATIONS IN QUATERNIONS
We shall need the treatmentof this equation to complete the discussion of (8).
each y with a subscriptbeing real. We subSuppose that y =yo+y1i+y2j+y3ij,
stitute in (9) and separate the result with respect to the linearly independent
units 1, i, j, and ij, to obtain
2
2
2
2
yoyl=YoY2
d,
y0-yl-y2-y3=
YoY3
= 0.
and the roots of (9) are ?VAd. If d <0, then yo=O,
If dO0, then Y1=Y2=Y3=O,
and we obtain an infinitudeof quaternion solutions of (9), correspondingto the
real solutionsof y 2+y +y2= -d. Henceforthwe take b and c to be not both real.
Applyingthe division algorithmof ?3 to (8), we obtain the followingvalues
forthe functionsf and g:
(10)
g=c-n.
f=b+t,
Hence equations (6) become
(11)
nt2+ nbb- cc + n(c + )
-n2
=
0,
and
(12)
t3+ tbb-2nt + t(c + c) + bc + cb = O,
since b+b vanishes. Following the theory of ?3, we see that if a real solution
t=to, n=no of these equations satisfiesf=0 and g=0, then we have b= -to and
c=no. But tois real, and b has zero trace, so that both are zero; also, c must be
real. Hence equation (8) reduces to one of type (9), contraryto hypothesis.
Consequentlythe solutionsof (11) and (12) do not satisfyf=0, and this, by ?3,
implies that each of these real solutions gives exactly one solution of (8); the
solution is given by the substitutionof the functions(10) in equation (4).
We introducethe notation
(13)
B = bb+ c + J,
C = cc,
D = bc + cb,
noting that B, C, and D are real. First we consider solutions of (11) and (12)
with t= 0, so that D = 0, by (12). Then the possible values of n are given by
(11), which reduces to
(14)
n2 - Bn + C = 0.
We want real roots; any real root will be positive because of the manner in
whichequations (11) and (12) were set up. Thus we obtain 0, 1, or 2 roots of (8)
according as B2-4C is negative, zero, or positive.
Finally, we search forsolutions of (11) and (12) with t 40. We solve (12) for
n; thus
(15)
n=
(t3
+ tB + D)/2t,
and we substitutethis value in (11) to obtain
(16)
t6 + 2Bt4+ (B2
-
4C)2-D2
=
0.
658
[December,
EQUATIONS IN QUATERNIONS
Each distinctreal root of this equation gives us, by use of (15), a root of (8). In
orderto findthe numberof real roots of (16), we prove the following:
LEMMA
1. If B is negative,so is B2-4C.
Proof.Since bbis not negative,c+c must be negative by the hypothesis.We
can write
B2
-
4C = bbB + bb(c+ c) + (c
If c has the formco+Cli+c2j+C3ij,
-)2.
thenthe last termon the rightside of this
Hence the threetermson the rightare real, and
equation equals -4(c2+c2+c2).
none of them is positive. They cannot all be zero, for that would imply that
b = 0 and c = c, contraryto hypothesis,and this proves the lemma.
We consider (16) as a cubic in t2,and look forpositive roots. If D-0, the
numberof positive roots is one by Descartes' rule of signs and Lemma 1. Thus
equation (16), consideredas a sextic, has two real roots when D#0.
If D =0, we divide the obvious zero roots out of (16), and have
(17)
t4
+ 2Bt2+ B 2- 4C = 0.
Consideringthis as a quadratic equation, we see that the discriminantis not
negative,so that the rootsare real. If B2- 4C is positive,B is positive by Lemma
1, and the quadratic (17) has negative roots. Hence the quartic (17) has no real
root. Similarly,if B2- 4C is zero, we findthat the quartic (17) has no real roots
other than zeros. In both these cases, all roots of (8) are obtained from (14).
Finally, ifB2- 4C is negative,the quartic (17) has exactly two real roots,giving
two solutions of (8). Note that in this case no solutions of (8) result from(14).
We summarize these resultsin the following:
THEOREM 2. Considerthequaternionequation (8), thetraceof b beingzero. If
b and c are real (so thatb = 0), theequationhas an infinitenumberof rootsorjust
theequationhas one or two
tworootsaccordingas c is positiveor negative.Otherwise,
rootsaccordingas thequantitiesdefinedin (13) satisfytherelationsD = B2 -4C=O
or not.
5. The number of roots of (1). We suppose firstthat no real solution of
equations (6) satisfiesf = g = 0, so that there is a one-to-onecorrespondencebetween the roots of (1) and the real solutionsof (6). We now need some information about the nature of the functionsf and g of equation (3).
LEMMA 2. Thefunctionsf and g of equation (3) are ofdegreem-1 in n and t;
moreover,
f has only one termof this degree,namelytm-i. Also, everytermof g is
divisibleby n, withtheexceptionof am.
PROOF. The proof is by induction on m. Equations (10) indicate the truth
of the lemma in case m = 2. We now obtain recurrencerelationsforf and g. The
polynomial of degree m+1 analogous to a (x) can be written in the form
1941]
659
IN QUATERNIONS
EQUATIONS
a(x) x+am?1, and correspondingto the algorithm(3) we have
a(x). x + am+1 = qx(x2
-
tx + n) + fx2 + gx + am+1
= (qx + f)(x2 -tx
+ n) + ftx-fn + gx +
am,+.
Calling the remainderin the last expressionabove Fx+G, we have the relations
F=ft+g and G = -fn+am,+. The induction is completed by noting that if the
functionsf and g have the propertiesstated in the lemma, so do F and G, m
being replaced by m+1.
It is a consequence of the above lemma that equations (6) are of degree
2m-1 in n and t. Now it is known [3 ] that the curves representedby (6) cannot
have more than (2m - 1)2 intersectionsprovided that the polynomials N and T
are relativelyprime; and this is the case when neitherof the two resultantsof
N and T vanishes identically.We now show that this is the case.
By Lemma 2, equations (6) can be writtenin the forms
N(t, n)
=
CO +
Clt +
T(t, n) = do+ dit +
+
C2m-3t2m-3
+
* + d2m-2t2m-2+
C2m-2t2m-2y
t2m-l,
the coefficients
cr and drbeing polynomialsin n. Then the resultantobtained by
eliminatingt is
Co
C1
0
Co
C1
0
do
O
0
d1
C2m-2 0
0
C2m-3 C2m-2 0
0
CO
do d1
O
C2m-2
Ci
d2m_21
0
1
d2m_2
do
d1
2m - 1 rows
0
0
2m-2
rows
1
d2m_2
This is a polynomialin n; in orderto show that it does not vanish identically,we
prove that it has a non-zeroconstant term. When we set n =0, Lemma 2 and
, 2m-2, whereas co assumes
equation (6) show that each cr= O forr = 1, 2,
the value - dmam;that is, all elements of the determinantabove the principal
diagonal vanish. Hence the value of the constant term of this resultant is
(dmam)
2m-1,
which is not zero.
Having shown that the polynomialsf and g have no common factorinvolving t, we now eliminate the possibilitythat they have a polynomial in n alone
as a commonfactor.We could show that this is not possible by provingthat the
resultantwhich eliminatesn does not vanish identically. But it is easier to proceed directly.If a polynomialin n divides T, it must divide the coefficient
of the
highestpower of t. But this coefficientis unity.
660
EQUATIONS IN QUATERNIONS
[December,
Having shownthat equation (1) cannot have morethan (2m -1) 2 roots when
no real solutionof (6) satisfiesf=g = 0, we turnto the case wherethese equations
are satisfiedby certain solutions of (6). Let there be s such real solutions. Then
we have s equations of type (7), each having eithertwo roots or an infinitenumber of roots (by Theorem 2). If the number of roots is finite,that is, if these
equations have two roots each, we divide them out of equation (1). Thus we
obtain an equation of degree m-2s, which has no factorsof the form(7), and
hencehas at most(2m- 4s - 1)2 quaternionroots.Adding2s to accountforthe
roots of the quadratic equations, we note that
2s + (2m - 4s -
1)2
<
(2m -
1)2,
when m > 2s and s >0. Hence we have shown that if the numberof roots of (1)
is finite,it cannot exceed (2m -1)2. Theorem 2, and in particular equation (9),
can be used now to give the followingresult:
THEOREM3. Equation (1) has an infinitenumberofquaternionrootsif and only
if a(x) is divisiblebyan expressionoftype(7), withthereal valuestoand nO satisfying theinequalityt' <4no. If thenumberis finite,it cannotexceed(2m- 1) 2.
6. The necessity of the conditionsof Theorem 1. Denote by u the order of
D over its centrumC. Thus thereexist u elementsin D which are linearlyindependent over C, but any u+1 elements in D are linearly dependent over C.
Given any element x in D, there exist thereforeelements ci in C such that
u- cix; = 0. Since D is a divisionalgebra, it followsnow that the sub-field
xu +
C(x) ofD whichis generatedby adjoining the elementx ofD to C, is a commutative field,finiteover C, and the irreducibleequation in C which is satisfiedby x
has a degreenot exceedingu. Since everyequation in C has a solutionin D, this
implies that the degrees of irreducibleequations in C do not exceed u.
LEMMA3. Let A denotetheessentiallyuniquelydetermined
algebraicallyclosed
commutative
fieldwhichcontainsC and is algebraicoverC. Then A is finiteoverC.
Proof. Suppose firstthat C is of characteristicp 0. We prove that there is
no element t in C such that the equation zP-t=O has no solution in C. For, if
therewere such an element,then none of the equations zP - t=0 would have a
solution in C. Since this last equation has the form(z-ti)P =O in the fieldA, it
has one and only one solution in A; since the pi-lth power of this solution is a
solution of zP- t =0, it followsthat each of these equations is irreduciblein C.
But this is impossible,since some pi is largerthan u. Hence every elementin C
is the pth power of an element in C. Consequently [4], A is separable over C;
and this resultis also true when the characteristicof C is zero.
If B is some fieldbetweenA and C, and ifB is finiteover C, then B is a simple extension of C since it is separable over C. Thus the degree of B over C is
equal to the degree of the irreducibleequation in C whose solution generates
B over C. Since the degreesof irreducibleequations in C do not exceed u, it follows that the degrees of finiteextensionsof C do not exceed u. Consequently,
1941]
EQUATIONS IN QUATERNIONS
661
thereexists a fieldM betweenA and C which is finiteof maximal degree over C.
If w is any elementof A, then M(w) is finiteover C. Since the degree of M over
C is as large as possible, it followsthat M and M(w) have the same degree over
C. Hence w is in M, and M=A, and the lemma is proved.
Now it followsfroma theoremby Artin-Schreier[5] that either C is a realclosed field,or A equals C. The latter is impossible since C, the centrumof a
non-commutativedivision algebra, cannot be algebraically closed. Hence C is a
real-closed field; but the only non-commutativedivision algebra over C is the
algebra of real quaternions [6], and this completes the proof.
References
1. Oystein Ore, Theory of non-commutativepolynomials,Annals of Math. (II), vol. 34, 1933,
pp. 480-508.
2. Outlined by Cayley in P. G. Tait's Quaternions, third edition, Cambridge Press, 1890, pp.
157-159.
3. Cf., B6cher's Introductionto Higher Algebra, Macmillan, 1927, problem 4 on p. 239, and
the theoremon p. 202. The theoremon p. 202 is not preciselywhat we want here, since we need a
propositionabout polynomialsin two variables. Two polynomialsf(x, y), g(x, y) have two resultants, R. obtained by eliminatingx, and R, by eliminatingy. B6cher's proofshows that the vanishing
of R. identicallyis a necessaryand sufficientcondition forthe two polynomials to have a common
factor involving x. But it is possible forf(x, y) and g(x, y) to have a common factor in y alone,
withR. not identicallyzero; in such a case R vanishes (e.g.,considerthe polynomialsxy-4x-3y+ 12,
xy+y2-4x-4y).
4. See E. Steinitz, Algebraische Theorie der K6rper, 1930, p. 50.
5. Artin-Schreier,Hamburger Abhandlung, vol. 5, 1927, p. 230, Theorem 4.
6. See A. A. Albert,Structureof Algebras,A.M.S. Coll. Publ. 24, Chap. 9, Theorem 28, p. 146.