LINEAR MAPS THAT PRESERVE COMMUTING PAIRS OF MATRICES OVER GENERAL BOOLEAN ALGEBRA
Seok-Zun Song and Kyung-Tae Kang
Abstract. We consider the set of commuting pairs of matrices and their preservers over binary Boolean algebra, chain semiring and general Boolean algebra. We characterize those linear operators that preserve the set of commuting pairs of matrices over a general Boolean algebra and a chain semiring.
1. Introduction
One of the most active and fertile subjects in matrix theory during the past one hundred years is the linear preserver problem, which con- cerns the characterization of linear operators on matrix spaces that leave certain functions, subsets, relations, etc., invariant. Although the linear operators concerned are mostly linear operators on matrix spaces over some fields or rings, the same problem has been extended to matrices over various semirings ([1]-[8]).
Let M
n(S) denote the set of n × n matrices over a nonempty set S.
A set of commuting pairs of matrices, C, is the set of (unordered) pairs of matrices (A, B) such that AB = BA. A linear operator T on M
n(S) is said to preserve C (or T preserves commuting pairs of matrices) if (T (A), T (B)) ∈ C for all (A, B) ∈ C.
Commutativity of matrices play a central role in the theory of matri- ces. There are many papers on the linear transformations that preserve the set of commuting pairs of matrices over various algebraic structure (see [1]-[3], [5]-[6], [8]). Song and Beasley [6] obtained characterizations of linear operators that preserve commuting pairs of matrices over the
Received October 1, 2004.
2000 Mathematics Subject Classification: 15A03, 15A04, 15A33.
Key words and phrases: linear operator, constituent, canonical form, commuting pair of matrices.
This work was supported by a grant from Chuongbong Academic Research Fund
of the Cheju National University Development Foundation.
nonnegative reals, R
+. Watkins [8] showed that if n ≥ 4 and S is an algebraically closed field of characteristic 0, and T is a nonsingular lin- ear operator on M
n(S) which preserves commuting pairs of matrices, then there exist an invertible matrix U , a nonzero scalar α and a linear functional f : M
n(S) → S such that either
(1) T (X) = αU XU
−1+ f (X)I
nfor all X ∈ M
n(S), or (2) T (X) = αU X
tU
−1+ f (X)I
nfor all X ∈ M
n(S).
Beasley [1] extended this to the case n = 3. The real symmetric and complex Hermitian cases were investigated by Chan and Lim [2].
Further extensions to more general fields were obtained by Radjavi [5]
and Choi, Jafarian and Radjavi [3].
In this paper, we study the semigroup of linear operators that pre- serve commuting pairs of matrices over a Boolean algebra and a chain semiring. In Theorem 3.3 we show that it is generated by transpositions and similarity operators over the binary Boolean algebra and a chain semiring. Also in Theorem 4.3 we extend these results to the general Boolean case and obtain an linear operator which is not generated by transpositions and similarity operators.
2. Preliminaries
A semiring is a nonempty set S on which operations of addition(+) and multiplication(·) have been defined such that the following condi- tions are satisfied :
(a) (S, +) is a commutative monoid with identity element 0;
(b) (S, ·) is a monoid with identity element 1;
(c) multiplication distributes over addition from either side;
(d) s0 = 0 = 0s for all s ∈ S.
For a fixed positive integer k, let B
kbe the Boolean algebra of subsets of a k-element set S
kand σ
1, σ
2, . . . , σ
kdenote the singleton subsets of S
k. Union is denoted by + and intersection by juxtaposition; 0 denotes the null set and 1 the set S
k. Under these two operations, B
kis a com- mutative antinegative semiring(that is, only 0 has an additive inverse);
all of its elements, except 0 and 1, are zero-divisors. In particular, if k = 1, B
1is called the binary Boolean algebra.
Let K be any set of two or more elements. If K is totally ordered by <
(i.e., x < y or y < x for all distinct elements x, y in K), then define x + y
as max(x, y) and xy as min(x, y) for all x, y ∈ K. If K has a universal
lower bound and a universal upper bound, then K becomes a semiring, and called a chain semiring. The following are interesting examples of a chain semiring.
Let H be any nonempty family of sets nested by inclusion, 0 = ∩
x∈Hx, and 1 = ∪
x∈Hx. Then S = H ∪ {0, 1} is a chain semiring.
Let α, w be real numbers with α < w. Define S = {β ∈ R : α ≤ β ≤ w}. Then S is a chain semiring with α = 0 and w = 1. It is isomorphic to the chain semiring in the previous example with H = {[α, β] : α ≤ β ≤ w}. Furthermore, if we choose the real numbers 0 and 1 as α and w in the previous example, then m × n matrices over F ≡ {β : 0 ≤ β ≤ 1}
is called fuzzy matrices.
In particular, if we take H to be a singleton set, say {a}, and denote
∅ by 0 and {a} by 1, the resulting chain semiring becomes the binary Boolean algebra B = {0, 1}, and it is a subsemiring of every chain semir- ing. Since any general Boolean algebra B
k(k ≥ 2) is not totally ordered under inclusion, it is not a chain semiring.
Hereafter, unless otherwise specified, S denote an arbitrary semiring, and B
ka general Boolean algebra, and K a chain semiring.
Let M
n(S) denote the set of all n × n matrices with entries in S.
Then algebraic operations on M
n(S) are defined as if S were a field.
The identity matrix and the zero matrix are denoted by I
nand 0
n. Since M
n(S) is a semiring, we can consider the invertible members of its multiplicative monoid. A matrix A ∈ M
n(S) is said to be invertible if there exists a matrix B ∈ M
n(S) such that AB = BA = I
n. It is well known [4] that the permutation matrices are the only invertible matrices on M
n(B
1).
For any matrix A = [a
ij] ∈ M
n(B
k), the l
thconstituent, A
l, of A is the n × n binary Boolean matrix whose (i, j)-th entry is 1 if and only if a
ij⊇ σ
l. Via the constituents, A can be written uniquely as
A = X
k l=1σ
lA
l,
which is called the canonical form of A (see [4]).
It follows from the uniqueness of the decomposition and the fact that the singletons are mutually orthogonal idempotents that for all matrices A, B ∈ M
n(B
k) and all α ∈ B
k,
(2.1) (AB)
l= A
lB
l;
(2.2) (A + B)
l= A
l+ B
l;
(2.3) (αA)
l= α
lA
l.
for all 1 ≤ l ≤ k. Thus each l
thconstituent operator, T
l, is a homomor- phism of M
n(B
k) onto M
n(B
1), and preserves the products of matrices in M
n(B
k).
Lemma 2.1. For any matrix A ∈ M
n(B
k) with k ≥ 1, A is invertible if and only if all its constituents are permutation matrices. In particular, if A is invertible, then A
−1= A
t, where A
tdenote the transpose of A.
Proof. If A is an invertible matrix in M
n(B
k), then there exists a matrix B ∈ M
n(B
k) such that AB = I
n. The equality (2.1) implies that (AB)
l= A
lB
l= I
nfor all l = 1, . . . , k. It follows that all constituents of A are permutation matrices. Conversely, assume that each l
thcon- stituent, A
l, of A is a permutation matrix. Then we have A
lA
tl= I
nfor all l = 1, . . . , k, and hence
AA
t= Ã X
kl=1
σ
lA
l!Ã X
kl=1
σ
lA
l!
t= X
k l=1σ
lA
lA
tl= X
kl=1
σ
lI
n= I
n. Therefore A is invertible.
Lemma 2.1 shows that all permutation matrices are only invert- ible matrices on M
n(B
1), while there exists an invertible matrix on M
n(B
k)(k ≥ 2) which is not a permutation. For example, consider
A =
σ
1σ
2σ
3σ
2σ
3σ
1σ
3σ
1σ
2
∈ M
3(B
3).
Then AA
t= I
3, but A is not a permutation matrix in M
3(B
3).
For each x ∈ K, define x
∗=
½ 0 if x = 0, 1 if x 6= 0.
Then the mapping x → x
∗is a homomorphism of K onto B
1= {0, 1}.
Its entrywise extension to a mapping A → A
∗of M
n(K) onto M
n(B
1) preserves matrix sums and products and multiplication by scalars.
Lemma 2.2. The permutation matrices are the only invertible mem- bers of M
n(K).
Proof. Let A be an invertible matrix in M
n(K). Then there exists a
matrix B ∈ M
n(K) such that AB = I
n. This implies A
∗B
∗= I
n, and
thus A
∗and B
∗are permutation matrices with B
∗= A
∗t. Since any
product of two element in K is their minimum, the nonzero entries in A
are 1’s. Therefore A is a permutation matrix.
3. The binary Boolean case
In this section we obtain characterizations of the linear operators that preserve the set of commuting pairs of matrices over the binary Boolean algebra B
1= {0, 1} and a chain semiring K.
The n×n matrix all of whose entries are zero except its (i, j)-th, which is 1, is denoted E
ij. We call E
ija cell. Let E
n= {E
ij| i, j = 1, . . . , n}
denote the set of all cells. If A = [a
ij] and B = [b
ij] are matrices in M
n(S), we shall use the notation A ≤ B(or B ≥ A) if b
ij= 0 implies a
ij= 0 for all i, j. This provides a reflexive and transitive relation on M
n(S). If A and B are matrices in M
n(S) with A ≥ B, it follows from the linearity of T that T (A) ≥ T (B) for any linear operator T on M
n(S).
Lemma 3.1. Let S = B
1or S = K. Then for a linear operator T on M
n(S), T is invertible if and only if T permutes E
n.
Proof. Suppose that T is invertible on M
n(S). Let E
ijbe any element in E
n. By invertibility of T , there exists at least one cell E
rsin E
nsuch that T (E
ij) ≥ E
rs. Thus we have E
ij≥ T
−1(E
rs) because T
−1is also linear. This implies that T
−1(E
rs) = αE
ijfor some nonzero scalar α ∈ S, equivalently E
rs= αT (E
ij).
If S = B
1, we have α = 1 so that T (E
ij) = E
rs. If S = K, the (r, s)-th entry of E
rsis 1, while that of αT (E
ij) is α. Thus we have α = 1, and hence T (E
ij) = E
rs.
Since E
ijis an arbitrary cell, T permutes E
n. The converse is imme- diate.
For S = B
1or S = K, let T be a linear operator on M
n(S) that preserves commuting pairs of matrices. Then the following example shows that T may be not invertible.
Example 3.2. For S = B
1, let T be an operator on M
n(S) defined by
T (X) = X + I
nfor all X ∈ M
n(S). Then we can easily show that T is linear and preserves commuting pairs of matrices. But Lemma 3.1 implies that T is not invertible.
Theorem 3.3. For S = B
1or S = K, let T be a linear operator on M
n(S). Then T is an invertible linear operator that preserves commuting pairs of matrices if and only if there exists a permutation matrix P such that either
(a) T (X) = P XP
tfor all X ∈ M
n(S), or
(b) T (X) = P X
tP
tfor all X ∈ M
n(S).
Proof. Let T be an invertible linear operator on M
n(S) which pre- serves commuting pairs of matrices. Note that if AX = XA for all X ∈ M
n(S), then we have A = 0
nor I
n. Thus we have T (I
n) = I
nbecause T is invertible.
By Lemma 3.1, T permutes E
n. It follows from T (I
n) = I
nthat there is a permutation σ of {1, . . . , n} such that T (E
ii) = E
σ(i)σ(i)for each i = 1, . . . , n. Define an operator L on M
n(S) by
L(X) = P
tT (X)P
for all X ∈ M
n(S), where P is the permutation matrix corresponding to σ so that L(E
ii) = E
iifor each i = 1, . . . , n. Then we can easily show that L is an invertible linear operator on M
n(S) which preserves commuting pairs of matrices. By Lemma 3.1, L permutes E
n. Therefore for any cell E
rsin E
n, there exists exactly one cell E
pqin E
nsuch that L(E
rs) = E
pq.
Suppose that r 6= s. Since L is injective, we have p 6= q because L(E
ii) = E
iifor each i = 1, . . . , n. Assume that p 6= r and p 6= s. Then
E
rs(E
rr+ E
ss+ E
pp) = (E
rr+ E
ss+ E
pp)E
rs, so that
L(E
rs)L(E
rr+ E
ss+ E
pp) = L(E
rr+ E
ss+ E
pp)L(E
rs), equivalently
E
pq(E
rr+ E
ss+ E
pp) = (E
rr+ E
ss+ E
pp)E
pq. It follows that q = r or q = s. Since
E
rs(E
rr+ E
ss) = (E
rr+ E
ss)E
rs, we have
L(E
rs)L(E
rr+ E
ss) = L(E
rr+ E
ss)L(E
rs), or equivalently
E
pq(E
rr+ E
ss) = (E
rr+ E
ss)E
pq. Since q = r or q = s, we have
E
pq(E
rr+ E
ss) = E
pror E
ps,
but (E
rr+ E
ss)E
pq= 0, a contradiction. Hence we have p = r or p = s.
Similarly we obtain q = r or q = s. Therefore, for each E
rs∈ E
n, we have
(3.1) L(E
rs) = E
rsor L(E
rs) = E
sr.
Suppose that L(E
rs) = E
rswith r 6= s and L(E
rt) = E
trfor some t 6= r, s. It follows from (3.1) and invertibility of L that L(E
st+ E
ts) = E
st+ E
ts. Let A = E
rr+ E
st+ E
tsso that L(A) = E
rr+ E
st+ E
ts. Then
(E
rs+ E
rt)A = A(E
rs+ E
rt), and hence
L(E
rs+ E
rt)L(A) = L(A)L(E
rs+ E
rt).
But L(E
rs+ E
rt)L(A) = E
rt+ E
tr, while L(A)L(E
rs+ E
rt) = E
rs+ E
sr. Thus we have t = s, a contradiction. It follows that if L(E
ij) = E
ijfor some E
ij∈ E
nwith i 6= j, then we have L(E
rs) = E
rsfor all E
rs∈ E
n. Similarly, if L(E
ij) = E
jifor some E
ij∈ E
nwith i 6= j, then we have L(E
rs) = E
srfor all E
rs∈ E
n. Consequently, we have established that L(X) = X or L(X) = X
tfor all X ∈ M
n(S).
Let L(X) = X for all X ∈ M
n(S). By the definition of L, we have P
tT (X)P = X,
or equivalently
T (X) = P XP
tfor all X ∈ M
n(S). Similarly, if L(X) = X
t, we obtain that T (X) = P X
tP
tfor all X ∈ M
n(S).
The converse is immediate.
4. The general Boolean case
In this section, we extend the results of the binary Boolean case to those of general Boolean case. Also we obtain another linear operator that preserves commuting pairs of matrices over a general Boolean alge- bra, which is neither a transposition operator nor a similarity operator.
If T is a linear operator on M
n(B
k) with k ≥ 1, for each 1 ≤ l ≤ k define its l
thconstituent operator, T
l, by
T
l(B) = (T (B))
lfor all B ∈ M
n(B
1). By the linearity of T , we have T (A) =
X
k l=1σ
lT
l(A
l) for any matrix A ∈ M
n(B
k).
Lemma 4.1. If T is an invertible linear operator on M
n(B
k) with
k ≥ 1, then each l
thconstituent operator, T
l, commutes E
n.
Proof. It follows from Lemma 3.1 and the definition of a constituent operator.
For any fixed invertible matrix U in M
n(S), the operator A → U AU
tis called a similarity operator. We can easily show that any similarity operator on M
n(S) is an invertible linear operator and preserves com- muting pairs of matrices. Also, we can restate Theorem 3.3 as follows:
the semigroup of linear operators that preserve commuting pairs of ma- trices over B
1is generated by transpositions and similarity operators.
But for general Boolean algebra B
kwith k ≥ 2, the following exam- ple shows that there exists another invertible linear operator preserving commuting pairs of matrices which is neither a transposition operator nor a similarity operator.
Example 4.2. Let U =
· σ
1σ
2σ
2σ
1¸
∈ M
2(B
2).
By Lemma 2.1, U is an invertible matrix in M
2(B
2) with U
−1= U
t. Define an operator T on M
2(B
2) by
T (X) = U (σ
1X
1+ σ
2X
2t)U
tfor all X = P
2l=1
σ
lX
l∈ M
2(B
2). Then we can easily show that T is a linear operator on M
2(B
2) which is neither a transposition nor a similarity.
Let T (X) = T (Y ). Since U is invertible, we have σ
1X
1+ σ
2X
2t= σ
1Y
1+ σ
2Y
2t,
and hence X
l= Y
lfor each l = 1, 2. By the uniqueness of canonical form of a matrix, X = Y and thus T is injective.
Let Y = P
2l=1
σ
lY
lbe any matrix in M
2(B
2). Then we can take the matrix X = σ
1Y
1+ σ
2Y
2t∈ M
2(B
2), so that T (X) = Y . This implies that T is surjective. Therefore T is invertible. It follows from the canon- ical form of a matrix in M
n(B
k) that T preserves commuting pairs of matrices.
Theorem 4.3. Let T be a linear operator on M
n(B
k) with k ≥ 1.
Then T is an invertible linear operator preserving commuting pairs of
matrices if and only if there exists an invertible matrix U in M
n(B
k)
such that
T (X) = U
³ X
kl=1
σ
lY
l´ U
tfor all X ∈ M
n(B
k), where Y
l= X
lor Y
l= X
ltfor each l = 1, . . . , k.
Proof. Assume that T is an invertible linear operator on M
n(B
k) pre- serving commuting pairs of matrices. Then we can easily show that all its constituent operators, T
l, are invertible linear operators on M
n(B
1) and preserve commuting pairs of matrices for each l = 1, . . . , k.
Let X = P
kl=1
σ
lX
lbe any matrix in M
n(B
k). Then we have T (X) = P
kl=1
σ
lT
l(X
l). By Theorem 3.3, each l
thconstituent operator, T
l, has the form
T
l(X
l) = P
lX
lP
ltor T
l(X
l) = P
lX
ltP
lt,
where each P
lis a permutation matrix for all l = 1, . . . , k. Thus we have T (X) =
X
k l=1σ
lP
lY
lP
lt,
where Y
l= X
lor Y
l= X
ltfor each l = 1, . . . , k, equivalently T (X) =
³ X
kl=1
σ
lP
l´³ X
kl=1
σ
lY
l´³ X
kl=1
σ
lP
l´
t.
If we let U =
³ P
kl=1