Also available at http://amc.imfm.si
ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.)
ARS MATHEMATICA CONTEMPORANEA 6 (2013) 247–252
A note on homomorphisms of matrix semigroup
Matjaž Omladič
Faculty of Mathematics and Physics, University of Ljubljana
Jadranska 19, SI-1000, Ljubljana Slovenia
Bojan Kuzma
University of Primorska, FAMNIT, Glagoljaška 8, SI-6000 Koper, Slovenia
and
Institute of Mathematics, Physics and Mechanics, Department of Mathematics,
Jadranska 19, SI-1000 Ljubljana, Slovenia
Received 12 August 2011, accepted 5 June 2012, published online 28 October 2012
Abstract
Let F be a field. We classify multiplicative maps from Mn(F) to M(nk)(F) which
annihilate a zero matrix and map rank-k matrix into a rank-one matrix.
Keywords: Matrix semigroup, Homomorphism, Representation.
Math. Subj. Class.: 20M15, 15A33, 20G05
1 Introduction and preliminaries
Let Mn(F) denote the semigroup of all n–by–n matrices with coefficients in a field F,
let Eij be its matrix units, and let Id = Idn :=
∑
Eii be its identity. In [5], Jodeit and
Lam classified nondegenerate semigroup homomorphisms π :Mn(F)→Mn(F), that is,
maps which are (i) multiplicative π(AB) = π(A)π(B) and (ii) their restriction on singular
matrices is nonconstant. It was shown that the semigroup of such maps is generated by
three simple types: (i) a similarity, (ii) a fixed field homomorphism applied entry-wise on
a matrix, and (iii) the map which sends A to a matrix of its cofactors. We refer below for
more precise definitions.
The complete classification of degenerate maps on Mn(F) is more involved. They
are all of the type A 7→ π1(A) ⊕ Idn−m for some integer m ∈ {0, . . . , n} and some
degenerate multiplicative π1 : Mn(F) → Mm(F) with π1(0) = 0 [5]. When m = 1,
Ðoković [2, Theorem 1] proved the following.
E-mail addresses: matjaz.omladic@fmf.uni-lj.si (Matjaž Omladič), bojan.kuzma@famnit.upr.si (Bojan
Kuzma)
Copyright c© 2013 DMFA Slovenije
248 Ars Math. Contemp. 6 (2013) 247–252
Lemma 1.1. Let F be a field, and n ≥ 2. If π :Mn(F) → F is multiplicative, then there
exists multiplicative φ : F→ F so that π(X) = φ(detX).
When m < n and the characteristic of F differs from 2, Ðoković [2, Theorem 2]
also showed π1 factors through determinant so that π1 = f ◦ det for some multiplicative
f : F → Mm(F). The classification of those seems to be difficult, and as far as we
know they are known only in case F = C is the filed of complex numbers, by the work
of Omladič, Radjavi, and Šemrl [8]. Later, Guralnick, Li, and Rodman [4], extended the
result of Ðoković to include also the case n = m.
Semigroup homomorphisms mapping into higher dimensional algebras are less known.
Kokol-Bukovšek [6, 7] classified them in case they are nondegenerate and map 2–by–2
matrices into 3–by–3 or into 4–by–4. Under additional assumption that a degenerate ho-
momorphism is a polynomial in matrix entries, the classification is well-known, see a book
by Weyl [9] (see also Fulton and Harris [3] for holomorphic homomorphisms over a field
of complex numbers).
It is our aim to show that all homomorphisms from n–by–n matrices to
(
n
k
)
–by–
(
n
k
)
matrices which map a rank-k matrix into a rank-one come from exterior product. Both
assumptions on the dimension of the target space as well as on the rank of the matrices
are essential; otherwise there are many more maps as we show in Remark 1.4 below. We
remark that the main idea, that rank-k idempotents are mapped into rank-1 idempotents, is
essentially due to Jodeit and Lam [5].
To be self-contained, we briefly repeat the basics about exterior products. Let e1, . . . ,
en be the standard basis of column vectors in Fn. Given a linear operator X on Fn, denote
by
∧k
(X) the k-th exterior product of X , acting on
∧k
(Fn), i.e., a k-th exterior product
of Fn. Recall [3] that, as a vector space,
∧k
(Fn) has a basis consisting of
(
n
k
)
elements
{ei1 ∧· · ·∧eik ; 1 ≤ i1 < i2 < · · · < ik ≤ n}, where x∧y = −y∧x and x∧x = 0 is the
alternating tensor. Then by definition,
∧k
(X) : ei1 ∧ · · · ∧ eik 7→ (Xei1) ∧ · · · ∧ (Xeik).
It follows easily that
∧k
(AB) =
∧k
(A)
∧k
(B). Also, in lexicographic order of a basis(
ei1 ∧ · · · ∧ eik
)
1≤i1<···<ik≤n
, the matrix of
∧k
(X) equals the
(
n
k
)
–by–
(
n
k
)
matrix of all
k–by–k minors ofX , where the element at position corresponding to (ei1∧· · ·∧eik , ej1∧
· · ·∧ejk) is the minor obtained by taking columns i1, . . . , ik and rows j1, . . . , jk of a matrix
X . In particular,
∧n
(X) = detX and
∧n−1
(X) is similar to a matrix of cofactors under
similarity S =
∑n
i=1(−1)i+1Ei (n−i+1).
Besides the (n − 1)-st exterior product there are at least two additional multiplicative
maps from Mn(F) to itself. One is an inner automorphism X 7→ SXS−1 where S ∈
Mn(F) is fixed, invertible. The other is induced by a field homomorphism φ : F→ F (i.e.
an additive multiplicative map) applied entry-wise, that is, it maps a matrix
∑
xijEij into∑
φ(xij)Eij . With a slight abuse of notation, we denote this map again by φ : X 7→ φ(X).
Theorem 1.2. Let F be a field, let n ≥ 2 be an integer, and let m =
(
n
k
)
for some integer
k = 1, . . . , n. If π : Mn(F) → Mm(F), π(0) = 0, is a multiplicative map such that
rk(π(A0)) = 1 for some matrix A0 of rank-k, then
π(X) = Sφ(
∧k
(X))S−1
where φ : F→ F is a multiplicative map and S ∈Mm(F) is invertible.
Moreover, if k < n then φ is also additive, hence a field embedding.
M. Omladič and B. Kuzma: A note on homomorphisms of matrix semigroup 249
Remark 1.3. Without the assumption π(0) = 0, there are more possibilities. Say, π(A) =
1⊕Sym2(
∧n−1
A)⊕ f(detA), where Sym2 is the second symmetric power (see [3]) and
f : F→Mm−1−(n+12 )(F) is multiplicative.
However, we remark that to classify multiplicative maps π it suffices to assume π(0) =
0. In fact, if π : Mn(F) → Mm(F) is multiplicative, then P := π(0) is an idempotent,
and from π(X)P = π(X0) = π(0) = P = π(0X) = Pπ(X) we deduce that, relative to
decomposition Fm = KerP ⊕ ImP we have
π(X) = π1(X)⊕ Idr,
where r := rkP and π1 :Mn(F)→Mm−r(F) is multiplicative with π1(0) = 0.
Remark 1.4. If m 6=
(
n
k
)
there are more possibilities, say π : Mn(F) → M(n24 )
(F),
defined by A 7→
∧4
(A ⊗ A), is multiplicative and maps a rank-2 matrix E11 + E22 into
matrix of rank-one but is not of the form in the Theorem. This is because if rkA = r then
rk(
∧k
A) =
(
r
k
)
, while π maps a rank-3 matrix E11+E22+E33 into a matrix whose rank
equals 126.
If rank(π(A0)) 6= 1 there are more possibilities as can be seen by the map π :
Mn(F)→M(nk)(F), defined by A 7→ A⊕ 0(nk)−n.
Proof of Theorem 1.2. If k = n then m = 1, so π : Mn(F) → F. Such multiplicative
maps were proven to be in accordance with our results by Lemma 1.1.
Hence, we may assume in the sequel that k < n. Clearly, π(Id) is an idempotent,
and from π(X)π(Id) = π(X · Id) = π(X) = π(Id)π(X) we deduce that, relative to
decomposition Fm = Imπ(Id)⊕Kerπ(Id) we have π(X) = π1(X)⊕0m−r ∈Mr(F)⊕
0m−r, where r := rkπ(Id) and π1 is multiplicative with π1(0) = 0 and π1(Id) = Idr. As
such, if X is invertible, then Idr = π1(Id) = π1(XX−1) = π1(X)π1(X−1), so π1(X) is
also invertible and π1(X)−1 = π1(X−1).
Let X be any matrix of rank-k. Then, there exist invertible S, T ∈ Mn(F) with
SXT = Idk ⊕0n−k, and in particular, there exist invertible S1, T1 such thatX = S1A0T1.
Consequently,
1 = rkπ(A0) = rk
(
π1(S
−1
1 XT
−1
1 )⊕ 0m−r
)
= rk
(
π1(S1)
−1π1(X)π1(T1)
−1 ⊕ 0m−r
)
,
wherefrom rkπ(X) = 1 for every X of rank-k. Consequently, π(Idk ⊕0n−k) is an idem-
potent of rank-1, and by appropriate similarity we may assume it equals E11.
Given X = X̂ ⊕ 0n−k ∈ Mk(F) ⊕ 0n−k, we have X = (Idk ⊕0)X(Idk ⊕0),
wherefrom π(X) = E11π(X)E11 ∈ FE11. Hence, π induces a multiplicative map
π̂ :Mk(F)→ F by
π̂(X̂)E11 := π(X̂ ⊕ 0n−k).
It follows by Lemma 1.1 that there exists a nonzero multiplicative map φ1 : F → F such
that
π̂(X̂) = φ1(det X̂).
In particular, if the rank of X = X̂ ⊕ 0n−k is smaller than k, then π(X) = π̂(X̂)E11 = 0.
By multiplicativity, π(X) = 0 for every X ∈Mn(F) with rkX ≤ k−1. Moreover, given
any A ∈ Mn(F), letting  be the compression of A to the upper-left k–by–k block, we
have
π
(
(Idk ⊕0)A(Idk ⊕0)
)
= π(Â⊕ 0n−k) = φ1(det Â).
250 Ars Math. Contemp. 6 (2013) 247–252
Next, among diagonal idempotent matrices, there exists exactly m :=
(
n
k
)
of them that
have k ones and n − k zeros on diagonal. We order them lexicographically according to
position of ones on diagonal, and denote them
P1 = (Idk ⊕0n−k), . . . , Pm = (0n−k ⊕ Idk).
Given two such diagonal idempotents Pi, Pj , we have rk(PiPj) ≤ k and the equality holds
only if Pi = Pj . Hence π(P1), . . . , π(Pm) are pairwise orthogonal idempotents of rank-
one. It is well-known (say, [1, Lemma 2.2]) that, by applying appropriate similarity, we
can achieve π(Pi) = Eii for i = 1, . . . ,m. Combined with π(Pi) = π1(Pi) ⊕ 0m−r ∈
Mr(F)⊕ 0n−r ⊆Mm(F) we see that r = 0. Hence, π = π1 is already unital.
As above for π(P1AP1) = φ1(det Â) we see that for each i = 1, . . . ,m there exist
nonzero multiplicative map φi : F→ F so that
π(PiAPi) = φi(detA
(ii))Eii, (1.1)
where, for a matrix X ∈ Mn(F), we denote X(ij) the k–by–k submatrix of X which lies
on the rows where Pi has nonzero entries and on the columns where Pj has nonzero entries.
Observe that a nonzero multiplicative φi satisfies φi(1) = 1.
Consider any A ∈Mn(F). Then,
π(A) = Idπ(A) Id =
( m∑
i=1
Eii
)
π(A)
( m∑
j=1
Ejj
)
=
∑
i,j
Eiiπ(A)Ejj =
∑
i,j
π(PiAPj).
Given indices i 6= j, there exists Bji ∈ Mn of rank-k such that Bji = PjBjiPi and
det (Bji)
(ji)
= 1; for instance, if Pi =
∑
t∈{t1,...,tk}Ett and Pj =
∑
s∈{s1,...,sk}Ess,
with t1 < · · · < tk and s1 < · · · < sk, we can take
Bji =
k∑
i=1
Esiti (1.2)
and then (Bji)
(ji)
= Idk. In particular then, π(Bji) = γjiEji 6= 0. It follows that
π(PiAPj)π(Bji) = π(Pi(APjPjBji)Pi) = φi
(
det (PiAPjPjBjiPi)
(ii))
Eii. (1.3)
Observe that
(PiAPjPjBjiPi)
(ii) = A(ij)B
(ji)
ji .
(This follows easily by writing Pi =
∑
t∈{t1,...,tk}Ett and Pj =
∑
s∈{s1,...,sk}Ess, t1 <
t2 < · · · < tk and s1 < s2 < · · · < sk, and observing that
PiXPj =
∑
(t,s)∈{t1,...,tk}×{s1,...,sk}
xtsEts
and
PjY Pi =
∑
(t,s)∈{t1,...,tk}×{s1,...,sk}
ystEst,
M. Omladič and B. Kuzma: A note on homomorphisms of matrix semigroup 251
and hence PiXPj · PjY Pi =
∑
t,t′∈{t1,...,tk}
∑
s∈{s1,...,sk} xtsyst′Ett′ .)
Hence,
φi(det(PiAPjPjBjiPi)
(ii)) = φi(detA
(ij))φi(det(Bji)
(ji)) = φi(detA
(ij)) · φi(1)
= φi(detA
(ij)).
On the other hand, π(PiAPj) = π(Pi)π(A)π(Pj) = Eiiπ(A)Ejj = αij(A)Eij where-
from π(PiAPj) · π(Bji) = αij(A)Eij · γjiEji = αij(A)γjiEii, and hence, by (1.3)
αij(A) =
1
γji
φi(detA
(ij)).
By similar arguments we also have that
γji αij(A)Ejj = π(Bji)π(PiAPj) = π(BjiPiAPj) = π(PjBjiPiAPj)
= φj(detA
(ij))Ejj
and since A was arbitrary, we see that φi = φj =: φ is independent of i, j. Hence,
π(X) =
∑
i,j
αij(X)Eij =
∑
i,j
1
γij
φ(detX(ij))Eij ,
where, in accordance with (1.1), we define γii = 1 for i = 1, . . . ,m. Recall also that
φ(1) = φi(1) = 1.
We only need to show that multiplicativity of π forces that φ is additive and that
γijγjv = γiv . To prove additivity of φ, choose a scalar α and consider a rank-(k + 1)
matrix Aα := ( 1 α0 1 )⊕ Idk−1⊕0n−k−1. It is easy to see that in Aα the number of k–by–k
submatrices of rank-k equals (k+2), and they are all obtained from the principal (k+1)–
by–(k + 1) block by deleting one of the following (i) the same row and column, or (ii)
second row and first column. Under (i) the resulting submatrix equals Idk, while under (ii)
it equals α⊕ Idk−1. Thus, there exist indices i1, . . . , ik+1 and i, j ∈ {i1, . . . , ik+1}, i 6= j,
so that
π(Aα) =
k+1∑
t=1
φ(1)Eitit +
1
γij
φ(α)Eij .
(A deeper analysis reveals that, in lexicographic order, i =
(
n−2
k−2
)
+ 1 and j =
(
n−1
k−1
)
+ 1).
As AαAβ = Aα+β , the multiplicativity of π together with φ(1) = 1 yields∑
t
Eitit +
1
γij
φ(α+ β)Eij = π(AαAβ) = π(Aα)π(Aβ) =
∑
t
Eitit +
φ(α)+φ(β)
γij
Eij ,
wherefrom φ is additive.
It remains to prove γijγjv = γiv . Take matrices Bij and Bjv defined in (1.2). Then,
det
(
(BijBjv)
(iv)
)
= det((Bij)
(ij)) det((Bjv)
(jv)) = 1 · 1 = 1. Hence,
1
γiv
Eiv =
1
γiv
φ(det(BijBjv)
(iv))Eiv = π(BijBjv) = π(Bij)π(Bjv) =
1
γijγjv
Eij ·Ejv,
wherefrom γiv = γijγjv.
Consider now an invertible diagonal matrix D = diag(γ11, . . . , γ1m). Then, π(X) =∑
i,j
1
γij
φ(detX(ij))Eij = D
−1∑
i,j φ(detX
(ij))EijD = D
−1φ(
∧k
(X))D.
252 Ars Math. Contemp. 6 (2013) 247–252
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