K . Neusser Economics Letters 67 2000 273 –281
279
3. Some algebraic properties
n 3n
From now on I view the set of all n 3 n matrices, R , as a semigroup with respect to the matrix
3 n 3 n
n 3n
multiplication. The identity element in R is I . For any given E [ R
consider the subset of
n E
n 3 n
R defined in the following way:
n 3n
u 5
h
A AE 5 EA 5 A and
B [ R : AB 5 BA 5 E
j
E n 3n
2
If is nonempty, there exists A [ and B [ R such that AB 5 E. Thus E 5 EAB 5 AB 5 E so
E E
that E is idempotent and E [ . It is also straightforward to see that is closed under
E E
n 3n
multiplication and that EBE [ is the inverse of A in . In other words is a subgroup of R
E E
E
with identity element given by E. The subgroup is maximal because every subgroup containing E is in fact a subgroup of . This can be seen by noting that E must be the identity of any group in
E
which it lies, so that any A[ satisfies the defining conditions for Drazin, 1958: 513. This leads
E 4
to a family of disjoint maximal subgroups h j where E runs over all idempotent matrices. This
E n 3 n
family does not form a partition of R because, as is shown in Theorem 4, matrices with index
strictly greater than one do not belong to any subgroup.
Theorem 4. Under Assumption 1, Ind
P 5 1 if and only if P belongs to some subgroup.
g
Proof.
⇐ : Suppose
P belongs to some subgroup . Then P must have an inverse P in . This
g g
g g
g g
g [
inverse satisfies PP 5 P P, P PP 5 P , PP P 5 P. Therefore P
is nothing but P , the
5
group inverse of P. Therefore IndP 51, given Assumption 1.
n 3n r 3r
⇒ : If Ind
P 51, there exist invertible matrices P [ R and I 2 J [ R
such that
r n 2r
21
P 5 P P
S D
I 2 J
r
Define the set of matrices as
[
PP 21
r 3r n 2r
u 5 P
P M [ R
, rank M 5 r
j
.
[
s d
H S D
PP
M It is easy to see that
P [ and that
is a group. h
[ [
PP PP
Theorem 5. 5
[ [
PP PP
[
Proof. Any element in satisfies the definition of
. Moreover, PP
is idempotent and
[ [
PP PP
[
PP [ , thus
is a subgroup of .
[ [
[
PP PP
PP
3
A semigroup is just a set together with an associative binary operation.
4
The fact that the groups are disjoint can be established as follows. Let G and G be two subgroups such that E ± F and
E F
G G ±[. Then for any A [ G G , there exists B and C such that EF 5 BAF 5 BA 5 E 5 AB 5 FAB 5 FE and
E F
E F
EF 5 EAC 5 AC 5 F 5 CA 5 CAE 5 FE. This implies that F [ G and E [ G . The two groups must therefore be equal
E F
because they are both maximal subgroups.
5 [
This is the justification of the name ‘group inverse’ for A .
280 K
. Neusser Economics Letters 67 2000 273 –281
21 q 3q
[ [
If A is any A [ , A 5 Q
Q with B [ R
and rank B 5 q. Because AA 5 PP
[
S D
PP
B these matrices are similar and therefore r 5 q. Furthermore there exists an invertible matrix R such
[ [
that Q 5 PR. Observing that the definition of implies
PP A 5 A and APP 5 A, it is easy to
[
PP
verify that A can be written as
21
A 5 P
21
P
S D
R BR
22 22
where R is the 2,2-element of the appropriately partitioned matrix R. Thus A [
. h
[
22 PP
[ 2
[ 2
Note that the set of matrices hE,P,P ,P ,P , . . . j constitutes a commutative Abelian
[ [
subgroup of . The identity is again the matrix E 5 P
P 5 PP . The subgroup is cyclic and
E
generated by the positive and negative powers of P where negative powers are interpreted as powers
[
of P .
Following Grillet 1995: 27, we introduce a partial order on the set of idempotent matrices:
Theorem 6. Partial order: The binary relation defined by E F
⇔ EF 5 FE 5 E represents a
partial order on the set of all idempotent matrices.
Proof. The reflexivity follows from E being idempotent. Symmetry is also immediate from the definition. Finally, if E F and F G then EG 5 EFG 5 EF 5 E 5 FE 5 GFE 5 GE. Thus E G
and is also transitive.
Let us translate this definition into our context. The definition of E F implies that E and F commute. Moreover, E and F are diagonizable matrices, consequently they must share the same
21
eigenvectors Strang, 1988: 259. E and F are therefore both of the form P P
. They may
S D
I
r
differ only with respect to r. Thus for any two idempotent matrices E and F, E F is equivalent to E and F share the same eigenvectors and r 5rank E r 5rank F. The cointegration space
E F
corresponding to E is therefore included in the one corresponding to F.
4. Conclusion