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2. A condition of equivalence of the generalized characteristic polynomial is obtained, i.e. four real high order polynomial equations of four variables, the problem of computing the left eigenvalues, then, is to solve special polynomial equations.

3. All the left eigenvalues of quaternion matrix are located in the in particular annulus connected with the right eigenvalues.

The outline of the paper is as follows. In Section 2, we describe the proposed method based on matrix representation. Section 3 gives some essential properties of left eigenvalue problems. Section 4 offers several illustrative examples. The last section 5 concludes.

2 Calculation of left eigenvalues

A total of 48 real matrix forms represent a quaternion [11]. Let E be the 4 4 identity matrix and for × k = 1, 2, , 48, Hk, Jk, Kk be the 4 4 matrices with real entries. A quaternion q can be written as · · · × various forms of real matrix

[q] E + [q] Hk + [q] Jk + [q] Kk , k = 1, 2, , 48. (1) 0 1 2 3 · · · Quaternion addition and multiplication correspond to matrix addition and multiplication provided that the matrices Hk, Jk, Kk satisfying the ”Hamiltonian conditions” (HkHk = JkJk = KkKk = HkJkKk = E). − First of all, we define two useful quaternion operators. For any quaternion q H, we define the 4×4∈ operator k to maps a quaternion to the 4 4 real matrix. That is, k : H R , one defines Q × Q → k(q) = [q] E + [q] Hk + [q] Jk + [q] Kk. k = 1, 2, , 48. (2) Q 0 1 2 3 · · · In this representation (2), the conjugate of quaternion corresponds to the transpose of the matrix k(q). The fourth power of the norm of a quaternion is the of the corresponding matrix Q k(q) (denote det k(q) ). Q Q m×n For any quaternion matrix A = (aij) H , we define the operator k to maps a m n quaternion ∈ m×n 4m×4n P × matrix to the 4m 4n real matrix. That is, k : H R , one defines × P → k(A) = ( k(aij)) k = 1, 2, , 48. (3) P Q · · · Since quaternion addition and multiplication correspond to matrix addition and multiplication pro- vided that the matrices E, Hk, Jk, Kk. Some properties of operators k and k are given as below. Q P 2 Proposition 2.1 k and k have the following properties Q P

1. q + q = q if and only if k(q )+ k(q )= k(q ), q ,q ,q H. 1 2 3 Q 1 Q 2 Q 3 ∀ 1 2 3 ∈

2. If q q = q if and only if k(q ) k(q )= k(q ), q ,q ,q H. 1 2 3 Q 1 Q 2 Q 3 ∀ 1 2 3 ∈

3. qA = B if and only if k(qI) k(A) = k(B), q H, I is m m identity matrix, A, P P P ∀ ∈ × ∀ B Hm×n. ∈

4. Aq = B if and only if k(A) k(qI) = k(B), q H, I is n n identity matrix, A, P P P ∀ ∈ × ∀ B Hm×n. ∈ m×r r×n m×n 5. If AB = C if and only if k(A) k(B)= k(C), A H , B H , C H . P P P ∀ ∈ ∀ ∈ ∀ ∈

m×n Theorem 2.1 For any matrix A = (aij) H and m n 1, there exist nonnegative integer s, ∈ ≥ ≥ with 0 s n, such that Rank( k(A)) = 4s. ≤ ≤ P

Proof. We show that the statement Rank( k(A)) = 4s by mathematical induction. P When m = 1, the statement Rank( k(A)) = 4s is straightforward for s = 0 and 1, since 4 P det Pk(A) = det Qk(a )= a 0. 11 k 11k ≥ Suppose that when m = t, statement holds, now we show that when m = t + 1, the statement still holds. This can be done as follows. By Eq.(3),

k(a ) k(a ) k(a n) Q 11 Q 12 ··· Q 1  k(a21) k(a22) k(a2n)  k(A)= Q . Q . ··· Q . . (4) P  . . .   · · ·   k(a ) k(a ) k(a )   Q (t+1)1 Q (t+1)2 ··· Q (t+1)n 

If k(A) is a zero matrix, then the result is obvious. If it is nontrivial, then there exist at least one P entry ai′j′ such that k(ai′j′ ) = 0. Without loss of generality (WLOG), let k(a ) = 0. Then there Q 6 Q 11 6 exist several block elementary transformations in row and column of the matrix such that

k(a11) 0 0 Q ⋆ · · · ⋆  0 k(a22) k(a2n)  k(A) . Q . ··· Q . (5) P →  . . .   . . .   ⋆ · · · ⋆   0 k(a(t+1)2) k(a(t+1)n)   Q ··· Q  and ⋆ ⋆ k(a22) k(a2n) Q . ··· Q . B =  . .  (6) ⋆ · · · ⋆  k(a ) k(a )   Q (t+1)2 ··· Q (t+1)n  where 0 = k(0) is the zero matrix. Using the induction hypothesis that for m = t holds, then there Q exist nonnegative s⋆ (0 s⋆ n 1) such that Rank(B) = 4s⋆. According to (5), we obtain ⋆ ≤ ≤ − Rank( k(A)) = 4(s +1) since k(a ) = 0. Thereby showing that indeed for m = t+1 holds since P Q 11 6 the block elementary transformations do not change the rank of matrix. And the proof is complete.

3 Theorem 2.2 For any matrix k(A) = [a1, a2, , a4n] and k(B) = [b1, b2, P m×n · · · P m×1 b3, b4] corresponding to A = (aij) H and B = (bij) H respectively, where as and bt are ∈ ∈ ′ the column vectors (1 s 4n and 1 t 4). If there was a column vector bt′ (1 t 4) can be ≤ ≤ ≤ ≤ ≤ ≤ lineally expressed by a , a , , a n, then the other three column vectors can also lineally expressed by 1 2 · · · 4 a , a , , a n. 1 2 · · · 4

T 4n×1 Proof. WLOG. let bt′ = b1, there exist x = [x1,x2, ,x4n] R such that b1 = k(A)x = 4n · · · ∈ P n×1 aixi. Obviously, we can find k(c) = [c , c , c , c ] corresponding to c = (cij) H such that P 1 2 3 4 ∈ iP=1 c = x. And we can get b = k(A)c , b = k(A)c and b = k(A)c . 1 2 P 2 3 P 3 4 P 4 Now we can obtain the following result according to the above Theorem 2.1 and Theorem 2.2.

m×m Theorem 2.3 For any nonzero matrix A = (aij) H , if Rank( k(A)) = 4n, (n = 1, 2,...,m), n×n ∈ P then there exist n order sub-matrix B H such that Rank( k(B)) = 4n. ∈ P WLOG, we take

0 1 0 0 0 0 1 0 0 0 0 1 − − −  1 0 0 0   00 01   0 0 1 0  H1 = J1 = K1 = −  0 0 0 1   10 00   0 1 0 0   −     −   0 0 1 0   0 1 0 0   10 0 0     −   −  then we obtain (a ) (a ) (a m) Q1 11 Q1 12 ··· Q1 1  1(a21) 1(a22) 1(a2m)  1(A)= Q . Q . ··· Q . (7) P  . . .   · · ·   (am ) (am ) (amm)   Q1 1 Q1 2 ··· Q1  m×m Theorem 2.4 For any matrix A = (aij) H , m > 1, then the following statements are equiva- ∈ lent:

(1). det (A) = 0. P1 (2). The of all 4m 3 order sub-matrix of (A) are zero. − P1 (3). Let = B : B is m 1 order submatrix of A and max Rank( (B)) : B = 4n S { − } { P1 ∈ S} (0 n m 1). There exist the determinants of sixteen 4m 3 order sub-matrix (denoted as Ct, ≤ ≤ − ′ − ′ t = 1, 2, , 16) of (A) are zero, where (B ) is a sub-matrix of Ct and Rank( (B )) = · · · P1 P1 P1 4n (B′ ). ∈ S Proof. 1 2: this can be obtained according to the above Theorem 2.1. ⇔ 1 3: this can be seen from 1 2. ⇒ ⇔ 3 1: if n

1(a22) 1(a2m) ′ Q . ··· Q . B =  . .  · · ·  (am ) (amm)   Q1 2 ··· Q1 

4 and C = ( (A))(1, 5 : 4m; 1, 5 : 4m) C = ( (A))(2, 5 : 4m; 2, 5 : 4m) 1 P1 2 P1 C = ( (A))(3, 5 : 4m; 3, 5 : 4m) C = ( (A))(4, 5 : 4m; 4, 5 : 4m) 3 P1 4 P1 C = ( (A))(1, 5 : 4m; 2, 5 : 4m) C = ( (A))(2, 5 : 4m; 1, 5 : 4m) 5 P1 6 P1 C = ( (A))(3, 5 : 4m; 4, 5 : 4m) C = ( (A))(4, 5 : 4m; 3, 5 : 4m) 7 P1 8 P1 (8) C = ( (A))(1, 5 : 4m; 3, 5 : 4m) C = ( (A))(2, 5 : 4m; 4, 5 : 4m) 9 P1 10 P1 C = ( (A))(3, 5 : 4m; 1, 5 : 4m) C = ( (A))(4, 5 : 4m; 2, 5 : 4m) 11 P1 12 P1 C = ( (A))(1, 5 : 4m; 4, 5 : 4m) C = ( (A))(2, 5 : 4m; 3, 5 : 4m) 13 P1 14 P1 C = ( (A))(3, 5 : 4m; 2, 5 : 4m) C = ( (A))(4, 5 : 4m; 2, 5 : 4m) 15 P1 16 P1 where ( (A))(r, 5 : 4m; c, 5 : 4m) is obtained by taking row r and row from 5 to 4n and column P1 c and column from 5 to 4n of (A). P1 If det Ct = 0, then C4(:, 1) can be lineally expressed by the columns of C4(:, 2 : 4n + 1) since ′ 4n×1 Rank( k(B )) = 4n, i.e. there exist column vector x H such that C (:, 1) = C (:, 2 : P 1 ∈ 4 4 4n + 1)x1. Similarly, there exist column vector x2 x3 x4 such that C7(:, 1) = C7(:, 2 : 4n + 1)x2, ′ C (:, 1) = C (:, 2 : 4n+1)x and C (:, 1) = C (:, 2 : 4n+1)x . And because Rank( k(B )) = 10 10 3 13 13 4 Q 4n, we get x = x = x = x . Therefore, ( (A))(:, 4) can be lineally expressed by the columns of 1 2 3 4 P1 ( (A))(:, 5 : 4m), i.e. det k(A) = 0. P1 P Theorem 2.5 For any quaternion q, then

(1). P(3, 4)P(1, 2)P(4( 1))P(1( 1)) (q)P(1( 1))P(4( 1))P(1, 2)P(3, 4) = (q) − − P1 − − P1 (2). P(2, 4)P(1, 3)P(4( 1))P(3( 1)) (q)P(3( 1))P(4( 1))P(1, 3)P(2, 4) = (q) − − P1 − − P1 (3). P(2, 3)P(1, 4)P(4( 1))P(2( 1)) (q)P(2( 1))P(4( 1))P(1, 4)P(2, 3) = (q) − − P1 − − P1 where the elementary matrix P(i, j) is obtained by swapping row i and row j of the 4 4 identity matrix × and P(i(c)) is a 4 4 diagonal matrix, with diagonal entries 1 everywhere except in the ith position, × where it is c. Using the Theorem 2.5, it is not hard to see the following

m×m Theorem 2.6 For any matrix A = (aij) H , m> 1, then ∈ det C = det C = det C = det C , det C = det C = det C = det C , 1 2 3 4 − 5 6 − 7 8 det C = det C = det C = det C , det C = det C = det C = det C . − 9 10 11 − 12 − 13 − 14 15 16 where Ct is defined in a similar way described in the Theorem 2.4. Using Theorems 2.4 and 2.6, it is not hard to see the following.

m×m Theorem 2.7 For any matrix A = (aij) H , m > 1, then the following statements are equiva- ∈ lent:

(1). det k(A) = 0 P (2). t 1, 2, 3, 4 , t 5, 6, 7, 8 , t 9, 10, 11, 12 , and t 13, 14, 15, 16 , such ∃ 1 ∈ { } ∃ 2 ∈ { } ∃ 3 ∈ { } ∃ 4 ∈ { } that detCt1 = detCt2 = detCt3 = detCt4 = 0

5 (3). t 1, 2, 3, 4 , t 5, 6, 7, 8 , t 9, 10, 11, 12 , and t 13, 14, 15, 16 , such ∀ 1 ∈ { } ∀ 2 ∈ { } ∀ 3 ∈ { } ∀ 4 ∈ { } that detCt1 = detCt2 = detCt3 = detCt4 = 0 where Ct is defined in a similar way described in the Theorem 2.4. m×m m Let λl H be a left eigenvalue of A = (aij) H with eigenvector v H , by the property ∈ ∈ ∈ (5) of Proposition 2.1, get

Av = λlv (A λlI)v = 0 (A λlI) (v)= (0). (9) ⇔ − ⇔ P1 − P1 P1 Since v Hm is nonzero, then the following homogenous linear equations ∈ 4m×1 (A λlI)y = 0, y R . (10) P1 − ∈ have nonzero solution, i.e. det (A λlI) = 0. And det (A λlI) is called the generalized P1 − P1 − characteristic polynomial of A. By the Theorem 2.7, We can solve and analyze the left eigenvalues from the specified four real 4m 3 order polynomial equations in four variables. (Theorem 2.7 (2) ). −

3 Some properties

m×m Theorem 3.1 Let λl H is a left (λr right) eigenvalue of matrix A = (aij) H , for any matrix m×m ∈ ∈ B = (bij) H , if there exist k such that k(A)= e(B), then γl is a left (γr right) eigenvalue of ∈ P Pk matrix B, if e(γl)= k(λl) ( e(γr)= k(λr)). Qk Q Qke Q

Proof. Let λl be a left eigenvalue of A with eigenvector v, then

( e(B) e(γlI)) (v) = ( k(A) k(λlI)) (v) = 0 Pk − Pk P P − P P by the Eq.9, the conclusion then follows immediately.

If λr be a right eigenvalue of A with eigenvector v, then we can take nonzero vector u such that e(u)= k(v). By the Eq.9, get Pk P

k(A) k(v)= k(v) k(λr) e(B) e(u)= e(u) e(γr) Bu = uγr P P P P ⇔ Pk Pk Pk Pk ⇔ The proof is complete.

Corollary 3.1 If k(A)= e(B), then A and B are similar. P Pk

m×m Theorem 3.2 Let λl H is a left eigenvalue of matrix A = (aij) H with eigenvector v, for any ∈ ∈ two nonzero quaternions a and b, then aλlb is a left eigenvalue of matrix aAb.

Proof. Since a, b are nonzero, then

−1 −1 (A λlI)v = 0 a(A λlI)b(b v) = 0 (aAb aλlbI)(b v) = 0 − ⇔ − ⇔ −

m×m Theorem 3.3 Let λl H is a left eigenvalue of matrix A = (aij) H , then there exist nonnegative ∈ ∈ real numbers m 0 and M 0 such that m λl M. ≥ ≥ ≤ k k≤

6 2 H 2 H H Proof. Let λl be a left eigenvalue of A with unit eigenvector v, then λl = v λl v= v λ λlv= k k k k l vH AH Av. Using Singular-value decomposition theorem [3], there exist unitary quaternionic matrices V such that A D2 0 vH AH Av = vH VH Vv  0 0  where D = diag d1, d2, , dr and the d’s are the positive singular values of A. Let y = Vv = T { · · · } [y ,y , ,yn] , since V is a and v is unit eigenvector, then y is unit vector and 1 2 · · · 2 r 2 H D 0 2 2 λ = y y = d yi k k  0 0  i k k Xl=1 if r = n, then r r 2 2 2 2 2 2 2 d = d yi λ d yi = d min min k k ≤ k k ≤ max k k max Xl=1 Xl=1 otherwise, may have 0 λ 2 d2 ≤ k k ≤ max where dmin = min d , d , , dr and dmax = max d , d , , dr . The conclusion then follows { 1 2 · · · } { 1 2 · · · } immediately. It was given that the left spectrum is compact [7] and all the left eigenvalues of quaternion matrix are located in the union of n Gersgorin balls [12]. In fact, all the left eigenvalues of quaternion matrix are located in the in particular annulus connected with the singular values.

Remark 3.1 the norm of left eigenvalues λl is dominated by the norm of the right eigenvalues λr, i.e.

α λl β ≤ k k≤ where α = min λr : Av = vλr, v = 0 and β = max λr : Av = vλr, v = 0 . {k k 6 } {k k 6 }

4 Examples

Example 4.1 Let 1 κ A = .  κ 1 

Then

1−[λl]0 [λl]1 [λl]2 [λl]3 0 0 0 −1 −[λl]1 1−[λl]0 [λl]3 −[λl]2 0 0 −1 0  −[λl]2 −[λl]3 1−[λl]0 [λl]1 0 1 0 0  (A λ I)= −[λl]3 [λl]2 −[λl]1 1−[λl]0 1 0 0 0 (11) 1 l  0 0 0 −1 1−[λl]0 [λl]1 [λl]2 [λl]3  P −    0 0 −1 0 −[λl]1 1−[λl]0 [λl]3 −[λl]2   0 1 0 0 −[λl]2 −[λl]3 1−[λl]0 [λl]1     1 0 0 0 −[λl]3 [λl]2 −[λl]1 1−[λl]0

Let B = (A λlI), From the Theorem 2.7, we can take P1 − 2 2 2 2 det B(1 : 5, 4:8)= [λl] + 2[λl] [λl] [λl] + [λl] 2 = 0. (12) − 0 0 − 1 − 2 3 −

det B(1 : 4, 6, 4:8)= 2[λl] [λl] = 0. (13) − 2 3 7 det B(1 : 4, 7, 4 : 8) = 2[λl]1[λl]3 = 0. (14)

det B(1 : 4, 8, 4:8)= [λl] ([λl] 1) = 0. (15) − 3 0 − If [λl]3 = 0, by the Eq.12, 13, 14 and 15, then λl need satisfy 2 2 2 det B(1 : 5, 4:8)= ([λl] 1) [λl] [λl] 1 = 0. − 0 − − 1 − 2 − Obviously, a real solution of this equation does not exist. If [λl] = 0, by the Eq.12, 13, 14 and 15, then 3 6 2 [λl] 1 = 0, [λl] = 0, [λl] = 0, [λl] = 1. 3 − 2 1 0 i.e. [λl] = 1, [λl] = 0, [λl] = 0, [λl] = 1. 3 ± 2 1 0 In conclusion, λl = 1 κ. ± Example 4.2 Let 1  A = − .   1 

Then

1−[λl]0 [λl]1 [λl]2 [λl]3 0 0 1 0 −[λl]1 1−[λl]0 [λl]3 −[λl]2 0 0 0 −1  −[λl]2 −[λl]3 1−[λl]0 [λl]1 −1 0 0 0  (A λ I)= −[λl]3 [λl]2 −[λl]1 1−[λl]0 0 1 0 0 (16) 1 l  0 0 −1 0 1−[λl]0 [λl]1 [λl]2 [λl]3  P −    0 0 0 1 −[λl]1 1−[λl]0 [λl]3 −[λl]2   1 0 0 0 −[λl]2 −[λl]3 1−[λl]0 [λl]1     0 −1 0 0 −[λl]3 [λl]2 −[λl]1 1−[λl]0 Let B = (A λlI), From the Theorem 2.7, we can take P1 − det B(1 : 5, 4:8)= 2[λl] [λl] = 0 (17) − 2 3 2 2 2 2 det B(1 : 4, 6, 4:8)= [λl] + 2[λl] [λl] + [λl] [λl] = 0 (18) − 0 0 − 1 2 − 3 det B(1 : 4, 7, 4:8)= 2[λl] [λl] = 0 (19) − 1 2 det B(1 : 4, 8, 4 : 8) = [λl] ([λl] 1) = 0 (20) 2 0 − If [λl]2 = 0, by the Eq.17, 18, 19 and 20, then λl need satisfy 2 2 2 ([λl] 1) [λl] [λl] +1=0 − 0 − − 1 − 3 i.e. 2 2 2 ([λl] 1) + [λl] + [λl] = 1 0 − 1 3 thus 2 2 2 λl = [λl] + [λl] ~ + [λl] κ, where ([λl] 1) + [λl] + [λl] = 1 0 1 3 0 − 1 3

If [λl] = 0, by the Eq.17, 18, 19 and 20, then 2 6 2 [λl]3 = 0, [λl]2 +1=0, [λl]1 = 0, [λl]0 = 1 Obviously, a real solution of this equation does not exist. 2 2 2 In conclusion, λl = [λl] + [λl] ~ + [λl] κ, where ([λl] 1) + [λl] + [λl] = 1. 0 1 3 0 − 1 3 In fact, the left eigenvalues of Example 4.1 and 4.2 can also be computed using the formula [5] and the same result is obtained.

8 Example 4.3 Let κ 0 1 A =  0 κ 0  . 1 0 κ   Then (A λlI)= P1 −

−[λl]0 [λl]1 [λl]2 −1+[λl]3 000 0100 0 −[λl]1 −[λl]0 −1+[λl]3 −[λl]2 000 0010 0  −[λl]2 1−[λl]3 −[λl]0 [λl]1 000 0001 0  1−[λl]3 [λl]2 −[λl]1 −[λl]0 000 0000 1  00 0 0 −[λl]0 [λl]1 [λl]2 −1+[λl]3 00 0 0     00 0 0 −[λl]1 −[λl]0 −1+[λl]3 −[λl]2 00 0 0  (21)  00 0 0 −[λl]2 1−[λl]3 −[λl]0 [λl]1 00 0 0     0 0 0 01−[λl]3 [λl]2 −[λl]1 −[λl]0 00 0 0   100 0000 0 −[λl]0 [λl]1 [λl]2 −1+[λl]3    010 0000 0 −[λl]1 −[λl]0 −1+[λl]3 −[λl]2   001 0000 0 −[λl]2 1−[λl]3 −[λl]0 [λl]1     000 1000 01−[λl]3 [λl]2 −[λl]1 −[λl]0 

Let B = (A λlI), From the Theorem 2.7, let λl = κ, we can take P1 − 6

det B(1 : 9, 4 : 12) = 2[λl] ([λl] 1)α = 0 (22) 0 3 −

det B(1 : 8, 10, 4 : 12) = 2[λl] [λl] α = 0 (23) − 0 2 det B(1 : 8, 11, 4 : 12) = 2[λl]1[λl]1α = 0 (24) 2 2 2 2 det B(1 : 8, 12, 4 : 12) = ( [λl] + [λl] + [λl] + [λl] 2[λl] + 2)α = 0 (25) − 0 1 2 3 − 3 and 2 2 2 2 α = ([λl] ) + ([λl] ) + [λl] + ([λl] 1) . 0 1 2 3 − Since λl = κ, thus 6 α = 0 6 If [λl] = 0, by the Eq.22, 23, 24 and 25, then 0 6 2 [λl] = 0, [λl] = 0, [λl] = 1 and [λl] 1 = 0 1 2 3 0 − i.e. [λl] = 0, [λl] = 0, [λl] = 1 and [λl] = 1 1 2 3 0 ± thus λl = 1+ κ ± If [λl]0 = 0, by the Eq.25, then

2 2 2 2 2 2 [λl] + [λl] + [λl] 2[λl] +2=[λl] + [λl] + ([λl] 1) +1=0 1 2 3 − 3 1 2 3 − Obviously, a real solution of this equation does not exist.

It is easy to see κ is a left eigenvalue of A. In conclusion, λl = 1+ κ or λl = κ. ±

9 Example 4.4 Let ~ 0 ~ A =  0  0  ~ 0 κ   Then (A λlI)= P1 −

−[λl]0 −1+[λl]1 [λl]2 [λl]3 0 0 0 00 −1 0 0 1−[λl]1 −[λl]0 [λl]3 −[λl]2 00 00100 0  −[λl]2 −[λl]3 −[λl]0 −1+[λl]1 00 00000 −1  −[λl]3 [λl]2 1−[λl]1 −[λl]0 00 00001 0  0 0 0 0 −[λl]0 [λl]1 −1+[λl]2 [λl]3 00 0 0     0 0 0 0 −[λl]1 −[λl]0 [λl]3 1−[λl]2 00 0 0  (26)  0 0 0 01−[λl]2 −[λl]3 −[λl]0 [λl]1 00 0 0     0 0 0 0 −[λl]3 −1+[λl]2 −[λl]1 −[λl]0 00 0 0   0 −10000 00 −[λl]0 [λl]1 [λl]2 −1+[λl]3    100000 00 −[λl]1 −[λl]0 −1+[λl]3 −[λl]2   0 0 0 −10 0 0 0 −[λl]2 1−[λl]3 −[λl]0 [λl]1     001000 001−[λl]3 [λl]2 −[λl]1 −[λl]0 

Let B = (A λlI), From the Theorem 2.7, let λl = , we can take P1 − 6

det B(1 : 9, 4 : 12) = ([λl] + [λl] 2[λl] [λl] 1)β = 0 (27) − 1 3 − 1 3 −

det B(1 : 8, 10, 4 : 12) = ([λl] [λl] + 2[λl] [λl] )β = 0 (28) − 0 − 2 1 2 2 2 2 2 det B(1 : 8, 11, 4 : 12) = ([λl] [λl] + [λl] + [λl] + [λl] [λl] + 1)β = 0 (29) − 0 − 1 1 2 3 − 3 det B(1 : 8, 12, 4 : 12) = ([λl] + [λl] 2[λl] [λl] )β = 0 (30) 0 2 − 0 1 and 2 2 2 2 β = [λl] + [λl] + ([λl] 1) + [λl] . 0 1 2 − 3 By λl = , we can get 6 β = 0 6 then

[λl] + [λl] 2[λl] [λl] 1 = 0 (31) 1 3 − 1 3 − [λl] [λl] + 2[λl] [λl] = 0 (32) 0 − 2 1 2 2 2 2 2 [λl] [λl] + [λl] + [λl] + [λl] [λl] +1=0 (33) 0 − 1 1 2 3 − 3 [λl] + [λl] 2[λl] [λl] = 0 (34) 0 2 − 0 1 From Eq.32 and 34, we get 2 [λl] (2[λl] 2[λl] + 1) = 0 2 1 − 1 i.e. [λl]2 = 0

And by the Eq.34, get [λl]0 = 0. Substituting [λl]0 = 0 and [λl]2 = 0 into Eq.33, we get

2 2 [λl] + [λl] + [λl] [λl] +1=0 (35) − 1 1 3 − 3

10 1 1 If [λl]3 = 2 , by the Eq.31, we get 2 = 0, this is contradictory. 1 If [λl] = , then 3 6 2 1 [λl]3 [λl]1 = − (36) 1 2[λl] − 3 Substituting the Eq.36 into the Eq.35, then

1 [λl]3 2 1 [λl]3 2 ( − ) + − + [λl]3 [λl]3 +1=0 − 1 2[λl] 1 2[λl] − − 3 − 3 i.e. 4 3 2 [λl] 8[λl] + 10[λl] 6[λl] +1=0 (37) 3 − 3 3 − 3 Solving the Eq.37 (Polynomial), we get two real roots

1 [λl] = (1 2+ √5 ) 3 2 ± q− and two complex roots is omitted

1 [λl]3 = (1 ~ 2+ √5 ) 2 ± q− by the Eq.36, get

1 2+ √5 1 2 √ [λl]1 = ∓ p− , when [λl]3 = (1 2+ 5 ) 2+ √5 2 ± q− ∓p− It is easy to see  is a left eigenvalue of A. In conclusion,

1 2+ √5 1 2 ~ √ λl = , λl = ∓ p− + (1 2+ 5 )κ 2+ √5 2 ± q− ∓p−

Example 4.5 Let a11 0 0 A =  0 a22 a23  . 0 a a  32 33 

a22 a23 It is not hard to see that aij and the left eigenvalues of are all the left eigenvalues of A.  a32 a33  Since a 2 2 quaternion matrix may have one, two or an infinite number of left eigenvalues [5], so × a 3 3 quaternion matrix can have one, two, three or an infinite number of left eigenvalues. Similarly, × a n n quaternion matrix may have 1, 2, ,n 1,n or an infinite number of left eigenvalues. In × · · · − addition, by Theorem 2.7 (2), by analysing the number of roots of specified four polynomial equations, obtains the number of left eigenvalues.

Remark 4.1 Numerical algorithms can be used to solve the specified four polynomial equations (The- orem 2.7 (2)) to find the left eigenvalues of m by m quaternion matrix.

11 5 Conclusion and Discussion

In this paper, we introduce a method to compute the left eigenvalues of quaternion matrix based on the matrix representation of quaternion. We obtain four real polynomial equations with four variables which are equivalent to the generalized characteristic polynomial, the left eigenvalues could be found via solving special polynomial equations. In addition, while A Hm×m may have infinite number of ∈ left eigenvalues, the norm of left eigenvalues λl of A is dominated by the norm of the right eigenvalues λr of A . Further theoretical analysis of the special polynomial equations about the left eigenvalues and finding much more potential applications of matrix representation will be left for future work.

References

[1] J. L. Brenner, et al., Matrices of quaternions., Pacific Journal of Mathematics 1 (3) (1951) 329–335.

[2] H. Lee, Eigenvalues and canonical forms of matrices with quaternion coefficients, in: Proceedings of the Royal Irish Academy. Section A: Mathematical and Physical Sciences, JSTOR, 1948, pp. 253–260.

[3] F. Zhang, Quaternions and matrices of quaternions, and its applications 251 (1997) 21–57.

[4] R. Wood, Quaternionic eigenvalues, Bulletin of the London Mathematical Society 17 (2) (1985) 137–138.

[5] L. Huang, W. So, On left eigenvalues of a quaternionic matrix, Linear algebra and its applications 323 (1-3) (2001) 105–116.

[6] L. Huang, W. So, Quadratic formulas for quaternions, Applied Mathematics Letters 15 (5) (2002) 533–540.

[7] E. Mac´ıas-Virg´os, M. Pereira-S´aez, A topological approach to left eigenvalues of quaternionic matrices, Linear and Multilinear Algebra 62 (2) (2014) 139–158.

[8] W. So, Quaternionic left eigenvalue problem., Southeast Asian Bulletin of Mathematics 29 (3).

[9] B. Kalantari, Algorithms for quaternion polynomial root-finding, Journal of Complexity 29 (3-4) (2013) 302–322.

[10] M. I. Falc˜ao, F. Miranda, R. Severino, M. J. Soares, Evaluation schemes in the ring of quaternionic polynomials, BIT Numerical Mathematics 58 (1) (2018) 51–72.

[11] R. W. Farebrother, J. Groß, S.-O. Troschke, Matrix representation of quaternions, Linear algebra and its applications 362 (2003) 251–255.

[12] F. Zhang, Gerˇsgorin type theorems for quaternionic matrices, Linear Algebra and its Applications 424 (1) (2007) 139–153.

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