31 The Jordan Canonical Form of a Product of Elementary S-unitary Matrices Erwin J. Gonda Agnes T. Paras* Institute of Mathematics and Natural Sciences Research Institute University of the Philippines Diliman ABSTRACT Let S be an n-by-n, nonsingular, and Hermitian matrix. A square complex matrix Q is said to be S-unitary if Q*SQ = S. An S-unitary matrix Q is said to be elementary if rank(Q — I) = 1. It is known what form every elementary S-unitary can take, and that every S-unitary can be written as a product of elementary S-unitaries. In this paper, we determine the Jordan canonical form of a product of two elementary S-unitaries. Keywords: elementary S-unitary matrix, Hermitian matrix, Jordan canonical form INTRODUCTION Let M n be the set of all n-by-n matrices with entries in the complex field ℂ and let GL n be the set of all nonsingular matrices in M n . Let S ∈ GL n be Hermitian. A Q ∈ M n is said to be S-unitary if Q*SQ = S, where Q* is the conjugate transpose of Q (Gohberg et al. 2005). If S = I, then the set of S-unitary matrices in GL n coincides with the set of unitary matrices. Let U S be the set of all S-unitary matrices. Observe that U S is nonempty since I ∈ US. If Q ∈ U S , then Q -1 is similar to Q*, |det Q| = 1, and αQ ∈ U S for all α ∈ ℂ with modulus 1. Moreover, U S is a group under multiplication and consists of all matrices in M n that preserve the scalar product [u,v] S = u*Sv for all u, v ∈ ℂn. An H ∈ US is called elementary if rank(H – I) = 1. Let H S be the set of all elementary S-unitary matrices. When S is Hermitian, H S = K S ∪ L S , where K S = {Kx,r = I + irxx*S ∶ x ∈ ℂ n\{0}, x*Sx = 0, and r ∈ ℝ\{0}} and LS = {Lx,φ = I + (eiφ – 1) x*Sx xx*S: x ∈ ℂn, x*Sx ≠ 0, φ ∈ ℝ, and eiφ ≠ 1} * Corresponding Author Science Diliman (January-June 2020) 32:1, 31-41 The Jordan Canonical Form of a Product of Elementary S-unitary Matrices 32 (Catbagan 2015). If v ∈ ℂn such that v*Sv ≠ 0, the Λ S -Householder matrix Sv = I – 2 v*Sv vv*S generalizes the Householder matrix of v, which is equal to Lv,π for S = I (Merino et al. 2011; Horn and Johnson 2013). If Kx,r ∈ KS, then Kx,r –1 = Kx,–r and Kx,r is similar to In–2 ⊕ J2 (1). If Lx,φ ∈ LS, then Lx,φ–1 = Lx,–φ and Lx,φ is similar to I n–1 ⊕ [eiφ]. Hence H ∈ H S if and only if H–1 ∈ H S . Thus, I is a product of two elements of H S . Moreover, if A ∈ U S , then A can be written as a product of elements of H S (Catbagan 2015). Thus, the elements of H S generate the group U S . Since there are two types of elements of H S , there are three types of products of two elements of H S up to similarity. We wish to determine which Jordan canonical forms arise for each possibility, since the Jordan canonical form of a matrix reveals a lot of information such as its rank, nullity, eigenvalues, and their algebraic and geometric multiplicities. Analogous results for symplectic matrices and J-Householder matrices can be found in de la Rosa et al. (2012). PRELIMINARIES If S = P*P for some P ∈ GL n , then x*Sx > 0, when 0 ≠ x ∈ ℂn; and Q ∈ U S if and only if PQP–1 ∈ U I . Hence when S is positive definite, H S = L S , and every S-unitary is similar to a unitary matrix. Since U S = U –S , from now on we only consider S that is *-congruent to I k ⊕–I n–k for 0 < k < n, that is S = P * (I k ⊕ –I n–k ) P, for some P ∈ GL n . Let n be a positive integer such that n ≥ 2, and T ⊆ ℂn be nonempty. Let TS be the S-perpendicular subspace of T defined by TS = { z ∈ ℂn | x * Sz = 0, for all x ∈ T } . Then dimTS = n – dim(spanT ) , since TS = (S(spanT))⊥, which is the orthogonal complement of S(spanT ) with respect to the usual inner product on ℂn, and ℂn = W ⊕ W⊥ for any subspace W of ℂn. Let Hx, Hy ∈ HS and A = Hx Hy . Then Hx = I + αxx*S and Hy = I + βyy*S, for some nonzero α, β ∈ ℂ. If {x, y} is linearly dependent, then y = δx, for some δ ∈ ℂ. This implies A = I + αxx*S + βyy*S + αβxx*Syy*S = I + (α + β|δ|2 + αβ|δ|2 x*Sx)xx*S. Hence A = I + μxx*S, where μ = α + β|δ|2 + αβ|δ|2 x*Sx ∈ ℂ. If μ = 0, then A = I, which implies Hx = Hy –1. If μ ≠ 0, then rank(A – I) = 1, and since A ∈ U S , we have A ∈ H S . E.J. Gonda and A.T. Paras 33 Suppose {x, y} is linearly independent. Let z ∈ ℂn be given. Suppose z ∈ ker (A – I), that is, Az = z. Then 0 = (A – I)z = (αx*Sz)x + (βy* Sz)y + αβ(x*Sy)(y*Sz)x. Since {x, y} is linearly independent, and α, β are nonzero, we have y*Sz = 0 and it follows that x*Sz = 0. Thus, z ∈ {x, y}S. Conversely, suppose z ∈ {x, y}S. Then x*Sz = y*Sz = 0 and so (A – I)z = α(x*Sz)x + β(y*Sz)y + αβ(x*Sy)(y*Sz)x = 0, that is, z ∈ ker(A – I). Therefore ker(A – I) = {x, y}S. Lemma 1. Let S ∈ GL n be Hermitian and let x, y ∈ ℂn be nonzero. Suppose H x , H y ∈ H S and A = H x H y . If {x, y} is linearly dependent, then A = I or A ∈ H S . If {x, y} is linearly independent, then ker(A – I) = {x, y}S. If {x, y} is linearly independent, an immediate consequence of Lemma 1 is that dim(ker(A – I)) = dim({x, y}S) = n – 2. Thus, there are n – 2 Jordan blocks corresponding to 1 in the Jordan canonical form of A. For completeness, we include the following lemma, which is used several times in the paper and can be readily verified. If A = [a ij ] ∈ M n , the trace of A is defined to be trA = Σn j = 1ajj. Lemma 2. Let A, B ∈ M 2 be given such that neither is a scalar matrix. Then A and B are similar if and only if trA = trB and det A = det B. Let {x, y} be a linearly independent subset of ℂn. We consider each of the three possibilities (i) H x , H y ∈ K S , (ii) H x , H y ∈ L S , or (iii) H x ∈ K S and H y ∈ L S , and determine the Jordan canonical form of the product H x H y . Hx, Hy ∈ KS Let {x, y} be a linearly independent subset of ℂn such that H x , H y ∈ K S , i.e., H x = I + ir x xx*S and H y = I + ir y yy*S, where x*Sx = y*Sy = 0, and rx, ry are nonzero real numbers. If A = H x H y , then A = I + ir x xx*S + ir y yy*S – r x r y (x*Sy)xy*S. Either x*Sy = 0 or x*Sy ≠ 0. The Jordan Canonical Form of a Product of Elementary S-unitary Matrices 34 Case 1: If x*Sy = 0, then A = I + ir x xx*S + ir y yy*S. Note that {x, y}S = {x}S ∩ {y}S, which is of dimension n – 2. If n > 2, then there exists z ∈ {y}S but z ∉ {x}S. Hence, (A – I ) z = ir x (x*Sz)x ≠ 0. Since x*Sx = y*Sy = x*Sy = 0, we have (A – I ) 2 = 0. Since A – I ≠ 0, the minimal polynomial of A is (x – 1)2 and so the largest Jordan block corresponding to 1 is of size 2. The number of Jordan blocks corresponding to 1 of size 1 is rank(A – I ) 0 –2 rank(A – I ) + rank(A – I )2 = n – 2(2) + 0 = n – 4. Since 1 is the only eigenvalue of A and there are n – 2 Jordan blocks corresponding to 1, A is similar to I n – 4 ⊕ J 2 (1) ⊕ J 2 (1). If n = 2, then x*Sy ≠ 0, otherwise x*Sx = y*Sy = x*Sy = 0 and {x, y} linearly independent imply ℂ2 = {x, y}S is of dimension n – 2 = 0, which is a contradiction. Case 2: Suppose x*Sy ≠ 0. We find any remaining eigenvalues of A. The images of x and y under A are Ax = x + ir x (x*Sx)x + ir y (y*Sx)y – r x r y (x*Sy)(y*Sx)x = (1 – r x r y |x*Sy|2) x + ir y (y*Sx)y and Ay = y + ir x (x*Sy)x + ir y (y*Sy)y – r x r y (x*Sy)(y*Sy)x = y + ir y (x*Sy)x . Hence span{x, y} is invariant under A. Consider the restriction of A to span{x, y} and its matrix representation M = 1 – rx ry |x*Sy| 2 ir y (x*Sy) iry (y*Sx) 1 with respect to the ordered basis {x, y}. Since x*Sx = y*Sy = 0 and x*Sy ≠ 0, we have ℂn = span{x, y} ⊕ {x, y}S. Thus A is similar to M ⊕ I n–2 and 1 is not an eigenvalue of M. Note that det(M) = 1 and tr(M) = 2 – r x r y |x*Sy|2 ∈ ℝ. Since A ∈ U S has determinant 1 and M is not a scalar matrix, we see that M is similar to one of the following: diag(eiθ, e–iθ), where θ ∈ ℝ such that eiθ ≠ ±1; J 2 (–1); or diag(λ, λ–1), where λ ∈ ℝ and |λ| > 1. We determine if the preceding three possibilities for the Jordan canonical form of M occur. Let θ ∈ ℝ such that eiθ ≠ ±1. If we choose rx, ry ∈ ℝ such that rx ry = 2(1–cosθ) |x*Sy|2 ≠ 0, then det (M) = 1 = det (diag(eiθ, e–iθ)) and tr (M) = 2 cosθ = tr (diag(eiθ, e–iθ)). By Lemma 2, M is similar to diag(eiθ, e–iθ). E.J. Gonda and A.T. Paras 35 If we choose rx, ry ∈ ℝ such that rx ry = 4 |x*Sy|2 , then tr(M) = –2 = tr(J 2 (–1)) and det (M) = 1 = det (J 2 (–1)). By Lemma 2, M is similar to J 2 (–1). Let λ ∈ ℝ such that |λ| > 1. If we choose rx, ry ∈ ℝ such that rx ry = (2–λ–λ–1) |x*Sy|2 ≠ 0, then we have det (M) = 1 = det (diag(λ, λ–1)) and tr(M) = –2 = tr(diag(λ, λ–1)). Since λ ≠ λ–1, we have that M is similar to diag(λ, λ–1). Theorem 3. Let S ∈ GL n be indefinite Hermitian and x, y ∈ ℂn be given. If {x, y} is linearly independent and Hx, Hy ∈ KS, then the product HxHy is similar to one of the following: a. I n–4 ⊕ J 2 (1) ⊕ J 2 (1) b. I n–2 ⊕ J 2 (–1) c. I n–2 ⊕ diag(eiθ, e–iθ), where θ ∈ ℝ such that eiθ ≠ ±1 d. I n–2 ⊕ diag(λ, λ–1), where |λ| > 1 and λ ∈ ℝ. Hx, Hy ∈ LS We now consider the product of two elements of L S . Let x, y ∈ ℂn such that {x, y} is linearly independent and Hx, Hy ∈ LS, that is, Hx = I + eiα –1 x*Sx xx*S and H y = I + eiβ –1 y*Sy yy*S, where x*Sx and y*Sy are nonzero, and α, β ∈ ℝ such that eiα ≠ 1 and eiβ ≠ 1. Since Hy = Hay for all nonzero a ∈ ℂ, we can assume that x*Sx, y*Sy ∈ {1, –1}. If A = Hx Hy, then A = I + eiα –1 x*Sx xx*S + eiβ –1 y*Sy yy*S + eiα –1 x*Sx eiβ –1 y*Sy (x*Sy)xy*S. Case 1: If x*Sy = 0, then A = I + eiα –1 x*Sx xx*S + eiβ –1 y*Sy yy*S. Observe that Ax = x + (eiα –1)x = eiαx. Hence, x is an eigenvector of A corresponding to eiα. Similarly, y is an eigenvector of A corresponding to eiβ. Since x*Sx and y*Sy are nonzero and x*Sy = 0, we have ℂn = span{x, y} ⊕ {x, y}S. Hence A is similar to I n–2 ⊕ diag(eiα, eiβ). The Jordan Canonical Form of a Product of Elementary S-unitary Matrices 36 Case 2: Suppose x*Sy ≠ 0. We find any remaining eigenvalues of A. The images of x and y under A are Ax = x + (eiα –1)x + eiβ –1 y*Sy (y*Sx)y + eiα –1 x*Sx eiβ –1 y*Sy (x*Sy)(y*Sx)x = �eiα + eiα –1 x*Sx eiβ –1 y*Sy |x*Sy|2�x + � eiβ –1 y*Sy (y*Sx)�y and Ay = y + eiα –1 x*Sx (x*Sy)x + eiβ –1 y*Sy (y*Sy)y + eiα –1 x*Sx eiβ –1 y*Sy (x*Sy)(y*Sy)x = eiβy + �eiβx*Sy eiα –1 x*Sx �x. Hence span{x, y} is invariant under A. Consider the restriction of A to span{x, y} and its matrix representation L = eiα + eiα –1 x*Sx eiβ –1 y*Sy |x*Sy|2 eiβx*Sy eiα –1 x*Sx eiβ –1 y*Sy ( y * S x ) eiβ with respect to the ordered basis {x, y}. Note that ax + by ∈ {x, y}S for some a, b ∈ ℂ if and only if x*S(ax + by) = 0 and y*S(ax + by) = 0, that is x*Sx x*Sy y*Sx y*Sy � � a b = 0 0 . Since x*Sx, y*Sy ∈ {1, –1}, we have {x, y}S ∩ span{x, y} = {0} if and only if x*Sx and y*Sy have opposite signs or |x*Sy| ≠ 1. If x*Sx = y*Sy ∈ {±1} and |x*Sy| = 1, then x, y ∉ {x, y}S and {x, y}S ∩ span{x, y} = span{(x*Sy)x – (x*Sx)y}. E.J. Gonda and A.T. Paras 37 Hence span{x} ⊕ {x, y}S is of dimension n–1 and contains span{x, y}. Now Ax can be rewritten as Ax = ei(α+β) x – (eiβ –1)(y*Sx)[(x*Sy)x – (x*Sx)y]. This implies that span{x} ⊕ {x, y}S is invariant under A. Since detA = ei(α+β) and rank(A – I) = 2, we have that A is similar to I n–3 ⊕ J 2 (1) ⊕ [ei(α+β)], if ei(α+β) ≠ 1; or I n–3 ⊕ J 3 (1), if ei(α+β ) = 1. If ℂn = span{x, y} ⊕ {x, y}S, then A is similar to I n–2 ⊕ L and 1 is not an eigenvalue of L. Observe that det L = ei(α+β) and tr L = eiα + eiβ + eiα –1 x*Sx eiβ –1 y*Sy |x*Sy|2. Since A ∈ U S and L is not a scalar matrix, then L is similar to one of the following: diag(eiθ, eiϕ), where θ, ϕ ∈ ℝ such that eiθ, eiϕ are distinct and both are not equal to 1; J 2 (λ), where |λ| = 1 but λ ≠ 1; or diag (λ, λ–1), where |λ| > 1. It suffices to determine whether the last two possibilities for the Jordan canonical form of L occur. But first we need to determine the possible nonzero values of x*Sy, when x*Sx, y*Sy ∈ {1, –1} and {x, y} is linearly independent. Let e i ∈ ℂn denote the column with ith entry 1 and 0 elsewhere. Suppose c ∈ ℂ is nonzero and S = P*(I k ⊕ – I n–k )P, for some nonsingular P and integer 0 < k < n. If |c| > 1, we can take x, y ∈ ℂn such that Px = e 1 and Py = ce 1 + |c|2 –1e k+1, so that x*Sx = 1, y*Sy = |c|2 – (|c|2 – 1) = 1, and x*Sy = c. Thus, if |c| > 1, there exists a linearly independent set {x, y} such that x*Sx = y*Sy and x*Sy = c. If c ∈ ℂ is nonzero and we take x, y ∈ ℂn such that Px = e 1 and Py = ce 1 + |c|2 –1e k+1, then x*Sx = 1, y*Sy = |c| 2 – (|c|2 + 1) = –1 and x*Sy = c. Hence every nonzero c ∈ ℂ can be realized as x*Sy by a linearly independent set {x, y} such that x*Sx = –y*Sy, when S is *-congruent to I k ⊕ –I n–k . Let α = β ∈ ℝ such that α ≠ kπ, for all k ∈ ℤ. If a = Re(eiα), then –4eiα (eiα – 1)2 = 2 1–a > 1. If we take x, y ∈ ℂn such that x*Sx = 1 = y*Sy and |x*Sy|2 = –4eiα (eiα – 1)2 , then tr L = 2eiα + (eiα–1)2 |x*Sy|2 = –2eiα and det L = ei2α. Since L is not a scalar matrix, it follows from Lemma 2 that L is similar to J 2 (–eiα), where eiα ≠ ±1. If we take eiα = e–iβ = i, and x, y ∈ ℂn such that x*Sx = 1 = –y*Sy and |x*Sy| = 1, then tr L = –2 and det L = 1. Since L is not a scalar matrix, L is similar to J 2 (–1). Let λ = reiθ, where r > 1 and θ ≠ 2kπ for all k ∈ ℤ. Then – eiθ(r –1)2 (eiθ–1)2r is positive. � � The Jordan Canonical Form of a Product of Elementary S-unitary Matrices 38 If we take α = β = θ, and x, y ∈ ℂn such that x*Sx = 1 = –y*Sy and |x*Sy|2 = – eiθ(r –1)2 (eiθ–1)2r , then tr L = 2eiθ– (eiθ–1)2 |x*Sy|2 = eiθ (r + r–1) = λ + λ–1 and det L = ei2θ = λ λ–1 . Hence L is similar to diag(λ, λ–1). Let λ = r, where r > 1. Let β = –α and α ∈ ℝ such that Re(eiα) = r–1. Since (r –1)2 r > 0, we have r –r –1 2(1–r –1) > 1. If we take x, y ∈ ℂn such that x*Sx = 1 = y*Sy and |x*Sy|2 = r –r –1 2(1–r –1) , then tr L = 2r –1 + 2(1–r –1) |x*Sy|2 = r + r –1 and det L = 1. Hence L is similar to diag(r, r –1). Theorem 4. Let S ∈ GL n be indefinite Hermitian and x, y ∈ ℂn be given. If {x, y} is linearly independent such that Hx, Hy ∈ LS, then the product HxHy is similar to one of the following: a. I n–2 ⊕ diag(eiθ, eiϕ), where θ, ϕ ∈ ℝ such that eiθ, eiϕ ≠ 1 b. I n–3 ⊕ J 2 (1) ⊕ [eiθ], where θ ∈ ℝ and eiθ ≠ 1 c. I n–3 ⊕ J 3 (1) d. I n–2 ⊕ J 2 (λ), where |λ| = 1 and λ ≠ 1 e. I n–2 ⊕ diag(λ, λ–1), where |λ| > 1. Hx ∈ KS and Hy ∈ LS Lastly, we consider the product of an element of K S and of L S . If x, y ∈ ℂn are nonzero such that Hx ∈ KS and Hy ∈ LS, then Hx = I + irxx*S, where r ∈ ℝ ∖ {0}, and x*Sx = 0, and Hy = I + e iα–1 y*Sy yy*S, where eiα ≠ 1. Note that {x, y} is linearly independent since x*Sx = 0 ≠ y*Sy. If A = Hx Hy, then A = I + irxx*S + e iα–1 y*Sy yy*S + ir e iα–1 y*Sy (x*Sy) xy*S. Case 1: If x*Sy = 0, then A = I + ir xx*S + e iα–1 y*Sy yy*S, Ax = x, and Ay = eiαy. Since x*Sy = 0 and x*Sx = 0, we have (A – I )2 = (e iα–1)2 y*Sy yy*S and (A – I )3 = (e iα–1)3 y*Sy yy*S . E.J. Gonda and A.T. Paras 39 Observe that rank(A – I ) = 2 and rank (A – I )2 = rank(A – I )3 = 1 , which imply that 2 is the size of the largest Jordan block corresponding to 1, and the number of Jordan blocks of size 2 corresponding to 1 is rank(A – I ) – 2rank(A – I )2 + rank(A – I )3 = 2 – 2(1) + 1 = 1. Since there are n–2 Jordan blocks corresponding to 1 and det A = eiα, we have that A is similar to I n–3 ⊕ J 2 (1) ⊕ [eiα]. Case 2: Suppose x*Sy ≠ 0. The images of x and y under A are Ax = x + e iα–1 y*Sy (y*Sx)y + ir e iα–1 y*Sy |x*Sy|2x = �1+ir e iα–1 y*Sy |x*Sy|2� x + e iα–1 y*Sy (y*Sx)y and Ay = y + ir(x*Sy)x + (eiα –1)y + ir(eiα – 1)(x*Sy)x = ireiα (x*Sy)x + eiαy. Hence span{x, y} is invariant under A. Consider the restriction of A to span{x, y} and its matrix representation K = 1 + ir eiα –1 y*Sy |x*Sy|2 ireiα(x*Sy) eiα –1 y*Sy ( y * S x ) eiα with respect to the ordered basis {x, y}. Since ℂn = span{x, y} ⊕ {x, y}S, A is similar to I n–2 ⊕ K and 1 is not an eigenvalue of K. Note that det K = eiα ≠ 1 and tr K = eiα + 1 + ir eiα –1 y*Sy |x*Sy|2. Since A is S-unitary, K is similar to one of the following: diag(eiθ, eiϕ), where θ, ϕ ∈ ℝ such that eiθ, eiϕ are distinct with both not equal to 1, and ei(θ + ϕ) = eiα; or diag(λ, λ–1 ), where |λ| > 1 and λ ≠ ±1. We now determine whether the three possibilities for the Jordan canonical form of K occur. The Jordan Canonical Form of a Product of Elementary S-unitary Matrices 40 Let θ, ϕ ∈ ℝ such that eiθ, eiϕ, and ei(θ+ϕ) are not equal to 1, and eiθ ≠ eiϕ. If α = θ + ϕ, choose r ∈ ℝ such that r(eiα –1) = (y*Sy)(1–e iθ)(eiϕ–1) i|x*Sy|2 . This has a solution since (1–eiθ)(eiϕ–1) eiα –1 = –1 + e iθ + eiϕ–2 eiα –1 is nonzero and the real part of eiθ + eiϕ–2 eiα –1 is 1. Then det K = ei(θ + ϕ) = det(diag(eiθ, eiϕ)) and tr K = eiθ + eiϕ = tr(diag(eiθ, eiϕ)). Thus K is similar to diag(eiθ, eiϕ). Let λ = teiγ, where t, γ ∈ ℝ such that t > 1 and ei2γ ≠ 1. Choose α = 2γ and r ∈ ℝ such that r(eiα –1) = (y*Sy) (1–te iγ) (t–1eiγ– 1) i|x*Sy|2 . This has a solution since (1–teiγ) (t–1eiγ– 1) eiα–1 = –1 + (t + t –1) eiγ– 2 ei2γ–1 is nonzero and the real part of (t + t –1) eiγ– 2 ei2γ–1 is 1. Then det K = λ λ–1 = det(diag(λ, λ–1 )) and tr K = (t + t–1) eiγ = tr (diag(λ, λ–1)). By Lemma 2, K is similar to diag(λ, λ–1). Let λ = eiβ, where β ∈ ℝ and λ ≠ ±1. Choose α = 2β and r ∈ ℝ such that r = (1–λ) y*Sy i(λ + 1)|x*Sy|2 . This has a solution since 1–λ λ + 1 = –1 + 2 λ + 1 and the real part of 2 λ + 1 is 1. Then det K = λ2 = det J 2 (λ) and tr K = 2λ = tr J 2 (λ). Since K is not a scalar matrix, K is similar to J 2 (λ). Theorem 5. Let S ∈ GL n be indefinite Hermitian and x, y ∈ ℂn be given. If {x, y} is linearly independent such that Hx ∈ KS and Hy ∈ LS, then the product Hx Hy is similar to one of the following: a. I n–3 ⊕ J 2 (1) ⊕ [eiα], for some α ∈ ℝ such that eiα ≠ 1 b. I n–2 ⊕ diag(eiθ, eiϕ), where θ, ϕ ∈ ℝ such that eiθ, eiϕ, ei(θ + ϕ) are all not equal to 1, and eiθ ≠ eiϕ c. I n–2 ⊕ diag(λ, λ–1), where |λ| > 1 and λ ∉ ℝ d. I n–2 ⊕ J 2 (λ), where |λ| = 1 but λ ≠ ±1. E.J. Gonda and A.T. Paras 41 ACKNOWLEDGMENTS The work of A.T. Paras was supported by the Natural Sciences Research Institute (NSRI) Project MAT-18-1-05. REFERENCES Catbagan K. 2015. On the ΛS-polar decomposition and the Λ S -elementary matrices [B.S. thesis]. Philippines: University of the Philippines Diliman. de la Rosa KL, Merino DI, Paras AT. 2012. The J-Householder matrices. Linear Algebra Appl. 436(5):1189-1194. Gohberg I, Lancaster P, Rodman L. 2005. Indefinite linear algebra and applications. Basel: Birkhäuser. Horn RA, Johnson CR. 2013. Matrix analysis. 2nd ed. New York: Cambridge University Press. Merino DI, Paras AT, Teh TD. 2011. The Λ S -Householder matrices. Linear Algebra Appl. 436:2643-2664. ______ Erwin J. Gonda is a B.S. Mathematics graduate of University of the Philippines Diliman and is currently working in the insurance industry. Agnes T. Paras is a professor of Mathematics at University of the Philippines Diliman