AbstractIt is proved that for n ⩾ 3, every n ×n matrix with integer entries and determinant zero is the product of 36n + 217 idempotent matrices with integer entries

Laffey, Thomas J.

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Factorizations of Integer Matrices 
as Products of Idempotents and Nilpotents 
Thomas J. Laffey 
Mathematics Department 
University College, Belfield 
Dublin 4, Ireland 
Submitted by Daniel Hershkowitz 
ABSTRACT 
It is proved that for n >/3, every n × n matrix with integer entries and determi- 
nant zero is the product of 36n +217 idempotent matrices with integer entries. 
INTRODUCTION 
Let F be a field, and M. (F)  the ring of n × n matrices with entries in F. 
Generalizing an earlier result of Erdos [3], Ballantine [1] proved that ff 
A ~ Mn(F ) is not invertible, then A is the product of n idempotents in 
Mn(F ). For complex matrices, P. Y. Wu [11] proved that under the same 
hypotheses, A is the product of two nflpotent matrices in M.(F) except 
when n = 2 and A is a nonzero nilpotent matrix, and his result can be 
extended to all fields using the methods of Sourour [9] or the author [6, (5.3)]. 
Let M.(Z) be the ring of n × n integer matrices, and let A ~ M.(Z) with 
det A = 0. We have shown [5] that A is the product of idempotent matrices 
in M.(Z), and it follows from this that A is the product of nilpotent elements 
in M.(Z). In this paper we consider whether the number of idempotent or 
nilpotent matrices required can be bounded as a function of n. We prove 
LINEAR ALGEBRA AND ITS APPLICATIONS 120:81-93 (1989) 81 
© Elsevier Science Publishing Co., Inc., 1989 
655 Avenue of the Americas, New York, NY 10010 0024-3795/89/$3.50 
82 THOMAS J. LAFFEY 
THEOREM 1. Given any positive integer N, there exists an element 
A ~ Ms(Z ) which is the product of  N but no fewer idempotents in M2(7), 
and there exists an element B E M2(Z ) which is the product of  N but no 
fewer nilpotent matrices in M~(Z). 
THEOREM 2. For n >t 3, every element A ~ M,(Z) with det A = 0 is the 
product of  36n+217 idempotents and 72n+434 nilpotent matrices in 
M2(Z). 
1. THE 2X2 CASE 
Let 
A=(  p 0 0 q)~M~(Z) ,  
where the highest common factor (p ,q )= 1. Suppose A = AlE  1, where 
E x = E~ and det A 1 = det E x = 0. Then A 1 has rank one, and thus its second 
row is l inearly dependent (over the field of rational numbers) on its first row 
[since (p q)  4= (0 0)], and thus 
° I0 
for some Pl, ql ~ Z, and since (p,  q) = 1, (Pl,  ql)  = 1 also. Let 
E=(  a~ b'l I 
c; d;/" 
Note that d~ = 1 - a~, since E~ = E 1 ~ I. There exist integers u, v with 
(u, v) = 1 and ua' x = Vb'l, uc~ = vd[. Note that p = pla~ + qlc;, q = plb'l + 
qld~, and a[=a[S+b~c~. Hence pb[=qa~. Hence a[=pa x, b'n=qal, 
c~ = pc 1, d~ = qc I for some integers a 1, c 1. Note also that pa 1 + qc 1 = a' 1 + 
d~ = 1 and p la l  + qlcx = 1. Since pa I + qc I = 1 and plal + qlcl = 1, we have 
P = Pl + rlcl, q = ql - rlal for some integer r1. So 
(P "/(Pl ql / 
- c~ a~ 1 - -  c l  a l  " 
IDEMPOTENT AND NILPOTENT FACTORIZATIONS 
Now repeating the argument, writing 
(01 01)=(02  O~)( a~ b~l c~ a~ t' 
83 
where d~ = 1 - a~ and a~d~ = b~c~, we get a~ = p la2 ,  b~ = qla~, c~ = pxca, 
d~ = qlc2 for some integers aa, c a, and 
P2 a 2 + q~c2 = 1, 
plaz + qlcz = 1. 
Since plal  + qlCl = 1 also, it follows that a I = a~ + sxq 1, c x = c 2 - slpx. 
Hence 
Thus 
-- Cl al S l  - -  C2 an " 
(L~ aq)=(10 - r~) (sl~ 0 1 1)( a":). 
Again there is an integer 2 such that 
Pl  = Pa + r~c2, 
and 
Suppose now that 
ql  = q2 -- r2a~, 
(pl ql / (1 )(p2 
- c  a aa = 0 I r2 -c  a a2 " 
for some k >t i and idempotents E 1 ..... E k with Ei 2 = E i ~: I. Note that E 1 is 
of the form 
(1 0) 
for some integer t. 
84 THOMAS J. LAFFEY 
Proceeding as above, we find that 
- -  c 1 a 1 
is the product of 2k - 1 elementary matrices (i.e. of the form 
1 a 1 0) 
for some integers a, b). Suppose 
(/gx q)y ~SL(2,Z) .  
Then from py - qx  = 1 and  pa  1 + qc  I = 1 we obtain 
y = a l  + qZ, x= - c l  + pz  
and thus 
Hence 
(Px q )=(1  0) (  / 9 q 
Z -- C1 a l ) '  
P q 
x yl 
is the product of 2k elementary matrices. But given any N/> 1, there exist 
elements in SL(2, Z) which cannot be expressed as the product of fewer than 
N elementary matrices--this may be proved using the theory of continued 
fractions (Stark [10]). So, given any N>~ 1, there exists A ~ M2(Z ) with 
det A = 0 which is not the product of fewer than N idempotents. Note that if 
E = E2~ M2(Z), then E is similar [over GL(2,Z)] to a diagonal matrix [5] 
and 
o o)(O o (o' °):(o lO) 
is the product of two nilpotent matrices. 
IDEMPOTENT AND NILPOTENT FACTORIZATIONS 
If B ~. M2(Z ) is nilpotent, then B is similar [over GL(2, Z)] to 
(0 
for some integer b, and 
(o to  o)(O o t 
85 
is the product of two idempotents. Hence every A ~ M2(Z ) with det A = 0 is 
the product of a finite number of nilpotent elements in Mz(Z) ,  but the 
minimum number required is not bounded independently of A. Theorem 1 
now follows immediately. 
2. THE CASE n >/3 
Let A ~ SL(n,Z),  n >/3. Then A is similar [via an element of GL(n,Z)]  
to a matrix B = (bij) where bii = 0 if j > i + 1. This follows easily, using 
induction, from the fact that if x l . . . . .  x m (m >~ 1) are integers with highest 
common factor 1, there exists an element of SL(m,Z) with first row (x I x 2 
• • • Xm). Assume then that A = (a i j ) ,  where aij  = 0 if j > i + 1. Since the 
highest common factor of at l ,  al2 is 1, we can choose integers x, g with 
a l lX  + amy = 1. Consider the product 
[i °-1 1 0 0 1 u 
A 0 0 = v' B 1 
0 In_ 3 
(say) 
lY ,°ll[0  °l]I xu] 0 In_ 1 ' 
where A 1 ~ SL(n - 1,Z). 
Note also that 
( i0  1 ) (00  1 ) (100)  1 o ~01 o y 10  
0 0 1 0 0 -x  0 1 
86 
and 
THOMAS J. LAFFEY 
1 1 0)( 0 1 " (o 1) (  )(1 
LEMMA 1. Let A~SL(n ,Z)  (n>~3). Then A is the product ( I+  
R1) . . . ( I  + Rk), where k = 5n + 31 and R 1 . . . . .  R k have trace 0 and rank at 
most one. 
Proof. Applying the identities of the last paragraph, we find 
1 0 )S2S3S455, A = S 1 0 A 1 
where S i is of the form I + T i in which trace T i is 0 and rank T i is at most one 
(i = 1,2,3,4,5). 
If n = 3, by Lemma 2 of Carter and Keller [2], 
(o Ol) 
is the product of 41 matrices of the desired form. If n > 3, then by induction 
A 1 is the product of 5(n - 3)+41 matrices of the desired form, and hence A 
is the product of 5(n - 2 )+41 = 5n +31 elements I + R where traceR = 0 
and rank R ~< 1. This completes the proof. • 
We next observe 
LEMMA 2. Let D = diag(dx, d g. . . . .  d n -1, O) ~ M n( Z ) be a singular diag- 
onal matrix. Then D is the product o f  n idempotents in Mn(Z). 
Proof. Note that 
0 
' o (d O) (o  o 0 0 )" 
Suppose n > 2 and that the result holds for n -  1. Suppose that 
IDEMPOTENT AND NILPOTENT FACTORIZATIONS 87 
diag(d2 . . . . .  dn_ l ,0  ) = EsE3... E., where E~ = Ei ~ M._  I(Z) with det E i = 
0 (i = 2 . . . . .  n), We have 
(d  1 0 ) (1  0 ) ( 1  0 )  
D= E 2 0 E 3 "'" 0 E .  " 
Now by [5], E2 is integrally similar to a diagonal matrix diag(e 2.. . . .  e . _ l ,0  ) 
where e i ~ {0,1} (i = 2 . . . . .  n - 1). Thus 
0 E 2 
is integrally similar to 
where E~ = E 1, and hence 
:)-1 
is the product of two idempotents. Hence D is the product of n idempotents 
in M.(Z),  as required. • 
Let A ~ Mn(Z ) (n >/3) with det A = 0. There exist P, Q ~ SL(n,Z)  with 
D = PAQ diagonal and having its last column zero. Then 
A = P-tDQ-1 
= p-1Q-~(QDQ-1) 
=(QP)-I(QDQ-1). 
By l_emma 2, QDQ -1 is the product of n idempotents and (Qp)-I is the 
product ( I  + R l ) .  • • ( I  + Rk) (k -- 5n +31), where traceR i = 0, rank R i ~< 1 
(i = 1, 2 . . . . .  k) by Lemma 1. We now prove. 
88 THOMAS J. LAFFEY 
LEMMa 3. Let E = E 2 ~ Mn(Z ) have rank n - 1, and let R have trace 0 
and rank one. Then (I  + R)E  is the product o f  eight idempotents in M,(Z).  
Proof• Using a similarity over GL(n ,Z) ,  we may assume 
and R = (rq). Write 
R l l  R12 
R= R21 R22 , 
where R u is (n -1 )×(n-1) .  Suppose Ru~0.  Then using a further 
similarity in GL(n ,Z) ,  we may assume R u = aE u + bE21 for some integers 
a, b. Since R has rank one, it follows that R12 has zero entries in positions 
3 . . . . .  n - 1 and R21 has zero entries in positions 2 . . . . .  n - 1. Thus 
R = 
a 0 0 . . .  0 Yl 
b 0 0 . . .  0 Y2 
0 0 0 . . .  0 0 
0 0 0 . . .  0 0 
c 0 0 . . .  0 -a  
where a 2 + cy 1 = 0. Hence 
( I+R)E= 
l+a  0 0 
b 1 0 
0 0 1 
0 0 0 
c 0 0 
Oi) oo  0 
I 
0 
Thus we are led to consider the factorization of 
l+a  0 ! )  
b 1 
c 0 
IDEMPOTENT AND NILPOTENT FACTORIZATIONS 
into idempotents. Observe that 
89 
and that 
(X+a 0 i)(1+° 0 0)(1 0 i) 
b 1 = 0 1 0 b 1 , 
c 0 c 0 0 0 0 
(l o i ) ( lO!)( lOO) 
b 1 = 0 1 0 1 0 
0 0 0 0 b 0 0 
is the product of two idempotents. Consider 
l+a  0 0 
0 1 0 
c 0 0 
This is similar over GL(3, 7) to 
(a+a 0 i)(l+a 0 !)(i 0 !) c 0 = c 0 0 • 
0 0 0 1 0 
Now, recalling that a 2 + cg i = 0, consider 
l+a  
C 
0 
o) gl 0 l+a  0 0 1 -a  c 0 
0 0 1 
(lO o (x a o 
= 0 0 0 1 -a  0 
0 1 0 0 1 
(lO o)(l+a o /
= 0 0 0 c 1 a 0 
0 1 0 0 0 0 
l+a  {11 O) 
c 1 -a  0 
0 0 1 
90 THOMAS J. LAFFEY 
Now 
l+a  Yl ) 
c 1 -a  
is similar over GL(2, Z) to 
l d  / 
I0 1 
for some integer d, since a ~ + cy  x = 0, and hence 
l+a  Yl 0) 
c 1 -a  0 
0 0 0 
is the product of two idempotents. Also 
ti0i)0x 
is the product of two idempotents. Hence 
l+a  0 ! )  
c 0 
0 1 
is the product of four idempotents, and thus 
l+a  0 i )  0 1 
c 0 
is the product of six idempotents, and 
l+a  0 i )  b 1 
c 0 
is the product of eight idempotents. 
IDEMPOTENT AND NILPOTENT FACTORIZATIONS 91 
If Rl l  = 0, then 
r 1 ra • • • r ,_ l  
which is idempotent. This completes the proof of Lemma 3. 
To prove Theorem 2, we find P ,Q~SL(n ,Z)  with PAQ= D diagonal 
with last column zero. Using Lemma 2, we factorize D as a product of n 
idempotents. Note that the first factor, E 1 say, can be assumed to have rank 
n - 1. Let S = (Qp) -  1. Then, by Lemma 1, S = ( I  + R1).  • • ( I  + Rk), where 
k = 5n +31.  Let E = QEtQ -1. By Lemma 3, (I + Rk)E is the product of 
eight (or fewer) idempotents, all of which can be taken to have rank n - 1. 
Taking the leading one, E 0 say, and applying Lemma 3 to ( I  + Rk_l)E0 etc., 
we eventually find A is the product of (n -  1 )+7(5n +31)+ 1 = 36n +217 
idempotents. If E ~ = E ~ Mn(Z ) with rank E = r, then E is similar over 
GL(n ,  Z) to 
:) 
But note that 
(01o ...... i//io 01/ 0 0 1 0 . . .  0 . . .  0 (:r 1 o 
0 0 . . .  1 
is the product of two nilpotent matrices. Hence the result for products of 
nilpotents follows from the result of idempotents. 
RV.MAaXS. 
(1) These bounds are not likely to be best possible. However, in the case 
of products of idempotents, a bound independent of n is impossible, since 
over the rational field, there are singular matrices which cannot be expressed 
as a product of fewer than n idempotents [1]. We do not know ff there is an 
92 THOMAS J. LAFFEY 
absolute bound (independent of n) on the nmnber of nilpotents required in 
Theorem 2. 
(2) A generalization of the results in this paper from the ring of integers 
to the ring of algebraic integers in an algebraic number field will be discussed 
elsewhere. 
(3) We have shown in [5] that if R is a Euclidean ring and A ~ Mn(R ) 
with detA =0,  then A is the product of idempotents in M,(R).  The 
question of the boundedness of the number equired when n > 2 and R is 
such that SL(n, R) does not have bounded word length for expressions of 
elements as products of elementary matrices is, as far as I am aware, still 
open. By a result of van der Kallen [4], if F is a field with infinite 
transcendence d gree over its prime subfield, then R = F[x] has this prop- 
erty (for every n). 
(4) The problem of writing an element A of SL(n,7/) [or in general 
of SL(n, R), where R is a commutative ring with identity] as a product of 
elementary matrices arising in K-theory. It is also related to the question 
of writing A as a product of commutators. The latter question is discussed in 
the recent paper of Newman [8]. 
This is an improved version of  the results presented at the conference; the 
original bounds were cubic in n. This improvement was carried out at 
the Technion after the conference, and I wish to express my gratitude to the 
members of  the Department of  Mathematics there for making my visit most 
fruitful and enjoyable. 
REFERENCES 
1 C.S. Ballantine, Products of idempotent matrices, Linear Algebra Appl. 19:81-86 
(1978). 
2 D. Carter and G. Keller, Elementary expressions for unimodular matrices, Comm. 
Algebra 12:379-389 (1984). 
3 J.A. Erdos, On products of idempotent matrices, G/asgow Math. 1. 8:118-122 
(1967). 
4 W. van der Kallen, SLa(C[x]) does not have bounded word length, in Algebraic 
K-theory, Lecture Notes in Math. Vol. 966, Springer-Verlag, 1982, pp. 357-381. 
5 T.J. Laffey, Products of idempotent matrices, Linear and Multilinear Algebra 
14:309-314 (1983). 
6 T.J. Laffey, Factorization of matrices involving symmetric matrices and involu- 
tions, in Current Trends in Matrix Theory, Proceedings of the Third Auburn 
Matrix Conference/( North Holland, New York, 1987. 
IDEMPOTENT AND NILPOTENT FACTORIZATIONS 93 
7 M. Newman, Integral Matrices, Academic, New York, 1972. 
8 M. Newman, Unimodular commutators, Proc. Amer. Math. Soc. 101:605-609 
(1987). 
9 A. R. Sourour, A factorization theorem for matrices, Linear and Multilinear 
Algebra 19:141-147 (1986). 
10 H.N. Stark, An Introduction to Number Theory, Markham, Chicago, 1970. 
11 Pei Yuan Wu, Products of nilpotent matrices, Linear Algebra Alrpl. 96:227-232 
(1987). 
Received 21une 1988; final manuscript accepted 21 November 1988 


Factorizations of integer matrices as products of idempotents and nilpotents 

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Factorizations of integer matrices as products of idempotents and nilpotents

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