On Left Eigenvalues of a Quaternionic Matrix

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Linear Algebra and its Applications 323 (2001) 105–116 www.elsevier.com/locate/laa

On left eigenvalues of a quaternionic matrix Liping Huang a ,∗,1,WasinSob,2 aInstitute of Mathematics and Software, Xiangtan Polytechnic University, Xiangtan, Hunan 411201, People’s Republic of China bDepartment of Mathematics and Computer Science, San Jose State University, San Jose, CA 95192-0103, USA Received 18 April 2000; accepted 15 August 2000 Submitted by C.-K. Li

Abstract In this paper, we introduce the concept of left and right eigenvalues for a quaternionic matrix, and investigate their properties, quantities and relationship. © 2001 Elsevier Science Inc. All rights reserved.

1. Introduction

Let R be the set of real numbers and H be the set of quaternions of the form a1 + a2i + a3j + a4k,whereai ∈ R and i2 = j2 = k2 = ijk =−1. For a = a1 + a2i + a3j + a4k ∈ H,leta = a1 − a2i − a3j − a4k be the conjugate of a,and √ q |a|= aa = a2 + a2 + a2 + a2 1 2 3 4 be the modulus of a.LetRe(a) = (a + a)/2bethereal part of a,andIm(a) = (a − a)/2betheimaginary part of a. Two quaternions a and b are similar if there exists a nonzero q ∈ H such that a = q−1bq. The similarity class of a is the set

∗ Corresponding author. E-mail addresses: [email protected] (L. Huang), [email protected] (W. So). 1 This research is partially supported by the National Natural Science Foundation of China and the NSF of Hunan Province. 2 This research was initiated during a visit to the Institute of Mathematics and Software in Xiangtan Polytechnic University.

{q−1aq: a/= 0}. Note that a is similar to b if and only if Re a = Re b and |a|=|b|. m×n Denote C the set of complex numbers of the form a1 + a2i,whereai ∈ R.LetX n be the set of all m × n matrices over a division ring X,andX be the set of vectors of n components over a division ring X,whereX = R, C or H. In this paper we are interested in the eigenvalues of a quaternionic matrix. Due to the noncommutativity of quaternions, there are two types of eigenvalues.

Deﬁnition 1.1. Given A ∈ Hn×n, λ ∈ H is called a right eigenvalue of A if AX = Xλ for some nonzero X ∈ Hn. The set of distinct right eigenvalues is called the right spectrum of A, denoted σr (A).

Deﬁnition 1.2. Given A ∈ Hn×n, λ ∈ H is called a left eigenvalue of A if AX = λX for some nonzero X ∈ Hn. The set of distinct left eigenvalues is called the left spectrum of A, denoted σl(A).

It is known that right spectrum is always nonempty, and indeed right eigenvalues are well studied in literature [1,5]. On the contrary, left eigenvalues are less known. Zhang [9] commented that left spectrum is not easy to handle, and few results have been obtained. In his thesis, Siu [6] spent a whole chapter on discussing left and right eigenvalues of quaternionic matrices. This paper is a follow up of Siu’s effort and another response to Zhang’s call on the investigation of left eigenvalues of a quaternionic matrix. Note that A − λI is a singular matrix if and only if AX = λX for some nonzero X ∈ Hn. Hence left eigenvalues are also known as singular eigenvalues in literature. Cohn [2] raised the question whether left eigenvalue always exists. In a footnote of his paper Lee [5] doubted the existence of left eigenvalues in general, but he provided no example. Wood [8] used a topological method to confirm that the left eigenvalue always exists and at once gave a negative answer to Lee’s speculation. However Wood’s proof is existential and provided no algorithm for computing left eigenvalues. At the end of his paper, Wood suggested an algebraic approach to find left eigenvalues. He demonstrated that the left eigenvalues of a 2 × 2 matrix can be found by solving a quaternionic quadratic equation, but he did not actually solve the quadratic equation. Later So [7] reduced the existence of a left eigenvalue of a 3 × 3 quaternionic matrix to the existence of a solution of a generalized quaternionic polynomial of degree 3, which is guaranteed by the Fundamental Theorem of Algebra for quaternions [3]. It is still an open problem whether this algebraic approach works for general n × n matrices for n > 4. The rest of the paper is organized as follows. Section 2 concerns the computation of left eigenvalues. Using results from Huang and So [4] on quaternionic quadratic formulas, we explain how to compute all left eigenvalues of a 2 × 2 matrix. In Section 3, we study the possible number of distinct left eigenvalues of a quaternion matrix. Section 4 is devoted to the conjecture that finiteness of both left and right spectra implies their equality. We also study those matrices with identical left and right spectra. L. Huang, W. So / Linear Algebra and its Applications 323 (2001) 105–116 107

2. Computation of left spectrum

We begin this section with some useful observation about a general left spectrum.

Lemma 2.1. For p and q ∈ H,σl(pI + qA) ={p + qt: t ∈ σl(A)}, where I is the identity matrix.

Right spectrum does not enjoy Lemma 2.1. On the other hand, right spectrum ∗ is unitarily invariant, i.e. σr (U AU) = σr (A),whereU is a unitary matrix, but left spectrum is not unitarily invariant.

Lemma 2.2. If A is a (upper or lower) triangular matrix, then σl(A) is the set of distinct diagonal elements of A.

The right spectrum of a triangular matrix is the union of distinct similarity classes of its diagonal elements. Next we focus on 2 × 2matrix.

Theorem 2.3. Let ab A = . cd

(a) If bc = 0, then σl(A) ={a,d}. − − (b) If bc= / 0, then σl(A) ={a + bλ: λ2 + b 1(a − d)λ − b 1c = 0}.

Proof. (a) A is a triangular matrix, and the result follows from Lemma 2.2. (b) Using Lemma 2.1, we have σ (A) = σ aI + b 01 l l b−1cb−1(d − a) = a + bσ 01. l b−1cb−1(d − a)

Note that λ ∈ σ 01 l b−1cb−1(d − a)

x if and only if there exists nonzero y such that x x 01 = λ b−1cb−1(d − a) y y x if and only if there exists nonzero y such that 108 L. Huang, W. So / Linear Algebra and its Applications 323 (2001) 105–116 y = λx, b−1cx + b−1(d − a)y = λy if and only if λ2 + b−1(a − d)λ − b−1c = 0.

Hence the computation of the left spectrum of a 2 × 2 matrix is reduced to solving a quaternionic quadratic equation, whose solutions can be obtained by the quaternionic quadratic formulas in [4]. For the sake of completeness, we include this result as follows.

Theorem 2.4. The solutions of the quadratic equation x2 + bx + c = 0 can be obtained by formulas according to the following cases: 1. If b, c, ∈ R and b2 < 4c, then x = 1 (−b + β + γ + δ ) 2 i j k for all β,γ,δ ∈ R with β2 + γ 2 + δ2 = 4c − b2. 2. If b, c, ∈ R and b2 > 4c, then √ −b b2 − c x = 4 . 2 3. If b ∈ R and c 6∈ R, then −b ρ c c c x = ∓ 1 i ∓ 2 j ∓ 3 k, 2 2 ρ ρ ρ where c = c0 + c1i + c2j + c3k,ci ∈ R, and v u q u 2 2 2 2 2 2 tb − 4c0 + (b − 4c0) + 16(c + c + c ) ρ = 1 2 3 . 2

4. If b 6∈ R, then − b x = Re − (b0 + T)−1(c0 − N), 2 where 0 0 Re b Re b b = b − Re b = Im b, c = c − b − , 2 2 and (T , N) is chosen√ as (a) T = 0,N= (B B2 − 4E)/2 provided that D = 0,B2 > 4E. p √ √ (b) T = 2 E − B, N = E provided that D = 0,B2 < 4E. √ (c) T = z, N = (T 3 + BT + D)/2T provided that D/= 0 where z is the unique positive root of the real polynomial z3 + 2Bz2 + (B2 − 4E)z − D2, where B =|b0|2 + 2Rec0,E=|c0|2, and D = 2Reb¯0c0. L. Huang, W. So / Linear Algebra and its Applications 323 (2001) 105–116 109

Example 2.5. Let " # 01+ i A = . 1 − i 0 Then n o 2 σl(A)= (1 + i)λ: λ + i = 0 ( √ ) 2 = (1 + i)λ: λ = (1 − i) 2 n√ √ o = 2, − 2 .

Example 2.6. Let A = 0 i . j 1 Then n o 2 σl(A)= iλ: λ + iλ + k = 0 1 1 = iλ: λ = (1 − i − j − k), (−1 − i − j + k) 2 2 1 1 = (1 + i + j − k), (1 − i − j − k) . 2 2

Example 2.7. Let A = 2 i . −i 2 Then n o 2 σl(A)= 2 + iλ: λ + 1 = 0 1 = 2 − iλ: λ = (βi + γ j + δk), β2 + γ 2 + δ2 = 4 n 2 o = 2 − β − δj + γ k: β2 + γ 2 + δ2 = 1 .

3. Possible number of distinct left eigenvalues

We begin with a few comments about the right eigenvalues of an n × n matrix A. − If λ ∈ σr (A),thenq 1λq ∈ σr (A) for q/= 0. Hence it can be shown that [9] σr (A) is inﬁnite if and only if σr (A) 6⊂ R. Moreover, σr (A) has at most n distinct real 110 L. Huang, W. So / Linear Algebra and its Applications 323 (2001) 105–116 parts and moduli. However if σr (A) is ﬁnite, then it is real and has at most n distinct elelments. In this section we investigate similar questions for left spectrum of a 2 × 2 matrix. Using Theorems 2.3 and 2.4, it is not hard to see the following.

× Theorem 3.1. Let A ∈ H2 2.Ifσl(A) is ﬁnite, then it has at most two distinct elements.

Theorem 3.2. If ab A = ∈ H2×2, cd − − then σl(A) is inﬁnite if and only if bc= / 0,b1(a − d),b 1c ∈ R, and D = [b−1(a − d)]2 + 4b−1c<0. , σ (A) , Moreover if l is inﬁnite then a + d b 2 σl(A) = + x: x =−x,|x| =|D| . 2 2

Lemma 3.3. Let α, β ∈ H, D ∈ R and C ={α + βx: x =−x,|x|2 =|D|}.Then the set C has inﬁnitely many or unique real part. Moreover, the set C has unique real part if and only if β ∈ R.

Proof. Let

β = b1 + b2i + b3j + b4k,x = x2i + x3j + x4k, where bi,xi ∈ R.Since

Re(C) = Re(α) − (b2x2 + b3x3 + b4x4), x ,x,x ∈ R x2 + x2 + x2 =|D| f(x ,x,x) = b x + b x + for 2 3 4 with 2 3 4 .Since 2 3 4 2 2 3 3 3 b4x4 is a continuous function on a sphere of R , thus the range of f(x2,x3,x4) is an interval or a singleton. Thus, the set C has inﬁnitely many or unique real part. Moreover, the set C has unique real part if and only if b2 = b3 = b4 = 0, i.e. β ∈ R.

Lemma 3.4. Let α, 0 =/ β ∈ H, D ∈ R and C ={α + βx: x =−x,|x|2 =|D|}. Then the set C has inﬁnitely many or unique modulus. Moreover, the set C has unique modulus if and only if β−1α ∈ R.

Proof. Let −1 β α = a1 + a2i + a3j + a4k,x = x2i + x3j + x4k, where ai,xi ∈ R.Then |α + βx|2 =|β|2[a2 + (a + x )2 + (a + x )2 + (a + x )2] 1 2 2 3 3 4 4 2 2 =|β| [|α| +|D|+2(a2x2 + a3x3 + a4x4)]. L. Huang, W. So / Linear Algebra and its Applications 323 (2001) 105–116 111 x ,x,x ∈ R x2 + x2 + x2 =|D| f(x ,x,x) = a x + a x + for 2 3 4 with 2 3 4 .Since 2 3 4 2 2 3 3 3 a4x4 is a continuous function on a sphere of R , thus the range of f(x2,x3,x4) is an interval or a singleton. Thus, the set C has inﬁnitely many or unique modulus. Moreover, the set C has unique modulus if and only if a2 = a3 = a4 = 0, i.e. β−1α ∈ R.

Theorem 3.5. Let ab A = ∈ H2×2. cd

If σl(A) is infinite, then σl(A) has infinitely many or unique real part. Moreover, σl(A) is infinite with unique real part if and only if b, c, (d − a) ∈ R such that (d − a)2 + 4bc < 0.

− − Proof. If σl(A) is infinite, by Theorem 3.2, we have bc= / 0, b 1(a − d),b 1c ∈ R, D =[b−1(a − d)]2 + 4b−1c<0, and a + d b 2 σl(A) = + x: x =−x,|x| =|D| . 2 2 Thus, by Lemma 3.3, it is clear that σl(A) has infinitely many or unique real part. Moreover, if σl(A) is infinite, then σl(A) has unique real part if and only if b ∈ R. Thus, σl (A) is infinite with unique real part if and only if b, c, (d − a) ∈ R such that (d − a)2 + 4bc < 0.

Similarly, by Theorem 3.2 and Lemma 3.4, we have:

Theorem 3.6. Let ab A = ∈ H2×2. cd

If σl(A) is infinite, then σl(A) has infinitely many or unique modulus. Moreover, − − − σl(A) is infinite with unique modulus if and only if b 1d,b 1a,b 1c ∈ R such that [b−1d − b−1a]2 + 4b−1c<0.

Corollary 3.7. Let ab A = ∈ H2×2. cd

Then σl(A) is inﬁnite but with unique modulus and real part if and only if a,b,c,d ∈ R such that (d − a)2 + 4bc < 0.

We now focus on n × n complex matrices. Recall the definition of a complex eigenvalue of a complex matrix. 112 L. Huang, W. So / Linear Algebra and its Applications 323 (2001) 105–116

Deﬁnition 3.8. Given A ∈ Cn×n, λ ∈ C is called a complex eigenvalue of A if AX = λX = Xλ for some nonzero X ∈ Cn. The set of distinct complex eigenvalues is called the spectrum of A, denoted σ(A).

n×n For A ∈ C , notice that σ(A) ⊂ σl(A) ∩ σr (A) and σ(A) has at most n elements.

−1 Lemma 3.9. If λ0 ∈ H but λ0 6∈ C, then the set {α λ0α:0=/ α ∈ C} is inﬁnite.

Proof. Let λ0 = a + bj,wherea,b ∈ C. Clearly, we have 2 − α {α 1λ α:0=/ α ∈ C}= a + b j:0=/ α ∈ C . 0 |α|2

Since b/= 0andtheset α2 j:0=/ α ∈ C |α|2

−1 is inﬁnite, thus the set {α λ0α:0=/ α ∈ C} is inﬁnite.

n×n Theorem 3.10. If A ∈ C , then σl(A) is ﬁnite if and only if σl (A) = σ(A).

n Proof. If λ0 ∈ σl(A), then there exists nonzero X ∈ H such that AX = λ0X. Clear- −1 −1 −1 −1 ly, for any nonzero α ∈ C,wehaveAα X = α AX = α λ0α(α X). Thus, −1 λ0 ∈ σl(A) if and only if the set {α λ0α:0=/ α ∈ C}⊆σl (A). If σl(A) is ﬁnite, then by Lemma 3.9 and above, we have σl(A) ⊂ C.Ifλ0 ∈ n σl(A), then there exists nonzero X = X1 + X2j ∈ H such that AX = λ0X,where n X1,X2 ∈ C . Clearly, we have AX1 = λ0X1 and AX2 = λ0X2.SinceX1 or X2 is nonzero, we have λ0 ∈ σ(A). Thus, σl (A) ⊆ σ(A).Clearly,wehaveσ(A) ⊆ σl(A), thus σl(A) = σ(A). Conversely, if σl(A) = σ(A), then it is clear that σl(A) is ﬁnite.

n×n Corollary 3.11. Let A ∈ C .Ifσl(A) is ﬁnite, then it must have at most n distinct elements.

It remains unknown whether the ﬁnite left spectrum of an n × n matrix has at most n elements.

4. Finiteness and equality of left and right spectra

Siu [6] suggested the following. L. Huang, W. So / Linear Algebra and its Applications 323 (2001) 105–116 113 n×n Problem 4.1. For A ∈ H with both σl(A) and σr (A) ﬁnite, is it true that σl(A) = σr (A)? We ﬁrst consider this problem for the special class of n × n complex matrices.

Theorem 4.2. Let A be an n × n complex matrix. If both σl(A) and σr (A) are ﬁnite, then σl (A) = σr (A).

Proof. Since σl(A) is ﬁnite, then by Theorem 3.10, σl(A) = σ(A). On the other hand, it follows from the ﬁniteness of σr (A) that σr (A) ⊂ R, thus we have σr (A) = σ(A) ⊂ R,andsoσl(A) = σr (A).

Hence the answer for Problem 4.1 is afﬁrmative for n × n complex matrices. However the problem is still open for general n × n matrices. In the following, we consider the class of 2 × 2 matrices. First we need a lemma.

Lemma 4.3. The quaternionic quadratic equation x2 + xab − bax = 0 either has a unique solution x = 0 or has inﬁnitely many solutions.

Proof. Since Re(ab) = Re(ba) and xRe(ab) − Re(ba)x = 0, thus without loss of generality, we assume that Re(ab) = Re(ba) = 0. Let y = x + ab.Theny satis- ﬁes y2 − (ab + ba)y + baab = 0, which has the same number of solutions as the equation x2 + xab − bax = 0.

Case 1. ab = 0. Then |ba|=|ab|=0, i.e. ba = 0. Hence the equation y2 − (ab + ba)y + baab = 0 becomes y2 = 0. It follows that y = 0 is the only solution.

Case 2. ab= / 0andab + ba = 0. Hence the equation y2 − (ab + ba)y + baab = 0 becomes y2 − (ab)2 = 0. Since Re(ab) = 0andab= / 0, one computes (ab)2 =−|ab|2 < 0. By Theorem 2.4(1), it follows that there are inﬁnitely many solutions.

Case 3. ab= / 0andab + ba= / 0. Then ab + ba 6∈ R,sinceRe(ab) = 0. Note that Re ab =Reba = 0and|ab|= |ba|, hence ab and ba are similar and so there exists nonzero q such that ba = q−1abq. Applying Theorem 2.4 to the equation y2 − (ab + q−1abq)y + q−1abqab = 0, we compute

− D =2Re(ab + q−1abq)q 1abqab − − =2Re(−ab − q 1abq)q 1abqab 114 L. Huang, W. So / Linear Algebra and its Applications 323 (2001) 105–116

− − =−2Re(abq 1abqab + q 1ababqab) − − =−2Re(ababq 1abq + ababqabq 1) − − =−2Re(−|ab|2q 1abq −|ab|2qabq 1) − − =2|ab|2 Re (q 1abq + qabq 1) − − =2|ab|2( Re q 1abq + Re qabq 1) =2|ab|2( Re ab + Re ab) =0, − − B =|ab + q 1abq|2 + 2Req 1abqab − − =|ab|2 +|q 1abq|2 + 2Reabq−1abq + 2Req 1abqab − − =2|ab|2 − 2Reabq 1abq + 2Req 1abqab =2|ab|2, E =|q−1abqab|2 =|ab|2.

Hence B2 = 4E and so there is only one solution by Theorem 2.4(4a).

Lemma 4.4. Let ab A = ∈ H2×2. cd

If c/= 0, then λ ∈ σl(A) if and only if − (λ − a)c 1(λ − d) − b = 0. (1)

Proof. λ ∈ σl(A) if and only if λI − A is a singular matrix. Since 1 −(λ − a)c−1 0 (λ − a)c−1(λ − d) − b λI − A = , (2) 01−cλ− d − if c/= 0, then it is clear that λ ∈ σl(A) if and only if (λ − a)c 1(λ − d) − b = 0.

× Theorem 4.5. Let A ∈ H2 2. If both σl (A) and σr (A) are ﬁnite, then σl(A) = σr (A).

Proof. Since σr (A) is ﬁnite, thus σr (A) ⊂ R and so σr (A) ⊂ σl(A).Sinceσl (A) is ﬁnite, by Theorem 3.1, σl(A) has at most two distinct elements. When |σr (A)|=2, it is clear that σl(A) = σr (A). Suppose that |σr (A)|=1. Let σr (A) ={λ1}⊂R.Then 2 {0}=σr (A − λ1I) and hence (A − λ1I) = 0. Let L. Huang, W. So / Linear Algebra and its Applications 323 (2001) 105–116 115 ab (A − λ I) = . 1 cd Then ab2 a2 + bc ab + bd = = . cd ca + dc cb + d2 0 Thus we have a2 + bc = 0, d2 + cb = 0, ab + bd = 0, ca + dc = 0. If c = 0, it is easy to see that a = d = 0andso b A − λ I = 0 . 1 00

Thus we have σl(A) ={λ1}=σr (A). If c/= 0, then b =−a2c−1, d =−cac−1.

−1 −1 2 −1 By Lemma 4.4, λ ∈ σl(A − λ1I) if and only if (λ − a)c (λ + cac ) + a c = 0, or, equivalently, if and only if − − − − − (c 1λ)2 + (c 1λ)ac 1 − c 1a(c 1λ) = 0. − Since σl (A) is ﬁnite, by Lemma 4.3, the equation only has a unique solution c 1λ = 0, thus |σl(A − λ1I)|=|σl(A)|=1, and clearly σl(A) = σr (A).

It is known that if A is a real matrix, then σl (A) = σr (A), see [9]. Moreover it is not hard to see that σl(A) ∩ R = σr (A) ∩ R.

× Theorem 4.6. For A ∈ H2 2, if σl(A) = σr (A) and σl(A) is inﬁnite, then A is a real matrix.

Proof. Let ab A = . cd

Since σl(A) = σr (A), σl(A) has ﬁnitely many real parts and moduli, by Theorems 3.5 and 3.6, the real part and modulus of σl(A) must be unique. Hence A must be a real matrix by Corollary 3.7.

× Theorem 4.7. If A ∈ H2 2, then σl(A) = σr (A) if and only if A is a real matrix or both σl(A) and σr (A) are ﬁnite. 116 L. Huang, W. So / Linear Algebra and its Applications 323 (2001) 105–116

Proof. Let σl (A) = σr (A).Ifσl(A) is infinite, by Theorem 4.6, A is a real matrix. If σl(A) is finite, then both σl(A) and σr (A) are finite. Conversely, if both σl(A) and σr (A) are finite, then by Theorem 4.5, it is clear that σl(A) = σr (A).IfA is a real matrix, then by Theorem 5.2 of [9], we have σl(A) = σr (A).

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