Invertible linear maps preserving {1}-inverses of matrices over PID

J. Appl. Math. & Computing Vol. 22(2006), No. 3, pp. 255 - 265

Website: http://jamc.net

INVERTIBLE LINEAR MAPS PRESERVING {1}-INVERSESOF MATRICES OVER PID

CHANGJIANG BU

Abstract. Let R be a PID, chR = 2, n > 1, Mn(R) be the n × n fullmatrix algebra over R. f denotes any invertible linear map preserving {1}-inverses from Mn(R) to itself. In this paper, we have proven that f is aninvertible linear map on Mn(R) preserving {1}-inverses if and only if fsatisfies any one of the following two conditions:

(i) there exists a matrix P ∈ GLn(R) such that

f(A) = PAP−1 for all A ∈ Mn(R),

(ii) there exists a matrix P ∈ GLn(R) such that

f(A) = PAtP−1 for A ∈ Mn(R).

AMS Mathematics Subject Classification : 15A04, 15A09.Key words and phrases : PID(principal ideal domain), {1}-inverse of ma-trix, linear preserver.

1. Introduction

Let R be a PID, Mn(R) be the full matrix algebra over R. For A ∈ Mn(R), ifthere exists X ∈ Mn(R) such that AXA = A, then we call X a {1}-inverse of A(see [1]-[2]). By [2], there may be no {1}-inverse for some A ∈ Mn(R). If thereexist some {1}-inverses for A, we denote the set of all the {1}-inverses by A(1).Suppose f is a linear map from Mn(R) to itself, for A ∈ Mn(R) and X ∈ A(1),if we always have f(X) ∈ f(A)(1), then we call f a {1}-inverse preserving linearmap.

The research of Linear Preserver Problems is an active area of matrix theory(see[3]-[6]). Many people take the generalized inverse of matrix as the invari-ant(see [4], [7]-[12]). As far as we know, there are only a few papers [8]-[10],[12]

Received May 21, 2005. Revised December 8, 2005.

This work is supported by the Fundamental Research Fundation of Harbin Engineering Univer-

sity(No:HEUF04019).

c© 2006 Korean Society for Computational & Applied Mathematics and Korean SIGCAM.

255

256 Changjiang Bu

treating the linear preservers of generalized inverses when the base field (ring)has characteristic equal 2. There are no results concerning additive maps pre-serving generalized inverses in the case of characteristic 2 because of its greatdifficulty and even for the case of invertible linear maps, results were obtainedunder some extra conditions on the base field. For instance, in [8]-[10],[12], thenumber of the invertible elements in R must be more than 4. In this paper,we study linear maps preserving generalized inverses by methods different from[9]. And we remove the constraint on the number of the invertible elements inR . This work generalized the work of [9] for any cases on PID of characteris-tic 2. In fact, when R = F2 we have characterized the linear maps preserving{1}-inverses.

In this paper, R is any PID of characteristic 2; Γ denotes the set of all theinvertible linear maps from Mn(R) to itself which preserve {1}-inverses; GLn(R)is the general linear group; Ei,j is the n × n matrix with 1 in the (i, j) entryand zero elsewhere; In is the unit matrix; Os is the s × s zero matrix; At is thetranspose of A; trA is the trace of A; and [1, n] is the set {1, 2, · · · , n}.

2. Some lemmas

For the proof of our main results, first we will present some lemmas. In thissection, R is any PID of characteristic 2 and the field F2 = {0, 1} .

Lemma 1. ([6, Lemma 1]) If A2 = A ∈ Mn(R), then there exists P ∈ GLn(R)such that A = P (Ir ⊕ On−r)P−1, where rankA = r.

Lemma 2. If A3 = A ∈ Mn(R), then there exist P ∈ GLn(R), Dp = diag(d1, d2, · · · , dp) ∈ Mp(R), di 6= 0 ( for any i ∈ [1, p]) and nonnegative p, q, ssuch that

A = P

( (Ip Dp

O Ip

)⊕ Iq ⊕ Os

)P−1,

where rankA = 2p + q, 2p + q + s = n.

Proof. It follows from A2 = A that A4 = A2. By Lemma 1, there existsP1 ∈ GLn(R) such that A2 = P1(Ir ⊕ On−r)P−1

1 , where rankA2 = r. FromA2A = A = AA2, we get

A = P1

(A1 OO O

)P−1

1 ,

where A21 = Ir .

Set rank(A1 − Ir) = p, we can write A1 − Ir = Q1

(B1 B2

O O

)Q−1

1 , where

B1 = ΛΣ1, B2 = ΛΣ2, and Λ is a p × p diagonal matrix with nonzero diagonalentries, Σ1 ∈ GLp(R),

(Σ1 Σ2

)is the first p rows of a r × r invertible matrix.

Invertible linear maps preserving {1}-inverses of matrices over PID 257

From A21 = Ir and chR = 2, we get (A1 − Ir)2 = 0, then B1Λ

(Σ1 Σ2

)= 0.

Hence B1 = 0, i.e.,

A1 − Ir = Q1

(Op B2

O O

)Q−1

1 .

Obviously, rankB2 = p, and by [2](see P32 [2]), there exist M ∈ GLp(R) andN ∈ GLr−p(R) such that B2 = M

(Dp O

)N , where Dp = diag(d1, d2, · · · , dp)

∈ Mp(R) and di 6= 0 for any i ∈ [1, p]. Thereby,

A1 − Ir = Q1

(M OO N−1

) (Op (DpO)O O

) (M−1 O

O N

)Q−1

1 .

Let Q2 = Q1

(M OO N−1

). Then

A1 = Q2

Ip Dp OO Ip OO O Iq

Q−1

2 .

Substituting A1 into A = P1

(A1 OO O

)P−1

1 , we can easily obtain the conclusion.

�

Lemma 3. For f ∈ Γ, we have the following(i) f(Eii)3 = f(Eii) for any i ∈ [1, n];(ii) f(Eii)f(Eks)f(Eii) = 0 for any i, k, s ∈ [1, n], Eii 6= Eks;(iii) f(In)3 = f(In);(iv) f(Eii)f(In)f(Eii) = f(Eii) for any i ∈ [1, n];(v) f(Eii)f(In)2 = f(In)2f(Eii) for any i ∈ [1, n];(vi) f(Eij)f(In)f(Eij) = 0 for any i, j ∈ [1, n], i 6= j;(vii) f(In)2f(Eij) + f(Eii)f(In)f(Eij) + f(Eij)f(In)f(Eii)

+ f(Eij)f(In)2 = f(Eij) for any i, j ∈ [1, n], i 6= j.

Proof. (i) By Eii ∈ E(1)ii (for any i ∈ [1, n]) and f ∈ Γ, we have f(Eii)∈ f(Eii)(1).

Hencef(Eii)3 = f(Eii).

(ii) By In ∈ E(1)ii (for any i ∈ [1, n]) and f ∈ Γ, we have f(In) ∈ f(Eii)(1),

and hencef(Eii)f(In)f(Eii) = f(Eii).

By (In + Eks) ∈ E(1)ii (for any i, k, s ∈ [1, n], Eii 6= Eks) and f ∈ Γ, we have

f(Eii)f(In + Eks)f(Eii) = f(Eii).

From the above equations, we get

f(Eii)f(Eks)f(Eii) = 0,

for any i, k, s ∈ [1, n], Eii 6= Eks.(iii) By In ∈ I

(1)n and f ∈ Γ, we have f(In)3 = f(In).

258 Changjiang Bu

(iv) See the proof of (ii).(v) By In ∈ (In + Eii)(1) (for any i ∈ [1, n]) and f ∈ Γ, we have

[f(In) + f(Eii)]f(In)[f(In) + f(Eii)] = f(In) + f(Eii),

i.e.,

f(In)3 + f(Eii)f(In)2 + f(In)2f(Eii) + f(Eii)f(In)f(Eii) = f(In) + f(Eii).

By (ii),(iv) and the above equation, we get

f(Eii)f(In)2 = f(In)2f(Eii).

(vi) By Eji ∈ E(1)ij , (In + Eji) ∈ E

(1)ij ( for any i, j ∈ [1, n], i 6= j) and f ∈ Γ,

we have the following two equations:

f(Eij)f(Eji)f(Eij) = f(Eij),

f(Eij)[f(In) + f(Eji)]f(Eij) = f(Eij).It follows from above two equations that f(Eij)f(In)f(Eij) = 0.

(vii) By In ∈ (In + Eii + Eij)(1) for any (i, j ∈ [1, n], i 6= j) and f ∈ Γ, we get

[f(In) + f(Eii) + f(Eij)]f(In)[f(In) + f(Eii) + f(Eij)]

= f(In) + f(Eii) + f(Eij),

that is,

[f(In) + f(Eii)]f(In)[f(In) + f(Eii)] + f(In)2f(Eij) + f(Eii)f(In)f(Eij)

+ f(Eij)f(In)2 + f(Eij)f(In)f(Eii) + f(Eij)f(In)f(Eij)

= f(In) + f(Eii) + f(Eij). (1)

By In ∈ (In + Eii)(1) and f ∈ Γ, we get

[f(In) + f(Eii)]f(In)[f(In) + f(Eii)] = f(In) + f(Eii). (2)

Substituting (2) into (1) yields

f(In)2f(Eij) + f(Eii)f(In)f(Eij) + f(Eij)f(In)2

+ f(Eij)f(In)f(Eii) + f(Eij)f(In)f(Eij) = f(Eij).(3)

By (vi), from (3) we finally have

f(In)2f(Eij)+f(Eii)f(In)f(Eij)+f(Eij)f(In)f(Eii)+f(Eij)f(In)2 = f(Eij).

�

Lemma 4. If f ∈ Γ, then rankf(Eii) = 1 and f(Eii)2 = f(Eii) for any i ∈[1, n].

Proof. Without loss of generality, we will prove rankf(E11) = 1 and f(E11)2 =f(E11). By (i) of Lemma 3, we have f(E11)3 = f(E11). From Lemma 2, thereexists a P ∈ GLn(R) such that

f(E11) = P ((

Ip Dp

O Ip

)⊕ Iq ⊕ Os)P−1,


where Dp = diag(d1, d2, · · · , dp), di 6= 0 for any i ∈ [1, n]. As f is invertible,2p + q ≥ 1.

Set f(E11) = P

(X OO O

)P−1 and f(Eks) = P

(Y1 Y2

Y3 Y4

)P−1 for any f(Eks)

∈ Mn(R), Eks 6= E11, where X =

Ip Dp OO Ip OO O Iq

∈ M2p+q(R) and Y1 ∈

M2p+q(R). By (ii) of Lemma 3, we get f(E11)f(Eks)f(E11) = 0 , therefore

f(E11)2f(Eks)f(E11)2 = 0 (4)

From f(E11)2 = P

(I2p+q O

O Os

)P−1 and (4), we have

f(Eks) = P

(O2p+q Y2

Y3 Y4

)P−1.

Hence,

f(A) = P

a11

0 a22 ∗...

. . . . . .0 . . . 0 all

∗ ∗

P−1

for any A ∈ Mn(R), where l = 2p + q. As f is invertible, l = 2p + q = 1, whichyields

f(E11) = P

(1 OO On−1

)P−1.

This equation shows that rankf(E11) = 1 and f(E11)2 = f(E11). �

Lemma 5. If f ∈ Γ, and f(In)2 = In, then the following equations hold:(i) f(In)f(Eii)+f(Eii)f(In)+f(In)f(Eii)f(In) = f(Eii) for any i ∈ [1, n];(ii) f(In)f(Eij)f(In) = f(Eij) and f(Eij)2 = 0 for any i, j ∈ [1, n], i 6= j.

Proof. (i) By (In + Eii) ∈ (In + Eii)(1) (for any i ∈ [1, n]) and f ∈ Γ, we have

[f(In) + f(Eii)]3 = f(In) + f(Eii).

From Lemma 4, we get f(Eii)2 = f(Eii). Substituting f(Eii)2 = f(Eii) andf(In)2 = In into the above equation, we have

f(In)f(Eii) + f(Eii)f(In) + f(Eii)f(In)f(Eii) + f(In)f(Eii)f(In) = 0.

From (iv) of Lemma 3, we have f(Eii)f(In)f(Eii) = f(Eii), and (i) has beenproven.

(ii) From (In + Eij) ∈ (In + Eij)(1) ( for any i, j ∈ [1, n], i 6= j) and f ∈ Γ, wehave

[f(In) + f(Eij)]3 = f(In) + f(Eij).

260 Changjiang Bu

From f(In)2 = In, (vi) of Lemma 3 and the above equation, we have

f(In)f(Eij)f(In) + f(In)f(Eij)2 + f(Eij) + f(Eij)2f(In) + f(Eij)3 = 0.(5)

Multiplying (5) on the right side by f(In)f(Eij), and from (vi) of Lemma 3 andf(In)2 = In, we get

f(In)f(Eij)2 + f(Eij)3 = 0. (6)

Substituting (6) into (5) gives

f(In)f(Eij)f(In) + f(Eij) + f(Eij)2f(In) = 0. (7)

Multiplying (7) on the right side by f(Eij), from (vi) of Lemma 3 , we get

f(Eij)2 = 0. (8)

Substituting (8) into (5) gives f(In)f(Eij)f(In) = f(Eij). �

Lemma 6. If f ∈ Γ, then f(In) = In.

Proof. By (iii) of Lemma 3, we have f(In)3 = f(In). By Lemma 2, without lossof generality, we can set

f(In) =

Ip Dp O OO Ip O OO O Iq OO O O Os

,

where Dp = diag(d1, d2, · · · , dp) ∈ Mp(R), di 6= 0 ( for any i ∈ [1, p]), 2p+q+s =n.

First, we will show s = 0. Set f(In) =(

A OO Os

), where A =

Ip Dp OO Ip OO O Iq

,

A2 = I2p+q . Set f(Eii) =(

B1 B2

B3 B4

)(for any i ∈ [1, n]), where B4 ∈ Ms(R).

Substituting the above f(Eii) and f(In) into (v) of Lemma 3 yields B2 = O, B3 =

O, i.e., f(Eii) =(

B1 OO B4

), and by (iv) of Lemma 3 , we have B4 = Os. Then

f(Eii) =(

B1 OO Os

)(9)

for any i ∈ [1, n].

Set f(Eij) =(

C1 C2

C3 C4

)(for any i, j ∈ [1, n], i 6= j), where C4 ∈ Ms(R).

Substituting f(Eij) =(

C1 C2

C3 C4

), f(In) =

(A OO Os

)and f(Eii) =

(B1 OO Os

)


into (vii) of Lemma 3, we get C4 = Os and

f(Eij) =(

C1 C2

C3 Os

). (10)

It follows from (9) and (10) that

f(A) =(

∆1 ∆2

∆3 Os

)

for any A ∈ Mn(R). As f is invertible, we get s = 0 , that is,

f(In) =

Ip Dp OO Ip OO O Iq

.

It remains to show that p = 0. Set

f(Eii) =

X1 X2 X3

X4 X5 X6

X7 X8 X9

for any i ∈ [1, n], where X1, X5 ∈ Mp(R). By substituting f(Eii) and f(In) =

Ip Dp OO Ip OO O Iq

into (i) of Lemma 5, we have X4 = Op and

f(Eii) =

X1 X2 X3

Op X5 X6

X7 X8 X9

(11)

for any i ∈ [1, n]. Set

f(Eij) =

Y1 Y2 Y3

Y4 Y5 Y6

Y7 Y8 Y9

for any i ∈ [1, n], i 6= j, where Y1, Y5 ∈ Mp(R). By substituting f(Eij) and

f(In) =

Ip Dp OO Ip OO O Iq

into (ii) of Lemma 5, we have Y4 = Op and

f(Eij) =

Y1 Y2 Y3

Op Y5 Y6

Y7 Y8 Y9

(12)

for any i ∈ [1, n], i 6= j.From (11) and (12), we have

f(A) =

∗1 ∗2 ∗3

Op ∗5 ∗6

7 ∗8 ∗9

262 Changjiang Bu

for any A ∈ Mn(R). As f is invertible, we have p = 0. It follows that f(In) =In. �

Lemma 7. If f ∈ Γ, then f preserves idempotence.

Proof. For any A2 = A ∈ Mn(R), we have A3 = A and (A + In)3 = A3 +A2 + A + In = A + In. So A ∈ A(1), (A + In) ∈ (A + In)(1). As f ∈ Γ, weget f(A + In)3 = f(A + In) and f(A)3 = f(A). From Lemma 6, f(In) = In,f(A)3 = f(A) and f(A+In)3 = f(A+In), we get f(A)2 = f(A), i.e. f preservesidempotence. �

Lemma 8. ([6, Theorem 3.2] and [13, Theorem]) Let R be a principal idealdomain of chR = 2, n > 1, f a invertible linear map on Mn(R) preservingidempotence, then f has one of the following forms:

(i) R 6= F2 and there exists a matrix P ∈ GLn(R) such that

f(A) = PAP−1 for all A ∈ Mn(R);

(ii) R 6= F2 and there exists a matrix P ∈ GLn(R) such that

f(A) = PAtP−1 for all A ∈ Mn(R);

(iii) R 6= F2, n is even and there exists a matrix P ∈ GLn(R) such that

f(A) = PAP−1 + (trA)In for all A ∈ Mn(R);

(iv) R 6= F2, n is even and there exists a matrix P ∈ GLn(R) such that

f(A) = PAtP−1 + (trA)In for all A ∈ Mn(R);

(v) R = F2 and there exist a matrix P ∈ GLn(R) and a linear functiong : Mn(R) → R with g(In) = 0 such that

f(A) = PAP−1 + g(A)In for all A ∈ Mn(R);

(vi) R = F2 and there exist a matrix P ∈ GLn(R) and a linear functiong : Mn(R) → R with g(In) = 0 such that

f(A) = PAtP−1 + g(A)In for all A ∈ Mn(R).

3. Main results

Theorem 1. If n > 1, then f ∈ Γ if and only if f satisfies one of the followingcases:

(i) there exists a matrix P ∈ GLn(R) such that

f(A) = PAP−1 for all A ∈ Mn(R),

(ii) there exists a matrix P ∈ GLn(R) such that

f(A) = PAtP−1 for all A ∈ Mn(R).


Proof. Obviously, if f satisfies one of the above two cases, then f ∈ Γ.Now we will prove for two cases that if f ∈ Γ then f satisfies (i) or (ii) of

Theorem 1.Case (1). R = F2. By Lemma 7, f preserves idempotence. So f satisfies one

of the cases (v),(vi) of Lemma 8.Let f be case (v) of Lemma 8. Then there exist a matrix P ∈ GLn(R) and

a linear function g : Mn(R) → R with g(In) = 0 such that

f(A) = PAP−1 + g(A)In (13)

for all A ∈ Mn(R).From f(Eij)2 = 0 in (ii) of Lemma 5 and (13), we have

g(Eij) = 0 (14)

for any i, j ∈ [1, n], i 6= j.1. When n > 2, by rankf(Eii) = 1 in Lemma 4 and (13), we get

g(Eii) = 0 (15)

for any i ∈ [1, n]. From (14) and (15), we have

f(A) = PAP−1 (16)

for all A ∈ Mn(R).2. When n = 2, by g(In) = 0, we get g(E11) = g(E22).If g(E11) = g(E22) = 0, by (14), we get

f(A) = PAP−1 (17)

for all A ∈ Mn(R).If g(E11) = g(E22) = 1, from (13) and (14), we have

f(A) = P1

(x1 x3

x2 x4

)P−1

1 = P1AtP−1

1 (18)

for all A =(

x1 x2

x3 x4

)∈ Mn(R), where P1 = P

(0 11 0

). From (16),(17) and

(18), then f has one of the forms of Theorem 1.Let f be case (vi) of Lemma 8, the discussion is the same as the above.Case (2). R 6= F2. From Lemma 7, we know that f is a linear map preserving

idempotence over Mn(R). So f satisfies one of the cases (i), (ii), (iii) and (iv)in Lemma 8.

Let f be the case (iii) of Lemma 8, then there exists a P ∈ GLn(R) such that

f(A) = PAP−1 + (trA)In (19)

for any A ∈ Mn(R), where n is even.1. When n ≥ 4 is even, then from (19), we have f(E11) = P (E11 + In)P−1.

This contradicts Lemma 4.2. When n = 2 , then from (19), we have

f(A) = P1AtP−1

1

264 Changjiang Bu

for any A =(

x1 x2

x3 x4

)∈ M2(R), where P1 = P

(0 11 0

). This shows that in

this case, f satisfies the case (ii) of Theorem 1.If f is the case (iv) of Lemma 8, the discussion is the same as the above. So,

if f ∈ Γ, then f is one of the formations (i) and (ii) of Theorem 1. �

Remark. Although when R = F2, an additive map is a linear map, but in thegeneral case the problem of additive map preserving {1}-inverses of matrices isstill meaningful and difficult even for the linear map without the assumption ofinvertibility. This is still an open problem.

The results we have obtained in this paper on the preserving problem of ma-trices over F2 may be helpful to the research of (0,1)-matrices in combinatorics.

Acknowledgements

The author would like to thank the referees for their valuable comments whichhave improve this paper significantly.

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Changjiang Bu received his BS from Heilongjiang University, china. Since 1985 he hasbeen at the Harbin Engineering University. In September of 2005, he received a professorfrom Heilongjiang Education Committee. His research interests focus on the theory ofmatrix algebra and the theory of graph.

Department of Applied Mathematics, School of Science, Harbin Engineering University,Harbin 150001, P. R. Chinae-mail: [email protected]

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Invertible linear maps preserving {1}-inverses of matrices over PID