This article was downloaded by: [University of Waterloo]On: 13 November 2014, At: 06:07Publisher: Taylor & FrancisInforma Ltd Registered in England and Wales Registered Number: 1072954 Registeredoffice: Mortimer House, 37-41 Mortimer Street, London W1T 3JH, UK

Linear and Multilinear AlgebraPublication details, including instructions for authors andsubscription information:http://www.tandfonline.com/loi/glma20

Maps preserving classical adjoint ofproducts of two matricesC.G. Cao a , Y.L. Ge a & H.M. Yao ba Department of Mathematics , Heilongjiang University , Harbin ,P.R. Chinab College of Science , Harbin Engineering University , Harbin , P.R.ChinaPublished online: 17 Jan 2013.

To cite this article: C.G. Cao , Y.L. Ge & H.M. Yao (2013) Maps preserving classical adjointof products of two matrices , Linear and Multilinear Algebra, 61:12, 1593-1604, DOI:10.1080/03081087.2012.753592

To link to this article: http://dx.doi.org/10.1080/03081087.2012.753592

PLEASE SCROLL DOWN FOR ARTICLE

Taylor & Francis makes every effort to ensure the accuracy of all the information (the“Content”) contained in the publications on our platform. However, Taylor & Francis,our agents, and our licensors make no representations or warranties whatsoever as tothe accuracy, completeness, or suitability for any purpose of the Content. Any opinionsand views expressed in this publication are the opinions and views of the authors,and are not the views of or endorsed by Taylor & Francis. The accuracy of the Contentshould not be relied upon and should be independently verified with primary sourcesof information. Taylor and Francis shall not be liable for any losses, actions, claims,proceedings, demands, costs, expenses, damages, and other liabilities whatsoever orhowsoever caused arising directly or indirectly in connection with, in relation to or arisingout of the use of the Content.

This article may be used for research, teaching, and private study purposes. Anysubstantial or systematic reproduction, redistribution, reselling, loan, sub-licensing,systematic supply, or distribution in any form to anyone is expressly forbidden. Terms &Conditions of access and use can be found at http://www.tandfonline.com/page/terms-and-conditions

Linear and Multilinear Algebra, 2013Vol. 61, No. 12, 1593–1604, http://dx.doi.org/10.1080/03081087.2012.753592

Maps preserving classical adjoint of products of two matrices�

C.G. Caoa, Y.L. Gea and H.M. Yaob∗

aDepartment of Mathematics, Heilongjiang University, Harbin, P.R. ChinabCollege of Science, Harbin Engineering University, Harbin, P.R. China

Communicated by M. Chebotar

(Received 14 June 2012; final version received 24 November 2012)

Let C be the complex field and n, m be integers with n ≥ m ≥ 2. Denote byMn(C) the set of all n × n complex matrices. In this paper, we address a generalform of maps ϕ : Mn(C)−→ Mm(C) satisfying ϕ[(AB)ad ] = [ϕ(A)ϕ(B)]ad ,∀A, B ∈ Mn(C), where Aad denotes the classical adjoint of A.

Keywords: classical adjoint; products of two matrices; preserver problemAMS Subject Classifications: 15A03; 15A04

1. Introduction

In recent years, many researchers are interested in the study of preserving maps withoutlinear and additive assumptions, see([1–5]). In paper [1], the authors considered the unitalsurjective maps on Banach space preserving the nonzero idempotency of products of twooperators in both directions. Correspondingly, in paper [4], the author studied surjectivemaps ϕ satisfying the condition A − λB is idempotent if and only if ϕ(A) − λϕ(B) isidempotent, where A and B are operators in Banach space and λ is a complex number.In 2010, Chooi and Ng [5] studied the maps between full matrix algebras over a field F

preserving classical adjoint of subtraction or linear combination of matrices, namely,

ϕ[(A − B)ad ] = [ϕ(A)− ϕ(B)]ad , ∀A, B ∈ Mn(F)

orϕ[(A + αB)ad ] = [ϕ(A)+ αϕ(B)]ad , ∀A, B ∈ Mn(F).

An immediate natural question arises: what is the general form of the maps between fullmatrix algebras preserving classical adjoint of products of two matrices?

Let Mm,n(C)(respectively, Mn(C)) be the set of all m ×n (respectively, n ×n) matricesover the complex field C and n and m be integers with n ≥ m ≥ 2. For simplicity, we alwayswrite Mn(C) by Mn . A map ϕ : Mn −→ Mm is said to be preserving classical adjont if

∗Corresponding author. Email: hongmeiyao@163.com�Supported by Education Department of Heilongjiang Province of China (No. 11551365), NaturalScience Foundation of Heilongjiang Province of China (No. A201013), National Science Foundationof China (No. 10671026), the Scientific Research Funds of Innovation for Postgraduates inHeilongjiang Province (No. YJSCX2012-272HLJ) and the Fundamental Research Funds for theCentral Universities.

© 2013 Taylor & Francis

Dow

nloa

ded

by [

Uni

vers

ity o

f W

ater

loo]

at 0

6:07

13

Nov

embe

r 20

14

1594 C.G. Cao et al.

ϕ[Aad ] = [ϕ(A)]ad , ∀A ∈ Mn ,where Aad is the classical adjoint of A, i.e. the transposed matrix of cofactors of the matrixA. More precisely, the (i, j)th entry of Aad is

(Aad)i j = (−1)i+ j det(A[ j | i]),where det(A[ j | i]) denotes the determinant of the (n − 1)× (n − 1) submatrix A[ j | i] ofA obtained by excluding its j th row and i th column. In 1960, Muir outlined the informationabout the early history of the notion of the classical adjoint in his book, The Theory ofDeterminants [6]. The classical adjoint is one of the important matrix functions on squarematrices, which has many applications in matrix theory. In particular, it has been used tovarious studies of generalized inverse matrices, see [7]. In preserver problems, there havebeen some papers that studied linear maps or additive maps preserving classical adjoint onvarious matrix spaces, see ([7–10]).

Inspired by the above, the purpose of this paper is to study maps between full com-plex matrix algebras preserving classical adjoint of products of two matrices, that is ϕ :Mn −→ Mm satisfying

ϕ[(AB)ad ] = [ϕ(A)ϕ(B)]ad , ∀A, B ∈ Mn . (1)

We end this section by introducing some notations that will be needed throughout ourdiscussion. Let GLn(F) be the general linear group over a field F and r(A) be the rankof matrix A. For any nonnegative integer p ≤ n,m, we set <p>= {0, 1, 2, · · · , p}.Denote by �p(respectively, �

′p), the subset of Mn(respectively, Mm) consisting of all rank

p matrices. When p is a positive integer, �<p>(respectively, �′<p>) denotes the subset

of Mn(respectively, Mm) consisting of all matrices having rank bounded above by p, and�<0> = {aEi j | a ∈ C, i, j ∈ {1, 2, · · · , n}}, where Ei j stands for the matrix with 1 at the(i, j)th entry and 0 otherwise. Let Ik be the k × k identity matrix over C and 0 be the zeromatrix which order is omitted in different matrices just for simplicity, At be the transposeof the matrix A. Let G = �n ∪ �1 ∪ �0, G1 = Mn\(�n ∪ �n−1), G

′ = �′n ∪ �′

1 ∪ �′0 and

G′1 = Mm\(�m ∪ �m−1). If δ is an injective field endomorphism of C, then Aδ = (aδi j ),

where A = (ai j ) ∈ Mn . The symbol ⊕ represents the direct sum and C∗ is the multiplicative

group of all nonzero elements of C.

2. Main results

Lemma 2.1 Let A, B ∈ Mn and n be an integer with n ≥ 2. Then the following resultshold:

(i) Aad A = AAad = (det A)In.

(ii) (Aad)ad = (det A)n−2 A.

(iii) r(Aad) =⎧⎨⎩

n A ∈ �n

1 A ∈ �n−10 A ∈ G1

.

(iv) (AB)ad = Bad Aad . Particularly, if P ∈ GLn(C), then (P AP−1)ad = P Aad P−1.

Proof By the results in Appendix D of [11], we can easily get the result of Lemma 2.1. �

Dow

nloa

ded

by [

Uni

vers

ity o

f W

ater

loo]

at 0

6:07

13

Nov

embe

r 20

14

Linear and Multilinear Algebra 1595

Lemma 2.2 Let n, m be integers with n,m ≥ 2. If a map ϕ : Mn −→ Mm satisfiescondition (1) and ϕ(S) = D, where S = In or S = 0, then D = 0 or D ∈ GLm(C) withD = (D2)ad .

Proof Since the proof of the case S = 0 is similar to the proof of the case S = In , we justgive the proof of the case S = In .

In (1), if A = B = In , then we have ϕ(In) = (ϕ(In))ad(ϕ(In))

ad , i.e.

D = (D2)ad . (2)

By (iii) of Lemma 2.1, we distinguish our proof into the following three cases:

Case 1 r(D2) = m. From (2), we have r(D) = m. Then D ∈ GLm(C).Case 2 r(D2) = m − 1. From (2), we get r(D) = r((D2)ad) = 1. Since r(D2) ≤ r(D),

we have m ≤ 2, and so m = 2.

Let D = P(1 ⊕ 0)Q, where P, Q ∈ GL2(C), by (2), we have

P(1 ⊕ 0)Q = [P(1 ⊕ 0)Q P(1 ⊕ 0)Q]ad .

Set

Q P =[

b11 b12b21 b22

],

where b11, b12, b21, b22 ∈ C. it follows that

P(1 ⊕ 0)Q = [P(b11 ⊕ 0)Q]ad .

Further

Q P(1 ⊕ 0)Q P = Q Qad(0 ⊕ b11)Pad P .

Hence [b2

11 b11b12b21b11 b21b12

]= det(P Q)

[0 00 b11

].

Then, by det(P Q) = b11b22 − b12b21 �= 0, we get b211 = 0, b11b12 = 0, b21b11 = 0 and

b21b12 = b11(b11b22 − b12b21). It follows that b11 = 0 and b21b12 = 0, which contradictsto P Q ∈ GL2(C). Thus, this case can not occur.

Case 3 r(D2) < m − 1. From (2), we have r(D) = r((D2)ad) = 0, i.e. D = 0.Therefore, we complete the Proof of Lemma 2.2. �

Lemma 2.3 Suppose that F is a field and that m, n are integers with n ≥ m ≥ 2. Letp ∈< n − 1 >, � be a subsemigroup of Mm(F), be a subsemigroup of Mn(F) containing�<p> and f : −→ � be a semigroup homomorphism. Then f has one of the followingforms:

(i) m = n, f (A) = P AδP−1, ∀A ∈ , where P ∈ GLn(F) and δ is an injective fieldendomorphism of F.

(ii) m = n = p + 1 ≥ 3, f (A) = P((Aad)t )δP−1, ∀A ∈ , where P ∈ GLn(F) andδ is an injective field endomorphism of F.

Dow

nloa

ded

by [

Uni

vers

ity o

f W

ater

loo]

at 0

6:07

13

Nov

embe

r 20

14

1596 C.G. Cao et al.

(iii) f (A) = P(Ir ⊕ g(A))P−1, ∀A ∈ , where P ∈ GLm(F), r ∈< m > andg : → {X ∈ Mm−r (F) | P(Ir ⊕ X)P−1 ∈ �} is a semigroup homomorphism suchthat g(�<p>) = {0}.

Proof By the Theorem 1 and the Corollary 1 of [12], we can easily get the proof ofLemma 2.3. �Lemma 2.4 Let B ∈ Mn with r(B) = r < n. Then there exist P1, Q1 ∈ GLn(C) withdetP1 = detQ1 = 1 such that B = P1(Ir ⊕ 0)Q1.

Proof Since r(B) = r , there exist P, Q ∈ GLn(C) such that

B = P(Ir ⊕ 0)Q

= P(In − Enn + (detP)−1 Enn)(Ir ⊕ 0)(In − Enn + (detQ)−1 Enn)Q.

Let P1 = P(In − Enn + (detP)−1 Enn), Q1 = (In − Enn + (detQ)−1 Enn)Q. Then detP1 =detQ1 = 1 and B = P1(Ir ⊕ 0)Q1. �Lemma 2.5 Suppose n,m are integers with n ≥ m ≥ 2. Let ϕ : GLn(C)→ GLm(C)

be a group homomorphism. Then there exist R ∈ GLn(C), a group endomorphism θ :C

∗ −→ C∗, an injective field endomorphism δ : C −→ C and a group homomorphism

π : C∗ −→ GLm(C) such that one of the following holds:

(i) m = n, ϕ(A) = θ(det A)R((Aad)t )δR−1, ∀A ∈ GLn(C).

(ii) m = n, ϕ(A) = θ(det A)R AδR−1, ∀A ∈ GLn(C).

(iii) ϕ(A) = π(det A), ∀A ∈ GLn(C).

Proof Using the Theorems 11.12, 11.13 and 11.14 of [11], we can easily get the results ofLemma 2.5. �Lemma 2.6 Suppose that A ∈ G. Then the matrix equation Xad = A has the followingsolutions:

(i) If A ∈ �n, then X = (det A)1

n−1 A−1, where (det A)1

n−1 denotes any (n − 1)th rootof det A.

(ii) If A = 0, then the set of solutions of Xad = A is G1.

(iii) If A = M Enn N ∈ �1 with detM = detN = 1, then X = N−1(X1 ⊕ 0)M−1, whereX1 ∈ Mn−1(C) with detX1 = 1.

Proof

(i) By Xad = A ∈ �n and X Xad = (detX)In , we derive

X = (detX)(Xad)−1. (3)

Then (detX)n−1 = det A, i.e. detX = (det A)1

n−1 . By (3), we obtain

X = (det A)1

n−1 A−1.

Dow

nloa

ded

by [

Uni

vers

ity o

f W

ater

loo]

at 0

6:07

13

Nov

embe

r 20

14

Linear and Multilinear Algebra 1597

(ii) It follows immediately from (iii) of Lemma 2.1.(iii) By Xad = A , we have 0 = (detX)In = X Xad = X A. Similarly, we obtain AX = 0.

Let X = N−1[

X1 X2X3 X4

]M−1, where M, N ∈ GLn(C) and X1 ∈ Mn−1, X2 ∈

Mn−1,1(C), X3 ∈ M1,n−1(C) and X4 ∈ M1(C). Then, by computing matrix equationAX = X A = 0, we get X2 = 0, X3 = 0, X4 = 0, and thus X = N−1(X1 ⊕ 0)M−1.From Xad = A, we have

(N−1(X1 ⊕ 0)M−1)ad = M Enn N . (4)Note that detM = detN = 1, thus Mad = M−1, N ad = N−1. By (4), we get detX1 = 1.Hence, the proof is completed. �Remark 1 If we consider the matrix equation Xad = A over the real field, then the solutionsof the equation may not exist. In fact, when n is odd with det A = −1, then (−1)

1n−1 is not

a real number.

Lemma 2.7 Suppose that D ∈ GLn(C). Then D = (Dad)2 if and only if D3 = (detD)2 In.

Proof Since D ∈ GLn(C), we have D = (Dad)2 ⇔ D3 = D(Dad Dad)D. So the resultfollows immediately from Lemma 2.1 (i). �

Using Lemma 2.1 and Lemma 2.7, it is not difficult to prove the following mapssatisfying (1).

Example 2.8 A map ϕ : Mn → Mn is defined as ϕ(A) = P AP−1, ∀A ∈ Mn , whereP ∈ GLn(C).

Example 2.9 A map ϕ : Mn → Mn is defined as ϕ(A) = Aδ, ∀A ∈ Mn , where δ is aninjective field endomorphism of C.

Example 2.10 A map ϕ : Mn → Mn is defined as ϕ(A) = x A, ∀A ∈ Mn , where x ∈ C

with x2n−2 = x .

Example 2.11 A map ϕ : Mn → Mn is defined as ϕ(A) = (Aad)t , ∀A ∈ Mn .

Example 2.12 A map ϕ : Mn → Mm is defined as ϕ(A) = (det A)k D, ∀A ∈ Mn , wherek is a fixed positive integer and D ∈ GLm(C) with D3 = (detD)2 Im .

Example 2.13 A map ϕ : Mn → Mm is defined as ϕ(A) = h A D, ∀A ∈ Mn , whereD ∈ GLm(C) with D3 = (detD)2 Im and h A ∈ C with hm−1

A = 1. In particular, if A ∈ G,then h A = 1.

Example 2.14 A map ϕ : Mn → Mm is defined as

ϕ(A) ={

0 ∀A ∈ Gf (A) ∀A ∈ Mn \ G

,

where f : Mn \ G → G′1 is a map.

Example 2.15 A map ϕ : Mn → Mm is defined as

ϕ(A) ={

0 ∀A ∈ Mn \ �n

(λωIs ⊕ λω2 It ⊕ λIm−s−t ) ∀A ∈ �n,

where λ, ω ∈ C, λ2m−3 = 1, 1 �= ω, ω3 = 1 and t ≡ s(mod3).

Dow

nloa

ded

by [

Uni

vers

ity o

f W

ater

loo]

at 0

6:07

13

Nov

embe

r 20

14

1598 C.G. Cao et al.

Theorem 2.16 Let m, n be integers with n ≥ m ≥ 2. A map ϕ : Mn −→ Mm satisfiesϕ(In) ∈ GLm(C) and ϕ(0) = 0, then ϕ satisfies condition (1) if and only if ϕ has one ofthe following forms:

(i) ϕ(A) =⎧⎨⎩λP AδP−1 ∀A ∈ G, m = nf (A) ∀A ∈ �<n−2> \ �<1>, m = n ≥ 4λP Mδ(X1 ⊕ 0)N δP−1 ∀A = M(In−1 ⊕ 0)N ,m = n ≥ 3

,

where λ ∈ C with λ2n−3 = 1, δ is an injective field endomorphism of C, f :�<n−2> \ �<1>→�<n−2> is a map, M, N ∈ Mn with detM = detN = 1,X1 ∈ Mn−1 with detX1 = 1 and P ∈ GLn(C).

(ii) ϕ(A) =⎧⎨⎩

0 ∀A ∈ �<1>ρ(A) ∀A ∈ �<n−1> \ �<1>, n ≥ 3ψ(A) ∀A ∈ �n

,

where ψ : �n →�′m is a map taking one of the following forms:

(a) ψ(A) = Dπ(det A), ∀A ∈ �n,(b) ψ(A) = λθ(det A)R AδR−1, ∀A ∈ �n, m = n,(c) ψ(A) = λθ(det A)R((Aad)t )δR−1, ∀A ∈ �n, m = n,

where λ ∈ C with λ2n−3 = 1, R ∈ GLn(C), ρ : �<n−1> \ �<1>→ G′1 is a map,

δ is an injective field endomorphism of C, θ : C∗ −→ C

∗ is a group endomorphism, π :C

∗ −→ GLm(C) is a group homomorphism with π(det Aad) = [π(det A)]ad for all A ∈ �n,and D ∈ GLm(C) satisfying D3 = (detD)Im and D(π(det A)) = (π(det A))D for allA ∈ �n.

Proof The ‘only if’ part: since ϕ(In) ∈ GLm(C), we first consider the special case

ϕ(In) = Im .

Let A = In , then by (1), we have

ϕ(Bad) = [ϕ(B)]ad , ∀B ∈ Mn . (5)

Then (1) can be rewritten as

ϕ(Bad Aad) = ϕ(Bad)ϕ(Aad), ∀A, B ∈ Mn . (6)

By (6) and Lemma 2.6, we know that ϕ |G : G → G′

is a semigroup homomorphism. Weset p = 1, = G and � = G

′. Clearly �<1> ⊆ G since p = 1, it follows from Lemma

2.3 that ϕ takes either Form (i) or Form (iii) in Lemma 2.3. We thus have(I) There exist P ∈ GLn(C) and an injective field endomorphism δ of C such that

ϕ(A) = P AδP−1, ∀A ∈ G.

Dow

nloa

ded

by [

Uni

vers

ity o

f W

ater

loo]

at 0

6:07

13

Nov

embe

r 20

14

Linear and Multilinear Algebra 1599

From (iii) of Lemma 2.3, we know ϕ(A) = P(Ir ⊕ g(A))P−1, ∀A ∈ G. Since ϕ(0) = 0,we get r = 0. Then

ϕ(A) = Pg(A)P−1, ∀A ∈ G,

where g : G → {X ∈ Mm(C)|P X P−1 ∈ G′ } = G

′is a semigroup homomorphism such

that g(�<1>) = {0}. Henceϕ(A) = 0, ∀A ∈ �<1>. (7)

If A ∈ �n , by ϕ(In) = Im and (6), then we obtain ϕ(A) ∈ �′m . Thus ϕ |�n : �n →�

′m is a

map. Applying (6) and Lemma 2.6 (i), we get

ϕ(B A) = ϕ(B)ϕ(A), ∀A, B ∈ �n .

Then ϕ |�n : �n →�′m is a group homomorphism. By Lemma 2.5 and (7), we have

(II) ϕ(A) =

⎧⎪⎪⎨⎪⎪⎩

0 ∀A ∈ �<1>π(det A) ∀A ∈ �n

θ(det A)R AδR−1 ∀A ∈ �n,m = nθ(det A)R((Aad)t )δR−1 ∀A ∈ �n,m = n

,

where δ is an injective field endomorphism of C, π : C∗ −→ GLm(C) is a group homo-morphism and θ : C∗ −→ C∗ is a group endomorphism, R ∈ GLn(C). From (5) and thedefinition of π , we have π(det Aad) = [π(det A)]ad for all A ∈ �n .

Next, we need to determine the images of Mn \ G under ϕ.When A ∈ �<n−2> \�<1>(n ≥ 4), by (5), we see (ϕ(A))ad = ϕ(Aad) = ϕ(0) = 0. If

m = n, then ϕ(A) ∈ �<n−2>. Let

ϕ(A) = f (A), ∀A ∈ �<n−2> \ �<1>,

where f : �<n−2> \ �<1>→�<n−2> is a map.If m �= n, then ϕ(A) ∈ G

′1. Let

ϕ(A) = f1(A), ∀A ∈ �<n−2> \ �<1>,

where f1 : �<n−2> \ �<1>→ G′1 is a map.

When r(A) = n−1, by Lemma 2.4, there exist M, N ∈ GLn(C)with detM = detN =1 such that A = M(In−1 ⊕ 0)N . Clearly, Aad = N−1 Enn M−1 ∈ �1.When (I) holds, we obtain ϕ(Aad) = P(Aad)δP−1. Then, by (5), we have

(ϕ(A))ad = P(Aad)δP−1 = P(N−1)δEnn(M−1)δP−1,

i.e.P−1(ϕ(A))ad P = (N−1)δEnn(M

−1)δ.

From Lemma 2.1 (iv), we obtain

(P−1ϕ(A)P)ad = (N−1)δEnn(M−1)δ.

Since detM = detN = 1 and δ is an injective field endomorphism of C, then det(M−1)δ =det(N−1)δ = 1 and ((M−1)δ)−1 = Mδ, ((N−1)δ)−1 = N δ . Applying (iii) of Lemma 2.6,we derive

ϕ(A) = P Mδ(X1 ⊕ 0)N δP−1, ∀A ∈ �n−1,

Dow

nloa

ded

by [

Uni

vers

ity o

f W

ater

loo]

at 0

6:07

13

Nov

embe

r 20

14

1600 C.G. Cao et al.

where X1 ∈ Mn−1with detX1 = 1.When (II) holds, we have ϕ(Aad) = 0. Then by (5), we get [ϕ(A)]ad = 0, let

ϕ(A) = ρ1(A), ∀A ∈ �n−1,

where ρ1 : �n−1 → G′1 is a map.

Now, f1 and ρ1 are together written as a map ρ : �<n−1> \ �<1>−→ G′1, then from

the above discussion, we have

(i′)ϕ(A) =

⎧⎨⎩

P AδP−1 ∀A ∈ G, m = nf (A) ∀A ∈ �<n−2> \ �<1>,m = n ≥ 4P Mδ(X1 ⊕ 0)N δP−1 ∀A = M(In−1 ⊕ 0)N ,m = n ≥ 3

,

where δ is an injective field endomorphism of C, f : �<n−2> \ �<1>→�<n−2>is a map, M, N ∈ Mn with detM = detN = 1, X1 ∈ Mn−1 with detX1 = 1 andP ∈ GLn(C).

(ii′)ϕ(A) =

⎧⎨⎩

0 ∀A ∈ �<1>ρ(A) ∀A ∈ �<n−1> \ �<1>ψ(A) ∀A ∈ �n

,

where ψ : �n →�′m is a map taking one of the following forms:

(a′) ψ(A) = π(det A) ∀A ∈ �n ,

(b′) ψ(A) = θ(det A)R AδR−1 ∀A ∈ �n, m = n,

(c′) ψ(A) = θ(det A)R((Aad)t )δR−1 ∀A ∈ �n, m = n,

where R ∈ GLn(C), δ is an injective field endomorphism of C, ρ : �<n−1> \ �<1>→ G′1

is a map, θ : C∗ −→ C

∗ is a group endomorphism, π : C∗ −→ GLm(C) is a group

homomorphism with π(det Aad) = [π(det A)]ad for all A ∈ �n .Finally, we consider the general case

ϕ(In) = D ∈ GLm(C).

Clearly, D3 = (detD)2 Im . Then ϕ[(AB)ad ] = Dad [ϕ(AB)]ad , ∀A, B ∈ Mn . FromLemma 2.1 and (1), we have

Dad [ϕ(AB)]ad = [ϕ(B)]ad [ϕ(A)]ad , ∀A, B ∈ Mn . (8)

By ϕ(Bad) = [ϕ(In)]ad [ϕ(B)]ad = [ϕ(B)]ad [ϕ(In)]ad , it implies that

Dad [ϕ(B)]ad = [ϕ(B)]ad Dad , ∀B ∈ Mn . (9)

From D ∈ GLm(C), (Dad)−1 = (D−1)ad and (9), we get

(D−1)ad [ϕ(B)]ad = [ϕ(B)]ad(D−1)ad , ∀B ∈ Mn .

From D = (D2)ad and (9), we get

D[ϕ(B)]ad = (D2)ad [ϕ(B)]ad = [ϕ(B)]ad(D2)ad = [ϕ(B)]ad D, ∀B ∈ Mn .

Dow

nloa

ded

by [

Uni

vers

ity o

f W

ater

loo]

at 0

6:07

13

Nov

embe

r 20

14

Linear and Multilinear Algebra 1601

Using (8), we obtain

(D−1)ad(D−1)ad Dad [ϕ(AB)]ad = [ϕ(B)]ad(D−1)ad [ϕ(A)]ad(D−1)ad , ∀A, B ∈ Mn .

(10)Further, by D−1 = (D−1)ad(D−1)ad , we derive that

D−1ϕ[(AB)ad ] = [D−1ϕ(B)]ad [D−1ϕ(A)]ad , ∀A, B ∈ Mn . (11)

Let φ : Mn → Mm be the map defined by φ(A) = D−1ϕ(A), ∀A ∈ Mn . Then (11) canbe rewritten as

φ[(AB)ad ] = [φ(B)]ad [φ(A)]ad , ∀A, B ∈ Mn .

Note that φ(In) = D−1ϕ(In) = D−1 D = Im and φ(0) = D−1ϕ(0) = 0. Then φ takes oneof the Form (i

′) or (ii

′). Since ϕ(A) = Dφ(A), ∀A ∈ Mn , we have

D[φ(A)]ad = [φ(A)]ad D, ∀A ∈ Mn . (12)

Suppose that φ(A) has Form (i′). Let A ∈ �n . Then φ(A) = P AδP−1 for some

P ∈ GLn(C) and injective field endomorphism δ of C. Using (12), we get

D[P AδP−1]ad = [P AδP−1]ad D, ∀A ∈ �n .

Applying Lemma 2.1 (iv) and Lemma 2.6, we get that D commutes with any invertiblematrix, then D = λIn , where λ ∈ C. So (D2)ad = D implies that λ2n−3 = 1.If A ∈ �<n−2> \ �<1>(n ≥ 4), then, in view of (i

′), we have ϕ(A) = Dφ(A) = g(A),

where g : �<n−2> \ �<1>→�<n−2> is a map.If A ∈ �n−1, then using (12), we have

D[P Mδ(X1 ⊕ 0)N δP−1]ad = [P Mδ(X1 ⊕ 0)N δP−1]ad D, ∀A ∈ �n−1.

where δ is an injective field endomorphism of C, M, N ∈ Mn with detM = detN = 1,X1 ∈ Mn−1 with detX1 = 1 and P ∈ GLn(C).Applying Lemma 2.1 (iv) and Lemma 2.6, we get that D commutes with any matrix withrank n − 1, then D = λIn , where λ ∈ C. So (D2)ad = D implies that λ2n−3 = 1.Therefore, (i) of Theorem 2.16 holds.

Suppose that φ(A) has Form (ii′), if φ(A) = π(det A), ∀A ∈ �n , then using (12), we

deriveD[π(det A)]ad = [π(det A)]ad D, ∀A ∈ �n . (13)

Set [π(det A)]ad = H , by Lemma 2.6 (i), we get π(det A) = (detH)1

n−1 H−1. Then, from(13), we have

D[π(det A)] = [π(det A)]D, ∀A ∈ �n .

If ϕ(A) = θ(det A)R AδR−1, ∀A ∈ �n, m = n, by (12), we have

D[θ(det A)R AδR−1]ad = [θ(det A)R AδR−1]ad D, ∀A ∈ �n .

Applying Lemma 2.1 (iv) and Lemma 2.6, we get that D commutes with any invertiblematrix, then D = λIn , where λ ∈ C. So (D2)ad = D implies that λ2n−3 = 1. For othercases of (ii

′), they are similar to the proofs of the cases of (i

′).

Hence, (ii) of Theorem 2.16 holds.As a result, (i) and (ii) of Theorem 2.16 hold.

Dow

nloa

ded

by [

Uni

vers

ity o

f W

ater

loo]

at 0

6:07

13

Nov

embe

r 20

14

1602 C.G. Cao et al.

The ‘if’ part: Since D is commutative with π(det A) and π(det Aad) = [π(det A)]ad , itis easy to check that (ii) of the Theorem 2.16 satisfies condition (1). Now, we only need toshow (i) of the Theorem 2.16 still satisfies condition (1).

In the following proof, we only consider the case A, B ∈ �n−1, for other cases, theproofs are similar.Let ϕ(A) = λP Mδ

1(X1 ⊕ 0)N δ1 P−1, ϕ(B) = λP Mδ

2(X2 ⊕ 0)N δ2 P−1,

where A = M1(In−1⊕0)N1, B = M2(In−1⊕0)N2, M1, M2, N1, N2 ∈ Mn with detM1 =detN1 = detM2 = detN2 = 1, X1, X2 ∈ Mn−1 with detX1 = detX2 = 1 andP ∈ GLn(C). Hence

AB = M1(In−1 ⊕ 0)N1 M2(In−1 ⊕ 0)N2.

Set N1 M2 =[

b1 b2b3 b4

], where b1 ∈ Mn−1, b2 ∈ Mn−1,1(C), b3 ∈ M1,n−1(C), b4 ∈ M1(C),

we have AB = M1(b1 ⊕ 0)N2. Then

(AB)ad = N ad2 (0 ⊕ detb1)M

ad1 .

On the other hand,

ϕ[(AB)ad ] = λP(N δ2 )

ad(0 ⊕ detbδ1)(Mδ1)

ad P−1,

and

[ϕ(A)ϕ(B)]ad = (λP Mδ1(X1 ⊕ 0)N δ

1 P−1λP Mδ2(X2 ⊕ 0)N δ

2 P−1)ad

= λ2n−2 P(N δ2 )

ad(0 ⊕ det(X1bδ1 X2))(Mδ1)

ad P−1.

Since λ2n−3 = 1 and detX1 = detX2 = 1, by comparing the two equations, we get theresult (1) holds.

Therefore, we complete the proof of Theorem 2.16. �Remark 2 In the third form of Theorem 2.16 (i), X1 and A are related in M and N . But fora fixed matrix A ∈ �n−1, X1 can be chosen from the set {X1 ∈ Mn−1|detX1 = 1}. Sincethe decomposition of the matrix in Lemma 2.4 is not unique, we suppose

A = M(In−1 ⊕ 0)N = M′(In−1 ⊕ 0)N

′, ∀A ∈ �n−1, (14)

where M, N ,M′, N

′ ∈ Mn with detM = detN = detM′ = detN

′ = 1. Then we want toshow the following fact: let X1, X

′1 ∈ Mn−1 and δ be an injective field endomorphism of

C, then

{Mδ(X1 ⊕ 0)N δ|detX1 = 1} = {(M ′)δ(X

′1 ⊕ 0)(N

′)δ|detX

′1 = 1}. (15)

Since δ is an injective field endomorphism of C, by (14), we get

Aδ = Mδ(In−1 ⊕ 0)N δ = (M′)δ(In−1 ⊕ 0)(N

′)δ, ∀A ∈ �n−1. (16)

Set [(M ′)δ]−1 Mδ =

[m1 m2m3 m4

], (N

′)δ(N δ)−1 =

[n1 n2n3 n4

],

where m1, n1 ∈ Mn−1,m2, n2 ∈ Mn−1,1(C),m3, n3 ∈ M1,n−1(C),m4, n4 ∈ M1(C), by(16), we have n2 = 0, m3 = 0, m1 = n1. Then

[(M ′)δ]−1 Mδ(X1 ⊕ 0)N δ[(N ′

)δ]−1 = (m1 X1m−11 ⊕ 0).

Dow

nloa

ded

by [

Uni

vers

ity o

f W

ater

loo]

at 0

6:07

13

Nov

embe

r 20

14

Linear and Multilinear Algebra 1603

Thus (15) holds, and then the third form of Theorem 2.16 (i) holds.

Theorem 2.17 Let m, n be integers with m, n ≥ 2 and ϕ : Mn −→ Mm be a mapsatisfying ϕ(In), ϕ(0) ∈ GLm(C). Then ϕ satisfies condition (1) if and only if there existsa D ∈ GLm(C) with D3 = (detD)2 Im such that ϕ(A) = h A D, ∀A ∈ Mn, where h A ∈ C

with hm−1A = 1. Particularly, h A = 1 when A ∈ G.

Proof Based on the Example 2.13, we only give the proof of the ‘only if’ part.From (1), we have ϕ(0) = [ϕ(0)]ad [ϕ(A)]ad = [ϕ(0)]ad [ϕ(0)]ad , ∀A ∈ Mn , then

[ϕ(0)]ad = [ϕ(A)]ad . Using Lemma 2.1, it implies ϕ(A) = detϕ(A)detϕ(0) ϕ(0). Set h A = detϕ(A)

detϕ(0) ,

D = ϕ(0), it is easy to see that ϕ(A) = h A D, D = (Dad)2 and hm−1A = 1. By Lemma

2.7, we know that D3 = (detD)2 Im . By combing (1) and [ϕ(0)]ad = [ϕ(In)]ad , this yieldsϕ(In) = [ϕ(In)]ad [ϕ(In)]ad = (Dad)2 = D, then from ϕ(Aad) = [ϕ(A)]ad [ϕ(In)]ad , weobtain h Aad D = [h A D]ad Dad = D, hence h Aad = 1, i.e. h A = 1, when A ∈ G. �

Theorem 2.18 Let m, n be integers with m, n ≥ 2 and ϕ : Mn −→ Mm be a mapsatisfying ϕ(In) = 0. Then ϕ satisfies condition (1) if and only if

ϕ(A) ={

0 ∀A ∈ Gf (A) ∀A ∈ �<n−1> \ �<1>

,

where f : �<n−1> \ �<1>→�′<m−1> is a map satisfying f (A) f (B) ∈ G

′1.

Proof The ‘if’ part is trivial, we only give the proof of the ‘only if’ part.Using (1) and ϕ(In) = 0, we derive ϕ(Aad) = [ϕ(A)]ad [ϕ(In)] = 0, i.e. ϕ(A) =

0, ∀A ∈ G. Hence 0 = ϕ[(AB)ad ] = [ϕ(A)ϕ(B)]ad , then ϕ(A)ϕ(B) ∈ G′1, ∀AB ∈

�<n−1> \ �<1>.Therefore, the result of Theorem 2.18 holds. �

Remark 3 Theorems 2.16–2.18 give all forms of maps satisfying condition (1). Specially,Theorem 2.17 and Theorem 2.18 even hold without the requirement m ≤ n. Maps givenin Examples 2.8–2.10 satisfy (i) of Theorem 2.16. Map given in Example 2.13 satisfiesTheorem 2.17. Map given in Example 2.14 satisfies Theorem 2.18. Maps given in Examples2.12 and 2.15 satisfy (ii) of Theorem 2.16. Map given in Example 2.11 satisfies (ii) ofTheorem 2.16 with n ≥ 3, and it satisfies (i) of Theorem 2.16 with n = 2. In fact, let

A =[

a bc d

], then

(Aad)t =[

d −c−b a

]=

[0 i−i 0

]A

[0 i−i 0

]−1

.

Remark 4 If we consider a map ϕ : Mn −→ Mm satisfying (1), when m > n, thecharacterization of ϕ is still an open problem.

Acknowledgements

The authors are grateful to the referee for his/her careful reading of the manuscript andvaluable comments which greatly improved the readability of the paper.

Dow

nloa

ded

by [

Uni

vers

ity o

f W

ater

loo]

at 0

6:07

13

Nov

embe

r 20

14

1604 C.G. Cao et al.

References

[1] Li F, Ji GX, Pang YF. Maps preserving the idempotency of products of operators. Linear AlgebraAppl. 2007;42:40–52.

[2] Li CK, Šemrl P, Sze NS. Maps preserving the nilpotency of products. Linear Algebra Appl.2007;424:222–239.

[3] Forstall V, Herman A, Li CK, Sze NS, Yannello V. Preservers of eigenvalue inclusion sets ofmatrix products. Linear Algebra Appl. 2011;434:285–293.

[4] Dolinar G. Maps on preserving idempotents. Linear Multilinear A. 2004;52:335–347.[5] Chooi WL, Ng WS. On classical adjoint-commuting mappings between matrix algebras. Linear

Algebra Appl. 2010;432:2589–2599.[6] Muir T. The theory of determinant. Vol. 1. New York (NY): Dover; 1960.[7] Sinkhorn R. Linear adjugate preservers on the complex matrices. Linear Multilinear A.

1982;12:215–222.[8] Cao CG, Zhang X, Tang XM. Additive adjugate preservers on the matrices over fields. Northeast.

Math. J. 1999;15:246–252.[9] Tang XM. Linear operators preserving adjoint matrix between matrix spaces. Linear Algebra

Appl. 2003;372:287–293.[10] Tang Xm, Zhang X. Additive adjoint preservers between matrix spaces. Linear Multilinear A.

2006;54:285–300.[11] Zhang X, Tang XM, Cao CG. Preserver problems on spaces of matrices. Beijing: Science Press;

2007.[12] Zhang X, Cao CG. Homomorphisms between multiplicative semigroups of matrices over fields.

Acta Math. Sci. 2008;28:301–306.

Dow

nloa

ded

by [

Uni

vers

ity o

f W

ater

loo]

at 0

6:07

13

Nov

embe

r 20

14