The Drazin inverses of infinite triangular matrices and their linear preservers
We consider the ring of all infinite $(N \times N)$ upper triangular matrices over a field $F$. We give a description of elements that are Drazin invertible in this ring. In the case where $F$ is such that $\mathrm{c}\mathrm{h}\mathrm{a}\mathrm{r}(F) \not = 2$ and $| F| > 4$, we find the fo...
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Ukrains’kyi Matematychnyi Zhurnal| _version_ | 1860507379090587648 |
|---|---|
| author | Słowik, R. Словік, Р. |
| author_facet | Słowik, R. Словік, Р. |
| author_sort | Słowik, R. |
| baseUrl_str | https://umj.imath.kiev.ua/index.php/umj/oai |
| collection | OJS |
| datestamp_date | 2019-12-05T09:19:04Z |
| description | We consider the ring of all infinite $(N \times N)$ upper triangular matrices over a field $F$. We give a description of elements
that are Drazin invertible in this ring. In the case where $F$ is such that $\mathrm{c}\mathrm{h}\mathrm{a}\mathrm{r}(F) \not = 2$ and $| F| > 4$, we find the form of
linear preservers for the Drazin inverses. |
| first_indexed | 2026-03-24T02:08:22Z |
| format | Article |
| fulltext |
UDC 512.5
R. Słowik (Inst. Math., Silesian Univ. Technology, Gliwice, Poland)
THE DRAZIN INVERSES OF INFINITE TRIANGULAR MATRICES
AND THEIR LINEAR PRESERVERS
ОБЕРНЕНI МАТРИЦI ДРАЗIНА ДЛЯ НЕСКIНЧЕННИХ
ТРИКУТНИХ МАТРИЦЬ ТА ЇХ ЛIНIЙНИХ ЗБЕРIГАЧIВ
We consider the ring of all infinite (\BbbN \times \BbbN ) upper triangular matrices over a field F. We give a description of elements
that are Drazin invertible in this ring. In the case where F is such that \mathrm{c}\mathrm{h}\mathrm{a}\mathrm{r}(F ) \not = 2 and | F | > 4, we find the form of
linear preservers for the Drazin inverses.
Розглянуто кiльце всiх нескiнченних (\BbbN \times \BbbN ) верхнiх трикутних матриць над полем F. Наведено опис елементiв,
що є оборотними за Дразiним у цьому кiльцi. У випадку, коли F є таким, що \mathrm{c}\mathrm{h}\mathrm{a}\mathrm{r}(F ) \not = 2 та | F | > 4, знайдено
форму лiнiйних зберiгачiв для обернених матриць Дразiна.
1. Introduction. Let R be a ring with identity. One knows that an arbitrary element x \in R does not
have to be invertible in R. However, sometimes elements of a ring have some generalized inverses.
The most commonly known (see, for instance, [1, 2, 7]) are the Moore – Penrose inverse and the
Drazin inverse. It was proved (in [17] and [11]) that in the ring of all n\times n, n \in \BbbN , matrices over a
field such inverses always exist and are unique.
In this paper we wish to focus on the Drazin inverses.
We say that xD is the Drazin inverse of x if the three following hold:
xxD = xDx, (1a)
xDxxD = xD, (1b)
\exists k \in \BbbN \cup \{ 0\} : xk+1xD = xk. (1c)
The existence and form of the Drazin inverse is particularly interesting for the case of matrix
rings. Such inverses have applications, for example in Markov chains, singular differential equations
and iterative methods.
The problem of expressing the Drazin inverse of a matrix was studied by many authors (to name
only [10, 14, 21, 24]), especially in the case where the given matrix is a sum or a difference of some
matrices whose Drazin inverses are known (see, for example, [9, 12, 13, 15]).
In this article we will develop the results about such inverses and consider the problem of
existence of the Drazin inverse in the ring of upper triangular infinite (\BbbN \times \BbbN ) matrices over a field.
A similar problem and its connection to Markov chains was considered in [6]. Let us also note that
the Moore – Penrose inverses of infinite matrices were investigated in [18, 19].
Our first result is the following.
Theorem 1. Let F be a field. Then x \in \scrT \infty (F ) is Drazin invertible if and only if there exists
k \in \BbbN such that the following condition is fulfilled. If xnn = 0 for some n \geq 2 and i(n) is a minimal
number for which xi(n)j \not = 0 for some j, then either
c\bigcirc R. SŁOWIK, 2018
534 ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 4
THE DRAZIN INVERSES OF INFINITE TRIANGULAR MATRICES AND THEIR LINEAR PRESERVERS 535
(1) (xk)pn = 0 for all i(n) < p \leq n,
or
(2) there exists i(n) < p \leq n such that (xk)pn \not = 0, but in this case (xk+1)p,n+1 \not = 0.
Next, we will move to the problem connected to the Drazin inverses and some linear maps.
We call \phi a Drazin inverses preserver if
xD = y \Rightarrow (\phi (x))D = \phi (y) (2)
for all x, y from a given ring. For the case of finite dimensional matrices such preservers were
studied in [5, 8, 23].
Before presenting the description of such preservers we need to introduce the following type of
maps.
We will say that a map \phi : \scrT \infty (F ) \rightarrow \scrT \infty (F ) is a separable sum, if there exist nonzero maps
\phi 1, \phi 2, . . . , \phi n, n \geq 2, or \phi 1, \phi 2, \phi 3, . . . , such that
\phi (x) =
\sum
i
\phi i(x) for all x \in \scrT \infty (F )
and
\phi i(x)\phi j(y) = \phi j(y)\phi i(x) = 0 for all i, j and x, y \in \scrT \infty (F ).
Now we can formulate the following theorem.
Theorem 2. Assume that F is a field such that \mathrm{c}\mathrm{h}\mathrm{a}\mathrm{r}(F ) \not = 2 and | F | > 4. If \phi : \scrT \infty (F ) \rightarrow
\rightarrow \scrT \infty (F ) is a linear map satisfying (2), then \phi is a separable map of \phi 1 and \phi 2, where \phi 1 and
- \phi 2 are idempotent preservers.
2. Proofs of results. We start with presenting the notation used in the paper.
We denote by \scrM n\times m(F ) the ring of all n\times m matrices over F.
We write enm for the infinite matrix such that
(enm)kl =
\left\{ 1, if k = n and l = m,
0, otherwise.
The symbols en and e\infty are used for, respectively, n\times n and infinite, identity matrices.
By xT we denote the matrix transposed to x.
By xy, where y is an invertible matrix, we mean y - 1xy.
If an infinite matrix x may have nonzero coefficients only in the positions (i, j) with i \in I,
j \in J, then we will write
x =
\sum
i\in I, : j\in J
xijeij .
In our considerations there will also appear some subrings of \scrT \infty (F ).
By \scrN \scrT \infty (F ) we understand the subring of \scrT \infty (F ) of all matrices x such that xnn = 0 for
all n \in \BbbN , whereas by \scrN \scrT k
\infty (F ) (where k \in \BbbN ) we mean a subring of \scrT \infty (F ) of all x such that
xn,n+i = 0 holds for all n \in \BbbN and 0 \leq i \leq k. Additionally we put \scrN \scrT 0
\infty (F ) = \scrN \scrT \infty (F ).
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 4
536 R. SŁOWIK
2.1. The Drazin invertibility. In this section we study the Drazin invertibility of elements of
\scrT \infty (F ).
The Drazin inverses of upper triangular finite matrices were studied in [3, 16]. As we have
already mentioned, it was proved that every (not only) triangular matrix posseses the Drazin inverse.
More precisely, if x is of the form \biggl(
a b
0 c
\biggr)
,
then
xD =
\Biggl(
aD B
0 cD
\Biggr)
with
B =
\bigl(
aD
\bigr) 2\left( ind(c) - 1\sum
i=0
(aD)ibci
\right) \bigl( en - k - ccD
\bigr)
+
\bigl(
ek - aaD
\bigr) \left( ind(a) - 1\sum
i=0
aib
\bigl(
cD
\bigr) i\right) \bigl( cD\bigr) 2 - aDbcD,
where \mathrm{i}\mathrm{n}\mathrm{d}(x) denotes the index of x, which is a finite number associated with x and will be
considered further later.
In the case of infinite matrices, it can be checked that this formula also holds, but, clearly, under
the assumption that the infinite matrix c is Drazin invertible. Obviously, this criterion does not solve
our problem.
First we will prove some results about the matrices from \scrN \scrT \infty (F ). We notice the following
remark.
Remark 1. Let F be any field and let k1, k2 \in \BbbN \cup \{ 0\} . Suppose that x \in \scrN \scrT k1
\infty (F ) \setminus
\scrN \scrT k1+1
\infty (F ), y \in \scrN \scrT k2
\infty (F ) \setminus \scrN \scrT k2+1
\infty (F ), z \in \scrT \infty (F ) \setminus \scrN \scrT \infty (F ). Then:
(1) xy \in \scrN \scrT k1+k2+1
\infty (F ),
(2) xz, zx \in \scrN \scrT k1
\infty (F ).
Proof. From our assumptions we have xnm = 0, if m - n \leq k1, and ynm = 0, if m - n \leq k2.
To prove the first point, notice that for n \in \BbbN and 0 \leq m \leq k1 + k2 + 1, we have
(xy)n,n+m =
m\sum
i=0
xn,n+iyn+i,n+m =
=
k1\sum
i=0
xn,n+iyn+i,n+m +
m - k1 - 1\sum
j=0
xn,n+m - jyn+m - j,n+m =
=
k1\sum
i=0
0 \cdot yn+i,n+m +
m - k1 - 1\sum
j=0
xn,n+m - j \cdot 0 = 0.
Analogously, for the matrix xz we have
(xz)n,n+m =
m\sum
i=0
xn,n+izn+i,n+m =
m\sum
i=0
0 \cdot zn+i,n+m = 0
for n \in \BbbN , 0 \leq m \leq k1. The case zx is almost the same, so we omit the calculations.
The above remark will be useful in the proof of the following proposition.
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THE DRAZIN INVERSES OF INFINITE TRIANGULAR MATRICES AND THEIR LINEAR PRESERVERS 537
Proposition 1. If F is any field and x \in \scrN \scrT \infty (F ), then x is Drazin invertible in \scrT \infty (F ) if
and only if it is nilpotent.
Proof. It is known that if an element of a ring is nilpotent, then it is Drazin invertible and its
Drazin inverse is 0. Therefore, it suffices to consider the elements that are not nilpotent.
Let x \in \scrN \scrT k
\infty (F ) \setminus \scrN \scrT k+1
\infty (F )
\bigl(
k \in \BbbN \cup \{ 0\}
\bigr)
be so. If xD existed, then one of the three cases
would hold:
(1) xD = 0,
(2) xD \in \scrN \scrT l
\infty (F ) \setminus \scrN \scrT l+1
\infty (F ) for some l \in \BbbN \cup \{ 0\} ,
(3) xD \in \scrT \infty (F ) \setminus \scrN \scrT \infty (F ).
If xD = 0, then by condition (1c), we would have xn = 0 for some n \in \BbbN . However, since x is not
nilpotent, this is impossible.
To discuss the second and third case we use condition (1b).
In the second case, from the first point of Remark 1, it follows that xDxxD \in \scrN \scrT 2l+k+2
\infty (F ).
As xD \in \scrN \scrT k
\infty (F )\setminus \scrN \scrT k+1
\infty (F ) and (1b) holds, this is possible only if k \geq 2l+k+2, i.e., l \leq - 1
— a contradiction.
In the third case, by the second point of Remark 1, we would have xDxxD \in \scrN \scrT k
\infty (F ) and
xD \in \scrT \infty (F ) \setminus \scrN \scrT \infty (F ) — a contradiction again. Concluding, x is not Drazin invertible.
Proposition 1 is proved.
Now we will discuss the general case.
Before we prove Theorem 1 we need to introduce two notions.
The first of them is the index of a matrix. The index of a matrix x \in \scrM n\times n(F ) is defined as the
minimal number n \in \BbbN \cup \{ 0\} for which we have
\mathrm{r}\mathrm{a}\mathrm{n}\mathrm{k}
\bigl(
an+1
\bigr)
= \mathrm{r}\mathrm{a}\mathrm{n}\mathrm{k}(an). (3)
We write then \mathrm{i}\mathrm{n}\mathrm{d}(x) = n. In particular, every invertible matrix has index 0. It is known that \mathrm{i}\mathrm{n}\mathrm{d}(x)
is the minimal number for which (1c) holds. (For more information about the index of triangular
matrices we refer to [3].)
The case of infinite matrices is more complicated. See, for example, the matrix J\infty defined by
J\infty =
\infty \sum
n=1
en,n+1 =
\left(
0 1 0 0 . . .
0 0 1 0
0 0 0 0
. . .
\right) . (4)
As all rows of J\infty are linearly independent, we have \mathrm{r}\mathrm{a}\mathrm{n}\mathrm{k}(J\infty ) = \infty . (By the rank we mean the
row rank.)
Notice that for every k \in \BbbN \cup \{ 0\} we have Jk
\infty =
\sum \infty
n=1
en,n+k, so \mathrm{r}\mathrm{a}\mathrm{n}\mathrm{k}(Jk
\infty ) = \infty . Hence,
according to the definition given above we should have \mathrm{i}\mathrm{n}\mathrm{d}(J\infty ) = 0, i.e., J\infty should be invertible
— a contradiction.
For this reason we can not consider an index of infinite matrix defined as in (3). However, we
let ourselves to introduce a notion of the Drazin inverse of an infinite matrix which will be denoted
by \mathrm{i}\mathrm{n}\mathrm{d}D . If an infinite matrix x is Drazin invertible, then we will say that its Drazin index is the
minimal number for which (1c) holds. Otherwise, we will write \mathrm{i}\mathrm{n}\mathrm{d}D(x) = \infty .
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 4
538 R. SŁOWIK
The second notion we want to define is connected to matrices from \scrM n\times 1(F ). For every such
matrix x there exists the minimal number k such that xl1 = 0 for all l \geq k. We will call this number
a nonzero-length of x and denote it nzl(x).
Now we will give a proof of our first main result.
Proof of Theorem 1. Let x \in \scrT \infty (F ) be of the form\biggl(
an bn
0 cn
\biggr)
for some an \in \scrT \infty (F ), cn \in \scrT \infty (F ).
If
xD =
\biggl(
a\prime n b\prime n
0 c\prime n
\biggr)
with a\prime n \in \scrT \infty (F ), c\prime n \in \scrT \infty (F ),
then aDn = a\prime n. Hence, if xD exists, then it fulfills the condition (xD)ij = a\prime ij for all 1 \leq i, j \leq n.
Notice that since \mathrm{i}\mathrm{n}\mathrm{d}(x) \geq \mathrm{i}\mathrm{n}\mathrm{d}(an), this means that there must exist such n that for all m \geq n we
have \mathrm{i}\mathrm{n}\mathrm{d}(am+1) = \mathrm{i}\mathrm{n}\mathrm{d}(am) = \mathrm{i}\mathrm{n}\mathrm{d}D(x). Consider then the matrices of the form
am+1 =
\biggl(
am b
0 c
\biggr)
where c \in F.
We need to investigate when \mathrm{i}\mathrm{n}\mathrm{d}(am+1) = \mathrm{i}\mathrm{n}\mathrm{d}(am). We divide this problem into two cases.
Case 1: c \not = 0.
Suppose that \mathrm{i}\mathrm{n}\mathrm{d}(am) = i. As am is finite, for all j \geq i we have \mathrm{r}\mathrm{a}\mathrm{n}\mathrm{k}
\bigl(
aj+1
m
\bigr)
= \mathrm{r}\mathrm{a}\mathrm{n}\mathrm{k}
\bigl(
ajm
\bigr)
.
Now observe that since c \not = 0, for all j \geq i we get
\mathrm{r}\mathrm{a}\mathrm{n}\mathrm{k}
\Biggl( \biggl(
am b
0 c
\biggr) j+1
\Biggr)
=
= \mathrm{r}\mathrm{a}\mathrm{n}\mathrm{k}
\Biggl( \Biggl(
aj+1
m b\prime
0 cj+1
\Biggr) \Biggr)
=
= 1 + \mathrm{r}\mathrm{a}\mathrm{n}\mathrm{k}(aj+1
m ) = 1 + \mathrm{r}\mathrm{a}\mathrm{n}\mathrm{k}(ajm) =
= \mathrm{r}\mathrm{a}\mathrm{n}\mathrm{k}
\Biggl( \Biggl(
ajm b\prime \prime
0 cj
\Biggr) \Biggr)
= \mathrm{r}\mathrm{a}\mathrm{n}\mathrm{k}
\Biggl( \biggl(
am b
0 c
\biggr) j
\Biggr)
for b\prime , b\prime \prime \in \scrM n\times 1(F ). Thus, \mathrm{i}\mathrm{n}\mathrm{d}(am+1) \leq \mathrm{i}\mathrm{n}\mathrm{d}(am) and we are done.
Case 2: c = 0.
First we notice that for any n \in \BbbN
anm+1 =
\biggl(
anm an - 1
m b
0 0
\biggr)
.
If \mathrm{i}\mathrm{n}\mathrm{d}(am) = i, then, for all j \geq i, \mathrm{r}\mathrm{a}\mathrm{n}\mathrm{k}
\bigl(
aj+1
m
\bigr)
= \mathrm{r}\mathrm{a}\mathrm{n}\mathrm{k}
\bigl(
ajm
\bigr)
. Therefore we have \mathrm{i}\mathrm{n}\mathrm{d}(am+1) =
= \mathrm{i}\mathrm{n}\mathrm{d}(am) if and only if
\mathrm{r}\mathrm{a}\mathrm{n}\mathrm{k}
\biggl( \biggl(
aj+1
m ajm
0 0
\biggr) \biggr)
= \mathrm{r}\mathrm{a}\mathrm{n}\mathrm{k}
\biggl( \biggl(
ajm aj - 1
m
0 0
\biggr) \biggr)
for all j \geq i.
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 4
THE DRAZIN INVERSES OF INFINITE TRIANGULAR MATRICES AND THEIR LINEAR PRESERVERS 539
Since \mathrm{r}\mathrm{a}\mathrm{n}\mathrm{k}
\bigl(
aj+1
m
\bigr)
= \mathrm{r}\mathrm{a}\mathrm{n}\mathrm{k}
\bigl(
ajm
\bigr)
, this happens if and only if nzl
\bigl(
aj - 1
m b
\bigr)
\leq nzl
\bigl(
ajmb
\bigr)
or
\mathrm{r}\mathrm{a}\mathrm{n}\mathrm{k}(ajm+1) = \mathrm{r}\mathrm{a}\mathrm{n}\mathrm{k}(ajm). These two are the situations described in our theorem.
Theorem 1 is proved.
It can be easily observed that in particular we have the following corollary.
Corollary 1. If F is a field and x \in \scrT \infty (F ) is a matrix such that xnn = 0 holds only for finitely
many n \in \BbbN , then x is Drazin invertible in \scrT \infty (F ).
2.2. The Drazin inverses preservers. In this section we will prove Theorem 2. We will use
some facts about tripotents, i.e., about elements x such that x3 = x.
Remark 2. Let R be a ring. If \phi : R \rightarrow R is a linear map satisfying (2), then \phi preserves
tripotents.
Proof. Let \phi satisfy our assumptions and let x \in R be tripotent. As x3 = x, we must have
xD = x. Hence, \phi (xD) = \phi (x) and from (2) we get (\phi (x))D = \phi (x). Substituing the latter into
(1b), we get (\phi (x))3 = \phi (x).
It is well-known that if x is n \times n tripotent matrix, then xy is a diagonal matrix for some y.
(Some more information about characterization of tripotents can be found in [22].) We wish to obtain
an analogous result for elements of \scrT \infty (F ). To do this, we first notice the following remark.
Remark 3. Let F be any field, n \in \BbbN , and let
x =
\biggl(
a b
0 0
\biggr)
,
y =
\biggl(
a b
0 1
\biggr)
,
z =
\biggl(
a b
0 - 1
\biggr)
be block matrices from \scrT n+1(F ).
1. The matrix x is tripotent if and only if a \in \scrT n(F ) is tripotent and a2b = b.
2. The matrix y is tripotent if and only if a \in \scrT n(F ) is tripotent and a2b+ ab = 0.
3. The matrix z is tripotent if and only if a \in \scrT n(F ) is tripotent and a2b = ab.
The proof is by elementary calculations, so we let ourselves to omit it.
We use the above remark in the proof of the following.
Lemma 1. Suppose F is a field of characteristic different from 2 and that n \in \BbbN . If x \in \scrT \infty (F )
is tripotent, then there exists an invertible y \in \scrT \infty (F ) such that xy is a diagonal matrix.
Proof. First we show that if
x =
\biggl(
a b
0 c
\biggr)
\in \scrT k+1(F ) (5)
is tripotent matrix with a \in \scrT k(F ) diagonal, then there exists y of the form
y =
\biggl(
ek \^y
0 1
\biggr)
such that xy is diagonal.
As x3 = x, the coefficient c in (5) is equal to 0, 1 or - 1. Let us discuss these three cases.
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 4
540 R. SŁOWIK
If c = 0, then by taking \^y = - ab we obtain
xy =
\biggl(
ek ab
0 1
\biggr) \biggl(
a b
0 0
\biggr) \biggl(
ek - ab
0 1
\biggr)
=
\biggl(
a - a2b+ b
0 0
\biggr)
,
so by the first point of Remark 3
xy =
\biggl(
a 0
0 0
\biggr)
.
Let now c = 1. Notice that from the second point of Remark 3 and the assumption \mathrm{c}\mathrm{h}\mathrm{a}\mathrm{r}(F ) \not = 2,
it follows that ai = 1 implies bi = 0. If
xy =
\biggl(
a 0
0 1
\biggr)
,
then \^y = (\^y1, \^y2, . . . , \^yk)
T must satisfy the system
(a1 - 1)\^y1 + b1 = 0,
(a2 - 1)\^y2 + b2 = 0,
. . . . . . . . . . . . . . . . . .
(ak - 1)\^yk + bk = 0.
As for those i for which we have ai = 1, the coefficients bi are equal to 0, we can choose \^yi
arbitrarily (say \^yi = 0). For the case where ai \not = 0, the coefficients ai - 1 are invertible, so we
simply have \^yi = (1 - ai)
- 1bi. Thus, the desired y exist.
The case where c = - 1 is analogous. Since \mathrm{c}\mathrm{h}\mathrm{a}\mathrm{r}(F ) \not = 2, from the third point of Remark 3 we
can deduce that if ai = - 1, then bi = 0. If
xy =
\biggl(
a 0
0 - 1
\biggr)
,
then \^y must satisfy the system
(a1 + 1)\^y1 + b1 = 0,
(a2 + 1)\^y2 + b2 = 0,
. . . . . . . . . . . . . . . . . .
(ak + 1)\^yk + bk = 0.
Analogously to the previous case, if for some i we have ai = - 1, then bi = 0 and \^yi may be
arbitrarily chosen, whereas if ai \not = - 1, then 1 + ai is invertible and we have \^yi = - (ai + 1) - 1bi.
Let x \in \scrT n(F ) be tripotent. Then
x2 :=
\biggl(
x11 x12
0 x22
\biggr)
is tripotent as well. According to what was done above, there exists y2 \in \scrT 2(F ) such that xy22 is a
diagonal matrix. Hence
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 4
THE DRAZIN INVERSES OF INFINITE TRIANGULAR MATRICES AND THEIR LINEAR PRESERVERS 541
x
\left( y2 0
0 en - 2
\right)
=
\biggl(
a2 b2
0 c2
\biggr)
where a2 \in \scrT 2(F ) is diagonal.
Suppose that we have proved that for some 2 \leq m < n we have
\left(
\left( x
\left( y2 0
0 en - 2
\right) \right)
\left( y3 0
0 en - 3
\right)
. . .
\right)
\left( ym 0
0 en - m
\right)
=
\biggl(
am bm
0 cm
\biggr)
with am \in \scrT m(F ) diagonal. Then, by the first part of the proof, there exists ym+1 \in \scrT m+1(F ) such
that \biggl(
am bm
0 cm
\biggr) \left( ym+1 0
0 en - m - 1
\right)
=
\biggl(
am+1 bm+1
0 cm+1
\biggr)
,
where am+1 \in \scrT m+1(F ) is diagonal.
Lemma 1 is proved.
From the above lemma we can easily obtain the following corollary.
Corollary 2. Let F be a field such that \mathrm{c}\mathrm{h}\mathrm{a}\mathrm{r}(F ) \not = 2. If x \in \scrT \infty (F ) is tripotent, then there exist
an invertible matrix y \in \scrT \infty (F ) such that xy is diagonal.
Proof. For any n \in \BbbN define xn \in \scrT n(F ) by the following rule: (xn)ij = xij for all 1 \leq i,
j \leq n.
As every xn is tripotent, by Lemma 1, for every n \in \BbbN there exists yn \in \scrT n(F ) such that xynn
is diagonal. Moreover, we can choose y1, y2, y3, . . . in such a way that (yn)ij = (yn+1)ij for all
1 \leq i, j \leq n. Define now y \in \scrT \infty (F ) according to the condition yij = (yn)ij for n \geq \mathrm{m}\mathrm{a}\mathrm{x}(i, j).
For every 1 \leq i < j we have
(xy)ij =
\sum
i\leq k\leq l\leq j
(y - 1)ikxklylj =
\sum
i\leq k\leq l\leq j
(y - 1
j )ik(xj)kl(yj)lj = (y - 1
j xjyj)ij = 0.
Thus, xy is a diagonal matrix.
Now we will investigate maps satisfying (2) that fulfil some extra conditions.
In the following lemmas we make use of ideas from [4].
Lemma 2. Let F be a field of characteristic different from 2 and containing more than 3
elements. If \phi : \scrT \infty (F ) \rightarrow \scrT \infty (F ) is a linear map preserving the Drazin inverses, then \phi (e\infty )\phi (x) =
= \phi (x)\phi (e\infty ) for all x \in \scrT \infty (F ).
Proof. We give a proof in 4 steps.
Step 1. For all n \in \BbbN we have \phi (e\infty )\phi (enn) = \phi (enn)\phi (e\infty ).
Fix \alpha \in F \ast \setminus \{ - 1, 1\} . As | F | > 3 such element exists.
It can be easily noticed that (e\infty +(\alpha - 1)enn)
D = e\infty +(\alpha - 1 - 1)enn. Hence, \phi (e\infty +(\alpha - 1)enn)
and \phi
\bigl(
e\infty + (\alpha - 1 - 1)enn
\bigr)
commute. From this and the linearity of \phi we have
(\alpha 2 - 1)(\phi
\bigl(
e\infty )\phi (enn) - \phi (enn)\phi (e\infty )
\bigr)
= 0.
As \alpha \not = \pm 1, we are done.
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 4
542 R. SŁOWIK
Step 2. For all n,m \in \BbbN , n < m we have \phi (e\infty )\phi (enm) = \phi (enm)\phi (e\infty ).
This time notice that (e\infty + enm)D = e\infty - enm, so \phi (e\infty + enm) and \phi (e\infty - enm) commute.
From this and the linearity of \phi , we obtain
2(\phi
\bigl(
e\infty )\phi (enm) - \phi (enm)\phi (e\infty )
\bigr)
= 0.
As \mathrm{c}\mathrm{h}\mathrm{a}\mathrm{r}(F ) \not = 2, the result follows.
Step 3. For any x \in \scrT \infty (F ) of the form\biggl(
a 0
0 0
\biggr)
with a \in \scrT n(F ) for some n \in \BbbN ,
we have \phi (e\infty )\phi (x) = \phi (x)\phi (e\infty ).
This is a consequence of the linearity of \phi .
Step 4. For all x \in \scrT \infty (F ) we have \phi (e\infty )\phi (x) = \phi (x)\phi (e\infty ).
Suppose that the claim of this point is not true.
Define x(n) to be a matrix from \scrT \infty (F ) such that
(x(n))ij =
\left\{ xij for 1 \leq i, j \leq n,
0, otherwise.
Clearly, by the preceding step, \phi (e\infty ) commutes with \phi
\bigl(
x(n)
\bigr)
for all n \in \BbbN . Hence, if \phi (e\infty )\phi (x) \not =
\not = \phi (x)\phi (e\infty ), then
\phi (e\infty )\phi
\bigl(
x - x(n)
\bigr)
\not = \phi
\bigl(
x - x(n)
\bigr)
\phi (e\infty )
for all n \in \BbbN . However, this means that if there exists x such that \phi (x) does not commute with
\phi (e\infty ), then there exists a matrix y that has an arbitrary number of first zero columns and such
that \phi (y) does not commute with \phi (e\infty ). Yet, the only matrix with the latter property is the zero
matrix, but ona can see that \phi (0) = 0 commutes with \phi (e\infty ) — a contradiction. Concluding,
\phi (e\infty )\phi (x) = \phi (x)\phi (e\infty ) for all x \in \scrT \infty (F ).
Lemma 2 is proved.
Lemma 3. Let F be a field of at least 4 elements. If \phi : \scrT \infty (F ) \rightarrow \scrT \infty (F ) is a linear map such
that \phi preserves the Drazin inverses and there exist N, N \subseteq \BbbN for which the following conditions
hold:
if i /\in N or j /\in N, then (\phi (x))ij = 0 for all x \in \scrT \infty (F ),
\phi (e\infty ) =
\sum
n\in N
enn,
then \phi preserves idempotents.
Proof. Let x \in \scrT \infty (F ) be an idempotent. Every idempotent is also a tripotent, so it is conjugate
to some diagonal matrix. Clearly, the map \phi preserves the Drazin inverses if and only if for every t
the map \phi \cdot \scrI nnt preserves them. Therefore, with no loss of generality, we can assume that x is a
diagonal matrix
\bigl(
more precisely, it is of the form
\sum
n\in M
enn for some \varnothing \subseteq M \subseteq N
\bigr)
.
Let us choose \alpha \in F \ast \setminus \{ 1, 2\} . One can see that
\bigl(
e\infty + (\alpha - 1 - 1)x
\bigr)
= e\infty + (\alpha - 1)x. As \phi
preserves the Drazin inverses, this means that
\bigl(
\phi (e\infty + (\alpha - 1 - 1)x)
\bigr) D
= \phi
\bigl(
e\infty + (\alpha - 1)x
\bigr)
. In
particular, condition (1b) must be satisfied. Hence
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THE DRAZIN INVERSES OF INFINITE TRIANGULAR MATRICES AND THEIR LINEAR PRESERVERS 543
\phi
\bigl(
e\infty + (\alpha - 1)x
\bigr)
\phi
\bigl(
e\infty + (\alpha - 1 - 1)x
\bigr)
\phi
\bigl(
e\infty + (\alpha - 1)x
\bigr)
= \phi
\bigl(
e\infty + (\alpha - 1)x
\bigr)
.
Using the linearity and fact that \phi (e\infty ) =
\sum
n\in N
enn commutes with \phi (y) for all y \in \scrT \infty (F ), we
get \Biggl[ \sum
n\in N
enn + (\alpha - 1)\phi (x)
\Biggr] \Biggl[ \sum
n\in N
enn + (\alpha - 1 - 1)\phi (x)
\Biggr] \Biggl[ \sum
n\in N
enn + (\alpha - 1)\phi (x)
\Biggr]
=
=
\sum
n\in N
enn + (\alpha - 1)\phi (x).
We multiply the above equality by \alpha and write it as follows:\Biggl[ \sum
n\in N
enn + (\alpha - 1)\phi (x)
\Biggr] \Biggl[ \sum
n\in N
enn + (\alpha - 1)
\Biggl( \sum
n\in N
enn - \phi (x)
\Biggr) \Biggr] \Biggl[ \sum
n\in N
enn + (\alpha - 1)\phi (x)
\Biggr]
=
=
\sum
n\in N
enn + (\alpha - 1)
\Biggl( \sum
n\in N
enn + \phi (x)
\Biggr)
+ (\alpha - 1)2\phi (x).
After evaluating we obtain
\alpha
\sum
n\in N
enn +
\bigl[
\alpha - 1 + 2(\alpha - 1)2
\bigr]
\phi (x) +
\bigl[
(\alpha - 1)3 - (\alpha - 1)2
\bigr] \bigl(
\phi (x)
\bigr) 2 - (\alpha - 1)3
\bigl(
\phi (x)
\bigr) 3
=
= \alpha
\sum
n\in N
enn + (\alpha - 1)\phi (x).
We simplify it and make use of the fact that
\bigl(
\phi (x)
\bigr) 3
= \phi (x) and have\bigl[
(\alpha - 1)3 - (\alpha - 1)2
\bigr] \bigl[
(\phi (x))2 - \phi (x)
\bigr]
= 0.
This is possible only if \alpha \in \setminus \{ 1, 2\} or (\phi (x))2 - \phi (x) = 0. As we have chosen \alpha \not = 1, 2, \phi (x)
must be an idempotent.
Lemma 3 is proved.
Lemma 4. Let F be a field of at least 5 elements. If \phi : \scrT \infty (F ) \rightarrow \scrT \infty (F ) is a linear map that
preserve the Drazin inverses and \phi (e\infty ) = 0, then \phi (x) = 0 for all x \in \scrT \infty (F ).
Proof. Again we divide the proof into steps.
Step 1. For all n \in \BbbN we have \phi (enn) = 0.
Analogously, as it was done in the proof of Lemma 3, we choose \alpha \in F \ast \setminus \{ 1\} and make use of
the fact that
\bigl(
e\infty + (\alpha - 1 - 1)enn
\bigr) D
= e\infty + (\alpha - 1)enn. Performing the same way and using the
fact that \phi (e\infty ) = 0 we obtain
\bigl[
(\alpha - 1)3 + (\alpha - 1)2 + \alpha - 1
\bigr]
\phi (enn) = 0. As \alpha \not = 1, this implies
(\alpha 2 - \alpha + 1)\phi (enn) = 0. (6)
Since | F | > 4, there exist \alpha \not = 0, 1 such that \alpha 2 - \alpha + 1 \not = 0. Substituing this \alpha into (6) we get
\phi (enn) = 0.
Step 2. For all n < m we have \phi (enm) = 0.
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544 R. SŁOWIK
Let \alpha \in F \ast . From (e\infty + \alpha enm)D = e\infty - \alpha enm, \phi (e\infty ) = 0 and condition (1b) we get
\alpha \phi (enm) + \alpha 3(\phi (enm))3 = 0. Now we choose \alpha 1 \in F \ast and \alpha 2 \in F \ast such that \alpha 1 \not = \pm \alpha 2 (as
| F | > 4, it is possible). For such coefficients the system of equations
\alpha 1\phi (enm) + \alpha 3
1
\bigl(
\phi (enm)
\bigr) 3
= 0,
\alpha 2\phi (enm) + \alpha 3
2
\bigl(
\phi (enm)
\bigr) 3
= 0
has unique solution and we have \phi (enm) = 0.
Step 3. For all x \in \scrT \infty (F ) of the form\biggl(
a 0
0 0
\biggr)
for some a \in \scrT n(F ) for some n \in \BbbN ,
we have \phi (x) = 0.
This is a consequence of the two preceding steps and linearity.
Step 4. For all x \in \scrT \infty (F ) we have \phi (x) = 0.
This can be proved by the same method as one used in step 4 in the proof of Lemma 2, so we do
not repeat the arguments.
Lemma 4 is proved.
Now we will prove our second main result.
Proof of Theorem 2. Suppose \phi preserves the Drazin inverses and is linear. Then, by Remark 2,
\phi (e\infty ) is tripotent. From Corollary 2 it follows that there exist t \in \scrT \infty (F ) such that (\phi (e\infty ))t is
diagonal. Consider then \phi \prime := \scrI nnt \cdot \phi instead of \phi . We have then \phi \prime (e\infty ) =
\sum
n\in \BbbN
anenn, where
an \in \{ - 1, 0, 1\} . By Lemma 2, we get \phi \prime (e\infty )\phi \prime (x) = \phi \prime (x)\phi \prime (e\infty ) for all x \in \scrT \infty (F ). Let n, m
be such natural numbers that n < m. We should obtain\bigl(
\phi \prime (e\infty )\phi \prime (x)
\bigr)
nm
= an
\bigl(
\phi \prime (x)
\bigr)
nm
=
\bigl(
\phi \prime (x)
\bigr)
nm
am =
\bigl(
\phi \prime (x)\phi \prime (e\infty )
\bigr)
.
As an \not = am and \mathrm{c}\mathrm{h}\mathrm{a}\mathrm{r}(F ) \not = 2, this is possible only if
\bigl(
\phi \prime (x)
\bigr)
nm
= 0 for any x. Concluding, we
have proved that if
\bigl(
\phi \prime (e\infty )
\bigr)
nn
\not =
\bigl(
\phi \prime (e\infty )
\bigr)
mm
, then
\bigl(
\phi \prime (e\infty )
\bigr)
nm
= 0 for all x \in \scrT \infty (F ). Hence,
\phi \prime is a separable sum of at most three ‘types’ of maps \psi 1, \psi 2, \psi 3 such that
all these maps preserve the Drazin inverses,
the maps of ‘type’ \psi 1 satisfy the condition \psi 1(e\infty ) =
\sum
n\in N1
enn for some \varnothing \subsetneq N1 \subseteq \BbbN ,
the maps of ‘type’ \psi 2 satisfy the condition \psi 2(e\infty ) = -
\sum
n\in N2
enn for some \varnothing \subsetneq N2 \subseteq \BbbN ,
the maps of ‘type’ \psi 3 satisfy the condition \psi 3(e\infty ) = 0.
Now we discuss these three ‘types’ of maps.
The map \psi 1 satisfies the assumptions of Lemma 3, so \psi 1 preserves idempotents.
The map - \psi 2 is of the same form as \psi 1, so - \psi 2 is an idempotent preserver.
The map \psi 3 fulfills the assumptions of Lemma 4, so \psi 3(x) = 0 for all x \in \scrT \infty (F ).
Summing up, \phi is a separable map of \psi 1 and \psi 2, where \psi 1 and - \psi 2 are idempotent preservers.
Theorem 2 is proved.
Since the problem of describing the Drazin inverses preservers is reduced to idempotent pre-
servers, we give a characterization of such maps. To do this we need to present a few more sorts of
linear maps.
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 4
THE DRAZIN INVERSES OF INFINITE TRIANGULAR MATRICES AND THEIR LINEAR PRESERVERS 545
If t is an invertible infinite upper triangular matrix, then \phi : \scrT \infty (F ) \rightarrow \scrT \infty (F ) such that
\phi (x) = t - 1xt is an inner automorphism of \scrT \infty (F ) will and is denoted by \scrI nnt.
For a family of infinite triangular matrices \{ bnm\} n<m such that
if i /\in \mu (n) or j /\in \mu (m), then (bnm)ij = 0,
for all n < p < m we have bnpbpm = bnm,
by \scrS pl\mu ,\{ bnm\} n<m
: \scrT \infty (F ) \rightarrow \scrT \infty (F ) we denote a map such that
\scrS pl\mu ,\{ bnm\} n<m
(x) =
\sum
n\in \BbbN
\sum
i\in \mu (n)
xnneii +
\sum
n<m
xnmbnm
for any x = (xij) \in \scrT \infty (F ).
In particular, if \tau is an increasing map on \BbbN , then \phi : \scrT \infty (F ) \rightarrow \scrT \infty (F ) defined by
\phi
\left( \sum
i\leq j
xij
\right) =
\sum
i\leq j
xije\tau (i)\tau (j)
is a special case of \scrS pl\mu ,\{ bnm\} n<m
. We call it a splashing map and denote it by \scrS pl\tau .
Suppose that N is a subset of \BbbN equal either to \{ 1, 2, . . . , n\} or \BbbN \setminus \{ n, n+ 1, . . . ,m\} , where
n,m \in \BbbN , n < m. By \scrC utN we will mean the map sending x \in \scrT \infty (F ) to\left(
0 . . . 0 0 0 0 . . .
. . .
...
0 0 0 0 . . .
xn+1,n+1 xn+1,n+2 xn+1,n+3 . . .
xn+2,n+2 xn+2,n+3
xn+3,n+3
. . .
\right)
in the first case, or to \left(
xnn . . . xnm 0 . . .
. . .
...
xmm 0
0
. . .
\right)
in the second case.
If x \in \scrT \infty (F ) is of the form\biggl(
x 0
0 0
\biggr)
, where x \in \scrT n(F ),
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 4
546 R. SŁOWIK
then by \scrJ (x) we will denote a matrix\biggl(
y 0
0 0
\biggr)
with y \in \scrT n(F ),
where yij = xn+1 - j,n+1 - i.
The defined above maps appear in the theorem below.
Corollary 3 [20]. Let F be a field whose characteristic is different from 2. If \phi : \scrT \infty (F ) \rightarrow
\rightarrow \scrT \infty (F ) is a linear map satisfying the condition
x is an idempotent \Rightarrow \phi (x) is an idempotent,
then \phi is a separable sum of any number of maps of one of the forms
\scrI nnt \cdot \scrS pl\mu ,\{ bnm\} n<m
,
\scrI nnt \cdot \scrS pl\mu ,\{ bnm\} n<m
\cdot \scrC utN ,
\scrI nnt \cdot \scrS pl\mu ,\{ bnm\} n<m
\cdot \scrJ \cdot \scrC utN ,
where N = \{ 1, 2, . . . , n\} or N = \BbbN \setminus \{ n, n+ 1, . . . ,m\} for some n < m.
2.3. Closing remarks. In this section we mention some problems and present some remarks
related to the considered issues.
Suppose that \phi has a following property: for every n \in \BbbN there exist
\bigl(
\phi (e\infty )
\bigr)
nn
\not = 0. Then it
can be observed that \phi preserves inverses, i.e., if x - 1 = y, then (\phi (x)) - 1 = \phi (y).
In this article we have considered the ring of upper triangular infinite matrices. It is worth
noting that this ring is contained in a more general one. More precisely, in \scrM Cf (F ) — the ring of
all \BbbN \times \BbbN matrices over F whose all columns contain only a finite number of nonzero entries.
Let us get back to the matrix J\infty defined in (4). Suppose that it has the Drazin inverse, but in
\scrM Cf (F ). Denote it by y. By (1a) we would then have
y =
\left(
y11 y22 y33 . . . ynn . . .
0 y22 y33 ynn
0 y22 y33 ynn
0 y22 y33 ynn
...
...
...
...
\right)
.
The condition y \in \scrM Cf (F ) forces y22 = y33 = y44 = . . . = 0. Thus, y = y11e11, i.e., y \in \scrT \infty (F ).
However, we have already proved that J\infty is not Drazin invertible in this ring. Summing up, J\infty is
not Drazin invertible not only in \scrT \infty (F ), but also in \scrM Cf (F ).
(3) At the end of the paper we prove the following result.
Proposition 2. Let F be a field and let n \in \BbbN . The matrix x \in \scrM Cf (F ) of the form\biggl(
a 0
0 b
\biggr)
with a \in \scrM n\times n(F ), b \in \scrM Cf (F )
is Drazin invertible if and only if b is so.
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 4
THE DRAZIN INVERSES OF INFINITE TRIANGULAR MATRICES AND THEIR LINEAR PRESERVERS 547
Proof. We use a result from [25] which says that if AB = BA, then A+B is Drazin invertible
if and only if 1 +ADB is.
As \biggl(
a 0
0 b
\biggr)
=
\biggl(
a 0
0 e\infty
\biggr)
+
\biggl(
0 0
0 b - e\infty
\biggr)
,
our matrix is Drazin invertible if and only if
e\infty +
\biggl(
a 0
0 e\infty
\biggr) D\biggl(
0 0
0 b - e\infty
\biggr)
=
= e\infty +
\biggl(
aD 0
0 e\infty
\biggr) \biggl(
0 0
0 b - e\infty
\biggr)
=
\biggl(
en 0
0 b
\biggr)
is.
Therefore, it suffices to prove that \biggl(
en 0
0 b
\biggr)
(7)
is Drazin invertible if and only if b is.
Suppose that the Drazin inverse of matrix given by (7) exists. Denote it by
y =
\biggl(
y1 y2
y3 y4
\biggr)
.
From (1a) we get
by3 = y3, (8)
whereas by (1c) we have
y2 = 0, (9)
bk+1y3 = 0. (10)
Now using (8) and (10) we obtain
bk(by3) = 0 \Rightarrow bky3 = 0 \Rightarrow bk - 1(by3) = 0 \Rightarrow bk - 1y3 = 0 \Rightarrow . . . \Rightarrow y3 = 0.
Hence, from the latter and (9)
y =
\biggl(
y1 0
0 y4
\biggr)
, i.e., y1 = eDn and y4 = bD.
Concluding, our matrix is Drazin invertible if and only if b is.
Proposition 2 is proved.
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Received 15.12.14
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 4
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| id | umjimathkievua-article-1574 |
| institution | Ukrains’kyi Matematychnyi Zhurnal |
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| language | English |
| last_indexed | 2026-03-24T02:08:22Z |
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| publisher | Institute of Mathematics, NAS of Ukraine |
| record_format | ojs |
| resource_txt_mv | umjimathkievua/9b/c3e06217584203665c31f29fb574639b.pdf |
| spelling | umjimathkievua-article-15742019-12-05T09:19:04Z The Drazin inverses of infinite triangular matrices and their linear preservers Оберненi матрицi Дразiна для нескiнченних трикутних матриць та їх лiнiйних зберiгачiв Słowik, R. Словік, Р. We consider the ring of all infinite $(N \times N)$ upper triangular matrices over a field $F$. We give a description of elements that are Drazin invertible in this ring. In the case where $F$ is such that $\mathrm{c}\mathrm{h}\mathrm{a}\mathrm{r}(F) \not = 2$ and $| F| > 4$, we find the form of linear preservers for the Drazin inverses. Розглянуто кiльце всiх нескiнченних $(N \times N)$ верхнiх трикутних матриць над полем $F$. Наведено опис елементiв, що є оборотними за Дразiним у цьому кiльцi. У випадку, коли $F$ є таким, що $\mathrm{c}\mathrm{h}\mathrm{a}\mathrm{r}(F) \not = 2$ та $| F| > 4$, знайдено форму лiнiйних зберiгачiв для обернених матриць Дразiна. Institute of Mathematics, NAS of Ukraine 2018-04-25 Article Article application/pdf https://umj.imath.kiev.ua/index.php/umj/article/view/1574 Ukrains’kyi Matematychnyi Zhurnal; Vol. 70 No. 4 (2018); 534-548 Український математичний журнал; Том 70 № 4 (2018); 534-548 1027-3190 en https://umj.imath.kiev.ua/index.php/umj/article/view/1574/556 Copyright (c) 2018 Słowik R. |
| spellingShingle | Słowik, R. Словік, Р. The Drazin inverses of infinite triangular matrices and their linear preservers |
| title | The Drazin inverses of infinite triangular matrices and their linear preservers |
| title_alt | Оберненi матрицi Дразiна для нескiнченних
трикутних матриць та їх лiнiйних зберiгачiв |
| title_full | The Drazin inverses of infinite triangular matrices and their linear preservers |
| title_fullStr | The Drazin inverses of infinite triangular matrices and their linear preservers |
| title_full_unstemmed | The Drazin inverses of infinite triangular matrices and their linear preservers |
| title_short | The Drazin inverses of infinite triangular matrices and their linear preservers |
| title_sort | drazin inverses of infinite triangular matrices and their linear preservers |
| url | https://umj.imath.kiev.ua/index.php/umj/article/view/1574 |
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