MATIMYAS´ MATEMATIKA Journal of the Mathematical Society of the Philippines ISSN 0115-6926 Vol. 35 Nos. 1-2 (2012) pp. 23-34

The Sum of Involutory Matrices

Dennis I. Merino Department of Southeastern Louisiana University Hammond, LA, USA [email protected]

Agnes T. Paras Institute of Mathematics University of the Philippines Diliman Quezon City, Philippines [email protected]

Diane Christine P. Pelejo Institute of Mathematics University of the Philippines Diliman Quezon City, Philippines [email protected]

Abstract

Let K = C, R, Q, Z, or Zk. An n-by-n A, with entries in K, is an 2 if A = In. We determine the necessary and sufficient conditions for an n-by-n matrix to be written as a sum of involutions over K. When K = C, we determine the possible Jordan Canonical Forms of a sum of two involutions.

1 Introduction

Decomposition of matrices into a sum of special matrices has been the topic of previous studies (see [2, 5, 6, 7, 9, 10]). Given a set G of n-by-n matrices, we wish to determine if every n-by-n matrix can be written as a sum of matrices from G. In particular, we are interested in the case when G is the set of matrices satisfying A2 = I. We are also interested in studying matrices with entries coming from different sets: complex numbers (C), real numbers (R), rational numbers (Q), (Z), and (Zk) integers modulo k. Let K ∈ {C, R, Q, Z, Zk}. We denote the set of m-by-n matrices with entries from K by Mm,n(K). If m = n, we set Mn(K) ≡ Mn,n (K). We denote the of a matrix A by det(A), the characteristic polynomial of A by pA(x), the minimal polynomial of A by mA(x), and the spectrum of A or set of characteristic values of A by σ(A).

2 Definition 1. Let A ∈ Mn(K). Then A is an involution or an involutory matrix if A = I. + + We denote by Γn the set of all involutions in Mn (K). Let Ωn be the set of matrices + which can be written as a sum of matrices in Γn . Proposition 2. 1. Any matrix that is similar to an involution is an involution.

23 24 Dennis I. Merino, Agnes T. Paras, Diane Christine P. Pelejo

2. The negative of an involution is an involution. 3. The direct sum of two involutions is an involution.

4. Let K be a field of characteristic not equal to 2. Let A ∈ Mn (K) be an involution. Then A is similar to Ik ⊕ −Ij, for some nonnegative integers k and j (where zero size means nonexistence).

5. If K is a field of characteristic 2, then an involution in Mn (K) is similar to M  1 0  I ⊕ , (1) j 1 1 k for some nonnegative integers k and j. Proof. Properties 1-3 are easy to verify. To show properties 4 and 5, notice that for an 2 involution A, mA(x) must divide x − 1 = (x − 1)(x + 1). If char (K) 6= 2, then A is diagonalizable and σ(A) ⊆ {1, −1} so that A is similar to Ik ⊕ −Ij. If char (K) = 2, then 2 2 2 x − 1 = (x − 1) and mA(x) is either x − 1 or (x − 1) .

1.1 The Trace of a Matrix Let p be a monic polynomial of degree n, and let C (p) be the of p. We look at the case when K is a field. Notice that any over K is similar to a direct sum of companion matrices of the elementary divisors of its characteristic polynomial. Moreover, if all the characteristic values of the matrix are contained in K, then the matrix is similar to a direct sum of Jordan blocks corresponding to its characteristic values (eigenvalues). We denote the upper triangular Jordan block corresponding to λ ∈ C of size k by Jk(λ). For a given M ∈ Mn(K), the trace of M, denoted tr(M), is the sum of the diagonal entries of M. It is known that the trace of a matrix over a commutative ring with unity 1K is similarity invariant, that is, if N = PMP −1, then tr(M) = tr(N). It follows that if K is a field, then tr(M) is the sum of the roots of the characteristic polynomial of M.

Set h1Ki ≡ {y · 1K : y ∈ Z}, that is, if y > 0, then y · 1K is the sum of y number of 1K; if y < 0, then y · 1K is the sum of −y number of −1K. We define parity in K.

Definition 3. Let a, b ∈ h1Ki. We say that a is even if a = 2x · 1K for some x ∈ Z. We say that a is odd if a = (2x + 1) · 1K for some x ∈ Z. If a and b are both even or both odd, then we say that a and b have the same parity and we write a ≡ b (mod 2).

Note that every element in Z2k+1 is both odd and even. Let z ∈ Z be given. Set zK ≡ z · 1K.

Proposition 4. Let K be a field. If A ∈ Mn(K) is an involution, then tr(A) ∈ h1Ki. Moreover, tr(A) and nK have the same parity.

Proof. Let A ∈ Mn(K) be an involution. If char(K) 6= 2, then Proposition 2(4) ensures that A is similar to Ij ⊕ −In−j for some nonnegative j. Hence, tr(A) = 2jK − nK ≡ nK (mod 2). If char(K) = 2, then Proposition 2(5) guarantees that tr(A) = nK.

We now look at the cases when K = Z or K = Z2k. Suppose that A = [aij] ∈ Mn(K). Define Amod2 ≡ [bij] ∈ Mn(Z2), where bij = aij (mod 2). Notice that ab (mod 2) = a (mod 2)b (mod 2) and that (a + b) (mod 2) = a (mod 2) + b (mod 2). It follows that The Sum of Involutory Matrices 25

k k 2 2 (A )mod2 = [Amod2] ∈ Mn(Z2). Hence, if A = I, then [Amod2] = Imod2 = I. That is, if A ∈ Mn (K) is an involution and if we reduce A modulo 2, then the resulting matrix is an involution over Z2.

Lemma 5. Let K = Z or Z2k and A = [aij] ∈ Mn(K). If A is an involution then Amod2 is an involution over Z2.

Let A ∈ Mn (Z) be an involution. Then A ∈ Mn (Q). The proof of Proposition 4 shows that tr(A) = 2k − n for some k ∈ Z. Notice that 2k − n ≡ n (mod 2).

Corollary 6. Let K = Z or Zk. If A ∈ Mn (K) is an involution, then tr(A) ≡ n (mod 2).

Proof. Let A ∈ Mn(Z2k+1) be an involution. Then tr(A) ≡ n (mod 2) since every element of Z2k+1 is both odd and even. Let A ∈ Mn(Z2k) be an involution. Set Amod2 ≡ [bij]. If k = 1, then Proposition n n P P 4 ensures that tr(A) = n (mod 2). Suppose k > 1. Then tr(A) = aii ≡ bii i=1 i=1 (mod 2) = tr(Amod2). Lemma 5 guarantees that Amod2 is an involution over Z2. Hence, tr(A) ≡ tr(Amod2) ≡ n (mod 2).

2 The Sum of Involutions

Let K ∈ {C, R, Q, Z, Zk} and E ∈ Mn (K) be an involution. Then tr(E) ∈ h1Ki. Let + A ∈ Ωn be given. Then tr(A) ∈ h1Ki. If K = Zk, then tr(A) ≡ n (mod 2), so that in this case, tr(A) is even whenever n is even. If K 6= Zk, then tr(E) has the form 2p − n, where p is an integer. Hence, tr(E) is even when n is even. It follows that in every case, tr(A) is even whenever n is even.

Theorem 7. Let K ∈ {C, R, Q, Z, Zk}. Let A ∈ Mn (K) be a sum of involutions. Then tr(A) ∈ h1Ki. Moreover, if n is even, then tr(A) is even.

We show that the necessary conditions given in Theorem 7 are also sufficient for a matrix + to be in Ωn . We begin with the following observations.

−1 Corollary 8. Let M ∈ Mn(K) be a sum of k involutions. Then PMP and −M are each a sum of k involutions. Moreover, if N ∈ Mm(K) is also a sum of k involutions, then M ⊕ N is also a sum of k involutions.

(n) Let Eij ∈ Mn (K) have (i, j) entry 1 and 0 elsewhere. When the context is clear, we n P (n) drop the superscript. Let A ∈ Mn(K) be given. Observe that A = aijEij . If i 6= j i,j=1 and if a ∈ K, then (n) (n) (n) (n) aEij = (I + aEij − 2Ejj ) + (2Ejj − I), (2)

+ + is a sum of two involutions. Thus, A ∈ Ωn if and only if diag(a11, ..., ann) ∈ Ωn . 26 Dennis I. Merino, Agnes T. Paras, Diane Christine P. Pelejo

2.1 K = Z or Zk

We take K = Zk or K = Z, and look at Mn (K). For each i = 1, ..., n, observe that 2Eii = I + (2Eii − I) is a sum of two involutions.

Proposition 9. Let n be a positive integer, and let K = Z or Zk. Then for each i = 1, ..., n, we have 2Eii ∈ Mn (K) can be written as a sum of two involutions.

We look at the case when the size of the matrix (n) is odd.

Theorem 10. Let A ∈ M2n+1(K), where K = Z or Zk. Then A is a sum of involutions. (p) Proof. Set p ≡ 2n + 1. Notice that it is sufficient to show that E11 is a sum of involutions (p) (p) as E11 is similar to Eii for all i = 1, . . . , p. Indeed,  !  !  ! (p) M 1 1 M −1 0 M 0 −1 E = [1] + [−1] + [1] , 11 0 −1 1 1 −1 0 n n n

(p) is a sum of three involutions. Thus, aEii is a sum of involutions for each i = 1, ..., p and every a ∈ Z or Zk.

We look at the case when the k in Zk is odd.

Theorem 11. Let A ∈ Mn(Z2m+1). Then A is a sum of involutions.

Proof. Proposition 9 guarantees that for each 1 ≤ i ≤ n, 2Eii is a sum of two involutions. Hence, for any a ∈ Z2m+1, aEii can be written as a sum of involutions as h2i = Z2m+1.

We now consider the case when the size of the matrix is even. If A ∈ M2n(K) is a sum of involutions, then Theorem 7 guarantees that tr(A) is even. We show that the converse is also true when K = Z or K = Zk.

Lemma 12. Let A ∈ Mn(K), where K = Z or Zk. Suppose tr(A) = 0. Then A is a sum of an even number of involutions.

Proof. Let K = Z or Zk. Suppose A ∈ Mn(K) has tr(A) = 0. We show by induction that if a1 + ··· + an = 0, where ai ∈ K, then diag(a1, . . . , an) can be written as a sum of an even number of involutions. For n = 1, we have [0] = p{[1] + [−1]} is a sum of 2p involutions for any positive integer p. For n = 2, 1 0   0 −1 1 1  −1 0 diag(a , −a ) = a + + + 1 1 1 0 −1 −1 0 0 −1 1 1 is a sum of an even number of involutions. Suppose n ≥ 3 and assume that the result holds for every integer l < n. Consider D = diag(a1, . . . , an), where a1 + ··· + an = 0. Set

D1 ≡ diag(a1, −a1) ⊕ 0n−2 and D2 ≡ diag(0, a1 + a2, a3, . . . , an) ≡ [0] ⊕ B, so that D = D1 + D2. Now, diag(a1, −a1) can be written as a sum of 2s involutions and B can be written as a sum of 2t involutions. Also notice that 0n−2 can be written as a sum of 2s involutions, and that 0 can be written as a sum of 2t involutions. Corollary 8 guarantees that D1 is a sum of 2s involutions and D2 is a sum of 2t involutions. The Sum of Involutory Matrices 27

Theorem 13. Let K = Z or Zk and A ∈ M2n(K). Then A is a sum of involutions if and only if tr(A) is even. Proof. The forward implication is done by Theorem 7. We show the converse. Suppose A ∈ M2n(K) has tr(A) = 2l, for some l ∈ Z. Write A = 2lE11 + B, so that tr(B) = 0. Lemma 12 guarantees that B is a sum of involutions. Proposition 9 guarantees that 2E11 is a sum of two involutions, so that 2lE11 is a sum of 2 |l| involutions.

2.2 K = C, R, or Q We prove the converse of Theorem 7 for K = C, R or Q. We follow arguments similar to the ones used in Section 2.1. Let E ∈ Mn (F) be an involution. Then tr(E) = 2k − n for some k ∈ Z. When n is odd, tr(E) is odd so that a sum of an even number of involutions has a trace that is an even integer; a sum of an odd number of involutions has a trace that is an odd integer. When n is even, tr(E) is even so that a sum of (any number of) involutions has a trace that is an even integer.

Lemma 14. Let A ∈ Mn(K), where K = C, R or Q and let tr(A) = 0. Then A can be written as a sum of an even number of involutions.

Proof. Let K = C, R, or Q and let A ∈ Mn(K) be such that tr(A) = 0. Arguing as in the proof of Lemma 12, it is sufficient to show that for any a ∈ K, diag(a, −a) is a sum of an even number of involutions over K. Notice that  a a2   a a2  2 1 − 4 2 4 − 1 diag(a, −a) = a + a 1 − 2 −1 − 2 is a sum of two involutions over K.

We now provide necessary and sufficient conditions so that a matrix can be expressed as a sum of involutions in Mn(K).

Theorem 15. Let A ∈ Mn(K), where K = C, R, or Q. Let tr(A) = k. Then A is a sum of involutions if and only if k ∈ Z and k is even whenever n is even. Proof. Corollary 8 guarantees that we may assume, without loss of generality, that k ≥ 0. Suppose that n is odd. The case n = 1 is trivial. Suppose n ≥ 3. Then,             (n) M 0 1 M 1 −1 M −1 0 E = [1]⊕ + [−1]⊕ + [1] ⊕ , 11  1 0   0 −1   −1 1  n−1 n−1 n−1 2 2 2

(n) so that kE11 is a sum of 3k involutions. Suppose that both k and n are even. Then k k kE(n) = I + (2E − I) 11 2 2 11 is a sum of k involutions. (n) Let A ∈ Mn (K) be given and suppose that tr(A) = k. Write A = kE11 + B, so that tr(B) = 0. Lemma 14 guarantees that B is a sum of involutions. When n is odd, then (n) (n) kE11 is also a sum of involutions. When n is even and k is even, then kE11 is also a sum of involutions. 28 Dennis I. Merino, Agnes T. Paras, Diane Christine P. Pelejo

3 The Sum of Two Involutions

Let Γn denote the set of n-by-n complex matrices which can be written as a sum of two C n + m involutions. Let E ∈ Γ , say E = E1 + E2, with E1, E2 ∈ Γn . Let F ∈ Γ , say C + C F = F1 +F2, with F1, F2 ∈ Γm. Corollary 8 guarantees that A = E1 ⊕F1 and B = E2 ⊕F2 are in Γ+ and E ⊕ F = A + B ∈ Γn+m. n+m C

Theorem 16. Let M = E⊕F , where E ∈ Mn (C) and F ∈ Mm (C) with σ(E)∩σ(−F ) = ∅. Then M ∈ Γn+m if and only if E ∈ Γn and F ∈ Γm. C C C Proof. Let M = E ⊕ F be such that σ(E) ∩ σ(−F ) = ∅. Suppose M = A + B, where A and B are involutions. Partition A and B conformal to M:  E 0   A A   B B  = 11 12 + 11 12 (3) 0 F A21 A22 B21 B22

2 2 2 From the equations (M − B) = A = I and B = I, we get EB12 + B12F = 0. Since σ(E) ∩ σ(−F ) = ∅, we have B12 = 0 [1, Sylvester’s theorem]. Similarly, B21 = 0 because B21E + FB21 = 0. Moreover, equation (3) shows that A12 = 0 and A21 = 0. Because A and B are involutions, A11, A22, B11, and B22 are also involutions.

It is known that every complex matrix is similar to a direct sum of Jordan blocks corre- k L sponding to its eigenvalues. Hence, a given B ∈ Mn (C) is similar to a direct sum Bi of i=1 blocks Bi such that σ(Bi) ⊆ {λi, −λi} and σ(Bj) ∩ σ(Bl) = ∅ whenever j 6= l. We inspect ni each Bi and determine the conditions that guarantee that Bi ∈ Γ . C We look at the case λi = 0. Notice that Bi is nilpotent. We look at a Jordan block  1 1  corresponding to 0. Set A(2) ≡ and notice that A(2) is an involution. 12 0 −1 12

Lemma 17. Let n be a given positive integer. Then J (0) ∈ Γn. n C Proof. If n = 2m is even, then ! ! M (2) M (2) Jn(0) = A12 + [−1] ⊕ A12 ⊕ [1] m m−1 is a sum of two involutions. If n = 2m + 1 is odd, then ! ! M (2) M (2) Jn(0) = A12 ⊕ [1] + [−1] ⊕ A12 m m is a sum of two involutions.

Let A ∈ Mn (C) be nilpotent. Then A is similar to a direct sum of Jordan blocks corresponding to 0. Lemma 17 guarantees that each of these blocks is a sum of two involutions.

Corollary 18. Let A ∈ M ( ) be nilpotent. Then A ∈ Γn. n C C The Sum of Involutory Matrices 29

Next, we consider Bi when λi 6= 0. We look at the cases when λi ∈ {±2} and when λi ∈/ {±2, 0}. Theorem 19. Let A ∈ M ( ). Suppose A ∈ Γn and tr(A) = 2m > 0 for some (positive) n C C integer m. Then 2 is an eigenvalue of A with geometric multiplicity at least m.

Proof. From Proposition 2(4), we may assume without loss of generality that

 I 0   I 0  A = P −1 k P + j (4) 0 −In−k 0 −In−j for some nonnegative integers k, j ≤ n. Note that tr(A) = 2(k+j−n). Suppose k+j−n > 0. We show that A − 2I is singular with nullity at least k + j − n. Notice that

 0 0   2I 0  A = P −1 P + j , 0 −2In−k 0 0 so that  0 0   0 0  A − 2I = P −1 P + . 0 −2In−k 0 −2In−j

Partition P = [Pij], where i, j ∈ {1, 2} and P11 ∈ Mk,j(C). Since P is nonsingular, rank(A− 2I) = rank[P (A − 2I)]. Now,

 0 −2P  P (A − 2I) = 12 . −2P21 −4P22

Since P12 is a k-by-(n − j) matrix, its rank is at most n − j. The rank of the (n − k)-by-n matrix [−2P21 − 4P22] is at most n − k. Thus,

rank (P (A − 2I)) ≤ n − j + n − k = 2n − j − k, which implies that nullity(P (A − 2I)) = n−rank(P (A − 2I)) ≥ n−(2n−j−k) = j+k−n > 0. Hence, P (A−2I) is singular and A−2I is also singular. Therefore, 2 is an eigenvalue of A and the geometric multiplicity of 2 is nullity(A−2I) = nullity(P (A − 2I)) ≥ j +k−n.

If A ∈ Γn is such that tr(A) = −2m, with m a positive integer, then −A is a sum of two C involutions and tr(−A) = 2m. Theorem 19, guarantees that 2 ∈ σ(−A) with geometric multiplicity at least m.

Corollary 20. Let A ∈ M ( ). Suppose that A ∈ Γn and that tr(A) = −2m for some n C C positive integer m. Then −2 is an eigenvalue of A with geometric multiplicity at least m.

We consider the case when a matrix has no eigenvalues equal to ±2 or 0. Note that for any complex matrix M ∈ Mn (C),  1 MI − 1 M 2   1 M −I + 1 M 2  M ⊕ −M = 2 4 + 2 4 ∈ Γ2n. (5) 1 1 C I − 2 M −I − 2 M

Theorem 21. Let A ∈ Mn(C) be such that σ(A) ⊆ {λ, −λ} for some λ ∈ C\{±2, 0}. Then A ∈ Γn if and only if A is similar to −A. C 30 Dennis I. Merino, Agnes T. Paras, Diane Christine P. Pelejo

Proof. Let A ∈ Mn(C) be such that σ(A) = {λ, −λ} for some λ ∈ C \ {±2, 0}. If A is nonsingular and similar to −A, then A is similar to a matrix of the form M ⊕−M for some nonsingular M. Equation (5) shows that M ⊕ −M ∈ Γn. Hence, A ∈ Γn. C C Now, suppose A ∈ Γn. Since ±2 ∈/ σ(A), Theorem 19 and Corollary 20 imply tr(A) = 0. C Hence, it is without loss of generality to assume that A is of the form  I 0   −I 0  A = P −1 k P + k . (6) 0 −In−k 0 In−k Suppose that σ(A) ⊆ {λ, −λ} and that ` is the algebraic multiplicity of λ as an eigenvalue of A. Then, the algebraic multiplicity of −λ as an eigenvalue of A is n − `. But since tr(A) = `λ − (n − `)λ = 0 and since λ 6= 0, we have n = 2`. In equation 6, partition −1 P = [Pij] and P = [Qij], where i, j ∈ {1, 2} and P11,Q11 ∈ Mk(C). Then  −Q P −Q P  A = 2 12 21 12 22 . (7) Q21P11 Q21P12 Multiplying equation (7) by P −1 on the right and multiplying equation (7) by P on the left, we get  0 −2Q   0 2P  AP −1 = 12 and PA = 12 . 2Q21 0 −2P21 0

Since A and P are nonsingular, we have that P12,P21,Q12,Q21 must be nonsingular and square. Thus k = `. Also,  Q P 0  A2 = AP −1PA = 4 12 21 . 0 Q21P12

2 Since ±2 ∈/ σ(A), 4 ∈/ σ(A ) and so (Q12P21 −I) and (Q21P12 −I) are nonsingular. Because PQ = QP = I, we have Q12P21(Q12P21 − I) = Q12P21(−Q11P11) = Q12(−P21Q11)P11 = Q12P22Q21P11. Since the left hand side is nonsingular, every factor on the right hand side is nonsingular. In particular, P11 and P22 are nonsingular. Define  0 −Q P  R = 12 22 , −Q21P11 0 and notice that R is nonsingular. Now,  −1 −1  −1 Q12P22Q21P12P22 Q12 Q12P22 RAR = 2 −1 −1 . −Q21P11 −Q21P11Q12P21P11 Q21

−1 −1 −1 −1 Moreover, Q12 (P22Q21) P12P22 Q12 = −Q12P21Q11P12P22 Q12 = Q12P21 since Q11P12 = −1 −1 −Q12P22. Similarly, Q21P11Q12P21P11 Q21 = Q21P12. Thus,  Q P Q P  RAR−1 = 2 12 21 12 22 = −A, −Q21P11 −Q21P12 and A is similar to −A.

k L Suppose A ∈ Mn(C) is similar to a direct sum of blocks Bi such that σ(Bi) ⊆ i=1 {λi, −λi} for some λi ∈ C \{0, ±2} and σ(Bi) ∩ σ(Bj) = ∅ whenever i 6= j. Theorem 16 guarantees that A ∈ Γn if and only if each B is a sum of two involutions. Theorem 21 C i ni n ensures that for each i, we have Bi ∈ Γ if and only if Bi is similar to −Bi. Thus A ∈ Γ C C if and only if for each i, we can find Ci such that Bi = Ci ⊕ −Ci. The Sum of Involutory Matrices 31

Corollary 22. Let A ∈ M ( ) be nonsingular such that 0, ±2 ∈/ σ(A). Then A ∈ Γn if and n C C only if A is similar to C ⊕ −C for some nonsingular matrix C.

We now deal with matrices whose spectrum contains 2 and −2. 2k−1 Lemma 23. Jk(2) ⊕ Jk−1(−2) ∈ Γ for any k ≥ 1. Consequently, Jk(−2) ⊕ Jk−1(2) ∈ C Γ2k−1. C Proof. Define C = [cij] ∈ Mk,k−1(C) and D = [dij] ∈ Mk−1,k (C) as follows:  (−1)i, if j = i   (−1)j, if j = i + 1 c = 1 (−1)i+1, if j = i + 1 d = . ij 4 ij 0, otherwise  0, otherwise One checks that  1   1  2 Jk(2) C 2 Jk(2) −C U ≡ 1 and V ≡ 1 D 2 Jk−1(−2) −D 2 Jk−1(−2) are involutions. Furthermore, Jk(2) ⊕ Jk−1(−2) = U + V is a sum of two involutions. Corollary 8 now guarantees that Jk(−2) ⊕ Jk−1(2) is also a sum of two involutions since it is similar to − (Jk(2) ⊕ Jk−1(−2)).

Theorem 24. For any C ∈ Mk,m(C) and D ∈ Mm,k(C) such that CD is nilpotent, the k+m matrix (2Ik + CD) ⊕ (−2Im + DC) ∈ Γ . C Proof. Let C ∈ Mk,m(C) and D ∈ Mm,k(C) be such that CD is nilpotent. Suppose that the sizes of the Jordan blocks of CD corresponding to zero are k1, k2, . . . , kj, with ki ≥ ki+1 for all i. Notice that DC is also nilpotent [8, Theorem 1]. Moreover, if the sizes of the Jordan blocks of DC corresponding to zero are l1, l2, . . . , lj such that li ≥ li+1 for all i, then |ki − li| ≤ 1. Let

A ≡ (2Ik + CD) ⊕ (−2Im + DC) ∈ Mk+m(C) (8) p p L L which is similar to Jki (2) ⊕ Jii (−2), for some nonnegative integer p. By Lemma 23 k=1 i=1 k+m and equation 5, Jk (2) ⊕ Jl (−2) is a sum of two involutions for all i. Thus, A ∈ Γ . i i C We prove that the converse of Theorem 24, that is, a matrix whose spectrum contains only 2 or −2 must be of the form in equation (8). The following is in [4, Problem 37 on pages 488 – 489.]. + 2 Lemma 25. If X ∈ Mn(C) is such that σ(X) ⊆ R , then X is a polynomial in X .

Proof. We show that if A ∈ Mn(C) has a positive spectrum, then A has a unique square root with positive spectrum. Furthermore, this unique square root, say B, is a polynomial in A, that is, there is a polynomial p(x) such that p(A) = B. The proof of the existence of such a B is in [4, Chapter 6]. To show uniqueness, assume + 2 2 + 2 that C ∈ Mn(C) is such that σ(C) ⊆ R and C = A = B . Now, if λ ∈ R , then (Jk (λ)) 2 −1 is similar to Jk λ . Hence, C and B are similar, say, C = SBS for some nonsingular S. Moreover, B = p(A) = p(C2) = p(SB2S−1) = Sp(B2)S−1 = SBS−1. Thus, BS = SB and C = SBS−1 = BSS−1 = B. 32 Dennis I. Merino, Agnes T. Paras, Diane Christine P. Pelejo

Theorem 26. Let A ∈ M ( ) be such that σ(A) = {2, −2}. If A ∈ Γn, then A is similar n C C to a matrix of the form (2Ik + CD) ⊕ (−2In−k + DC) for some nonnegative k, and matrices T C,D ∈ Mk,n−k(C) such that CD is nilpotent. −1 Proof. Suppose A = P (Ik ⊕−In−k)P +(Ij ⊕−In−j) and assume without loss of generality that k ≥ j. Partition P = [Pij], where i, j ∈ {1, 2, 3}, P11 ∈ Mj(C) and P22 ∈ Mk−j(C). Then   2P11 0 0 PA =  2P21 0 0  . 0 −2P32 −2P33 T T T If σ(A) = {2, −2}, then A is nonsingular. So [2P11 2P21] and [−2P32 − 2P33] should be nonsingular, hence square. Then k = j. Thus,

−1 A = P (Ik ⊕ −In−k)P + (Ik ⊕ −In−k).

−1 Partition P = [Pij] and P = [Qij] where i, j ∈ {1, 2} and P11,Q11 ∈ Mk(C). Then,

 2P 0   2Q 0  PA = 11 , AP −1 = 11 0 −2P22 0 −2Q22  2Q P −2Q P  and A = 11 11 12 22 . (9) 2Q21P11 −2Q22P22 Now,  Q P 0  A2 = AP −1PA = 4 11 11 . 0 Q22P22

Since σ(A) = {2, −2}, we have that σ(Q11P11) = {1} = σ(Q22P22). Thus, I − Q11P11 and −1 −1 I − Q22P22 are nilpotent matrices. Since P P = I = PP ,

−1 I − Q11P11 = Q12P21 and P22(I − Q22P22)P22 = I − P22Q22 = P21Q12.

1 We now claim that 2 A is similar to  Q P 0  11 11 , 0 −Q22P22 from which the desired conclusion follows. Note that  −1 −1 −1 −1      1 P11 Q11 −P11 Q11 Q12 I −Q12 Q11P11 −Q12P22 A = −1 · · , 2 0 −P22 P21 −I 0 −P22   1 I −Q12 that is, 2 A is similar to . Consider the matrix P21 −I

 −1  SQ12T (I + P22Q22) K = −1 , P21S(I + Q11P11) T where S is a nonsingular k-by-k matrix such that

−1 2 S (Q11P11)S = (Q11P11) (10) and T is a nonsingular (n − k)-by-(n − k) matrix such that

−1 2 T (P22Q22)T = (P22Q22) . (11) The Sum of Involutory Matrices 33

The existence of such S and T are guaranteed by the fact that σ(Q11P11) = {1} = σ(P22Q22). By equations 10, 11 and the equation [Pij][Qij] = I,

 I −Q   Q P 0  12 · K = K · 11 11 . P21 −I 0 −Q22P22

If K is nonsingular, the proof is done. Indeed, K = ΛΣΓ, where

 IQ T (I + P Q )−1T −1  Λ = 12 22 22 0 I

 I − Q T (I + P Q )−1T −1P S(I + Q P )−1S−1 0  Σ = 12 22 22 21 11 11 0 I  S 0  Γ = −1 P21S(I + Q11P11) T Clearly, Λ and Γ are nonsingular. We prove that Σ is also nonsingular by showing that −1 −1 −1 −1 Q12T (I + P22Q22) T P21S(I + Q11P11) S is nilpotent. By Lemma 25, there exist 2 2 polynomials p(x) and q(x) such that Q11P11 = p (Q11P11) and P22Q22 = q (P22Q22) . This means that p(1) = q(1) = 1. From equations (10) and (11), we have

2 Q11P11S = S(Q11P11) (12) 2 P22Q22T = T (P22Q22) (13)

Also, since I − P22Q22 = P21Q12 and I − Q11P11 = Q12P21,

Q12P22Q22 = Q11P11Q12 (14)

Q12P21Q11P11 = Q11P11Q12P21 (15)

Because of equations (14) and (15), we have that for any polynomial f(x),

Q12f(P22Q22) = f(Q11P11)Q12 (16) and f(Q11P11)Q12P21 = Q12P21f(Q11P11). (17) Since q(1) + 1 = 2 6= 0, we can find a polynomial f(x) that interpolates (1 + q(x))−1 and its derivatives at the roots of pQ11P11 (x) = 0, where pQ11P11 is the characteristic polynomial of Q11P11. Thus

−1 −1 −1 −1 Q12T (I + P22Q22) T = Q12[T (I + P22Q22)T ] −1 −1 = Q12(I + TP22Q22T ) 2 −1 −1 = Q12(I + T q((P22Q22) )T ) . (18)

Using equation (13), the right hand side of the above equation is equal to

−1 −1 −1 Q12(I + q(P22Q22)TT ) = Q12(I + q(P22Q22)) .

We now use equations (16) and (17):

−1 −1 Q12(I + q(P22Q22)) = Q12f(P22Q22) = f(P11Q11)Q12 = (I + q(Q11P11)) Q12. 34 Dennis I. Merino, Agnes T. Paras, Diane Christine P. Pelejo

Similarly, we use equation (12):

−1 −1 −1 −1 P21S(I + Q11P11) S = P21(I + SQ11P11S ) 2 −1 −1 = P21(I + Sp((Q11P11) )S ) (19) −1 = P21(I + p(Q11P11)) . (20) Hence, we get −1 −1 −1 −1 Q12T (I + P22Q22) T P21S(I + Q11P11) S −1 −1 = (I + q(Q11P11)) Q12P21(I + p(Q11P11)) . (21) −1 −1 Note that (I + q(Q11P11)) and (I + p(Q11P11)) are polynomials in Q11P11, and that they commute with Q12P21. Since Q12P21 is nilpotent, the matrix in equation (21) is also nilpotent and therefore, Σ is nonsingular. The following is a consequence of Lemma 17 and Theorems 21, 26 and 16. Theorem 27. Let A ∈ M ( ). Then A ∈ Γn if and only if A is similar to n C C

[B ⊕ −B ⊕ N ⊕ (2Ik + CD) ⊕ (−2Im + DC)] (22) for some nonsingular matrix B; N; and C ∈ Mk,m(C) and D ∈ Mm,k(C) such that CD (and consequently DC) is nilpotent. Theorem 27 characterizes all complex matrices that can be written as a sum of two involutions.

References

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