1. Introduction
Several of the useful properties of the group, Drazin and Koliha-Drazin inverses can be related to their spectral characterizations. Some of these can be traced to the property of being commuting outer inverses.
Let
be the set of bounded linear operators on a Banach space
, and let
. We denote the range by
and the null space of
by
.
Let
and
be closed subspaces of
. The outer inverse with prescribed range
and null space
, denoted
is the unique operator
which satisfies:

There is some advantage in prescribing the null space and range of an outer inverse by means of a third operator. In doing so, we will use the notion of invertibility along an element introduced by X. Mary ([1]). We say
is invertible along
if there exists
such that

In this case, the inverse along
is unique and we write
.
From
we have that BA and AB are projections such that
and
. Thus, we are effectively prescribing the range of the projection
and the null space of the projection
.
One of the useful properties of a generalized inverse is that, although the operator is not invertible, there is a subspace for which the reduction of the operator to that subspace is indeed invertible:
Theorem 1. ([2, Theorem 2]) Let
be nonzero operators. The following statements are equivalent.
1.
is invertible along
.
2.
is a closed and complemented subspace of
,
is closed such that
and the reduction
is invertible.
Recall an operator
is said to be group invertible if there exists
such that

In this case, such
is unique and we write
for the group inverse of
.
Proposition 2. ([2, Theorem 3]) If
is invertible along
, then AT and TA are group invertible and
.
Example 3. Let
the space of square-summable sequences. Let
be defined by
and
.
Then
is invertible along
with
.
In the following section, we study an operator
such that
is invertible along
with
. Then, in Section 3 we study some projections related to the outer inverse with prescribed range and null space. Finally, in Section 4 we specialize to spectral projections, covering results from Dajić and Koliha ([3]).
2. Invertibility along a Commuting Operator
Proposition 4. Let
be invertible along
. If
, then
.
Proof. From Proposition 2 we have:

If
is invertible along
, then we have the following matrix form ([2]):

where
is invertible and
is a complement of
, that is,
.
When
and
commute, we can say a little more:
Theorem 5. Let
be invertible along
and
. Then there exist an invertible operator
on
and an operator
on
such that

and 
Proof. Suppose
is invertible along
and
. Then by Proposition 4
. Thus, since
is a projection, we have that
.
Since

and

we also have

and hence we can consider the following matrix decomposition of
:

In this case,
,
, is invertible. Indeed, to see that it is onto note that since
and
, we have
and hence
. To see that it is also 1-1, let
. Since
, there exists
such that
. Then,
.
Moreover, since
, subspaces
and
are
-invariant and
maps
onto
, we get
. Thus,
and clearly
.
Example 6. Let
the space of square-summable sequences. Let
be defined by


Then it is easy to verify that
is the operator such that

It is clear that
, but we have

3. Projections
Commuting outer inverses are naturally linked to projections.
Proposition 7. Let
. If
is invertible along
and
, then there exists a bounded projection
such that
is invertible along
.
Proof. From Theorem 5 we have that if
is invertible along
and
, then

Thus, there exists a bounded projection
such that
and
. Hence,
is invertible along
.
For a sort of converse, we give a necessary condition in Theorem 9.
Example 8. An operator
and a projection
such that
is invertible along
and
.
Let
be the set of two by two matrices with real entries. Let
be the (rotation) matrix defined by

and let
be the projection defined by

Then, it is easy to check that
is group invertible, with group inverse
.
We have

Thus,
is invertible along
but since
, we see
. Notice that
.
Theorem 9. Let
be a projection and suppose
is invertible along
. If
, then
.
Proof. Since
is invertible along
,
has the following matrix form [2, Corollary 1]:

where
is invertible and
is the complement of
, that is,
.
Since
is a complement for
, we can take
. Then, we have

Now, suppose
. From Theorem 1 we know that
is invertiblewhich implies
. Thus,

It follows
.
Note that the theorem above together with Proposition 4 implies that if
, then
. However, we can prove:
Proposition 10. Let
be a projection, and suppose
is invertible along
. Then
.
Proof. Using Proposition 4,

Example 11. An operator
and a projection
such that
but
is not invertible along
.
Let
and let
be defined by


Then


However, the reduction
can not be invertible, and by Theorem 1
is not invertible along
.
Theorem 12. Let
be a projection and
be such that
. Then,
is invertible along
if and only if
and
.
Proof. Suppose
is invertible along
. Since
and from Proposition 4 and the definition of
we have

Similarly, from
Proposition 4 and the definition of
we get

Conversely, suppose
and
. We use Theorem 1. The reduction
is clearly onto. From
and
we get that it is also 1-1.To see that
is closed and
we will show
. It is clear that
. For the other inclusion, let
. Since
, there exists
such that
. Then, from
it follows that
. Thus,
. Finally, it is clear that
is closed and complemented. Hence
is invertible along
.
4. Spectral Projections
Recall the spectrum
of an operator
is the set 
Suppose
is invertible along
and
. Then we have the matrix form
.
From
and
are invariant under
, we have:

Since
is invertible but
is not, we know that
.
Note that from
and the matrix form of
, there exists a projection
such that
,
and
. Thus, without loss of generality, we can suppose
is invertible along a projection
. A very important class of projections which commutes with
is a class of spectral projections, which we now discuss.
The resolvent set
of
is
and for
the resolvent function
is

A subset
is said to be a spectral set of
if
and
are both closed in
. For a spectral set
of
,
the spectral projection associated with
is defined by

where
is a Cauchy contour that separates
from
.
If
is a point of the resolvent set or an isolated point of the spectrum
, then the operator
is called quasipolar. Let
be quasipolar and let
be the spectral projection associated with the spectral set
, then ([4]):

Let
. If
, then we say that
is a quasinilpotent operator. Recall that
is nilpotent if
for some
, and nilpotent operators are quasinilpotent.
Quasipolar operators are generalized invertible in the sense of Koliha: an operator
is Koliha-Drazin invertible if there exists
such that
is quasinilpotent, 

In this case, by Lemma 2.4 of [5], Koliha-Drazin inverse is unique and we write
.
An operator
is Koliha-Drazin invertible if and only if 0 is an isolated point of
. If
is a pole of the resolvent of order
, then
is Drazin invertible with Drazin index
. If 0 is a simple pole then it is group invertible.
As noted above, the Koliha-Drazin is a particular case when we consider the spectral set
. For the general case when
is a spectral set such that
, Dajić and Koliha have defined a generalized inverse and studied its properties ([3]).
Theorem 13. Let
and
be a spectral set for
. If
then
is invertible along
.
Proof. Let
. Then
and
are closed and
. Now, since 
is
-invariant,
and
we have that
is invertible. Thus,
is closed,
and
is invertible. Therefore, by Theorem 1
is invertible along
.
Corollary 14. Let
and
be a spectral set for
. If
then
is invertible along
.
Proof. If
, then
. From the theorem above,
is invertible along
.
5. Acknowledgements
This research is partially supported by Grant-in-Aid Scientific Research No.24540195.