����������� ������ 51 (1999), 69–76

UDK 517.983.2

���������� ������ ��� research paper

GENERAL REPRESENTATIONS OF PSEUDOINVERSES

Dragan S. -Dor - dević and Predrag S. Stanimirović

Abstract. In this paper we investigate general representations of various classes of generalized inverses for bounded operators over Hilbert and Banach spaces. These representations are expressed by means of the full-rank decomposition of bounded operators and adequately selected operators.

1. Introduction

Let X_{1} and X_{2} denote arbitrary Banach spaces and B(X_{1}, X_{2}) denote the set of all bounded operators from X_{1} into X_{2}. For an arbitrary operator A ∈ B(X_{1}, X_{2}), we use N (A) to denote its kernel, and R(A) to denote its image.

For A ∈ B(X_{1}, X_{2}) we say that an operator X ∈ B(X_{2}, X_{1}) is a generalized inverse of A, provided that some of the following equations are satisfied:

(1) AXA = A, (2) XAX = X

If X satisfies the equation (1), then X is called a g-inverse of A. If X satisfies the equations (1) and (2), then it is called a reflexive g-inverse of A.

It is well-known that an operator A ∈ B(X_{1}, X_{2}) has a g-inverse if and only if R(A) is closed, and N (A) and R(A), respectively, are complemented subspaces of X_{1} and X_{2}.

The notion of the full-rank decomposition for complex matrices is well-known and frequently used. Recall the definition of the full rank factorization for a bounded operator acting on Banach spaces from [2] and [3]:

Let A ∈ B(X_{1}, X_{2}). If there exist: a Banach space X_{3} and operators Q ∈ B(X_{1}, X_{3}) and P ∈ B(X_{3}, X_{2}), such that P is left invertible, Q is right invertible and

A = P Q,  (1.1)

then we say that (1.1) is the full-rank decomposition of A.

AMS Subject Classification: 47 A 05, 15 A 09

Keywords and phrases: Full-rank factorization, reflexive generalized inverses, Moore–Penrose inverse, Drazin inverse.

It is well-known that an operator A ∈ B(X_{1}, X_{2}) has the full-rank decomposition, if and only if A is g-invertible. In this case X_{3} is isomorphic to R(A), and R(A) = R(P) [3].

We say that A ∈ B(X) has the Drazin inverse, if there exists an operator A^{D} ∈ B(X), such that A^{D} satisfies the equation (2) and the equations

(1^{k}) A^{k+1}A^{D} = A^{k}, (5) A^{D}A = AA^{D},

for some non-negative integer k. Let us mention that the Drazin inverse, if it exists, is unique. The smallest k in the previous definition is called the index of A and denoted by ind(A). In the case ind(A) = 1 the Drazin inverse is known as the group inverse of A, denoted by A^{#}.

Recall that asc(A) (respectively des(A)), the ascent (respectively descent) of A, is the smallest non-negative integer n, such that N (A^{n}) = N (A^{n+1}) (respectively R(A^{n}) = R(A^{n+1})). If no such n exists, then asc(A) = ∞ (respectively des(A) = ∞) [4]. It is well-known that A has the Drazin inverse, if and only1 if the ascent and descent of A are finite (hence, equal to ind(A)) [3], [4].

In the case when H_{1} and H_{2} are Hilbert spaces, it is well-known that an operator A ∈ B(H_{1}, H_{2}) has a g-inverse if and only if R(A) is closed. Among the equations (1), (2) we also consider the following equations in X:

(3) (AX)^{∗} = AX, (4) (XA)^{∗} = XA.

For a subset S of the set {1, 2, 3, 4}, the set of operators obeying the conditions contained in S is denoted by A{S}. An operator in A{S} is called an S-inverse of A and is denoted by A^{(S)}. If R(A) is closed, the set A{1, 2, 3, 4} consists of a single element, the Moore-Penrose inverse of A, denoted by A^{†}.

We also consider the following equations, which define the weighted Moore-Penrose inverse:

\begin{aligned}(3M)  (MAX)^{∗} & = MAX  (4N)  (NXA)^{∗} = NXA; \\ (3M′)  (AX)^{∗}M & = MAX  (4N′)  (XA)^{∗}N = NXA, \end{aligned}

where M ∈ B(H_{2}), N ∈ B(H_{1}) are positive or invertible. Any solution of the equations (1), (2), (3M) and (4N), when it exists, will be denoted by A_{M,N}^{†}. Similarly, any solution of the equations (1), (2), (3M’) and (4N’), when it exists, will be denoted by A_{M′,N′}^{†}.

We investigate general representations and conditions for the existence of generalized inverses of bounded linear operatos on Hilbert spaces, arising from the factorization (1.1). As a related result we investigate some representations of a generalized inverse A_{T,S}^{(2)}. Obtained representations are generalizations of the analogous results available in the literature for marices. We also introduce a general representation and conditions for thye existence of the Drazin inverse of a bounded operator on a Banach space. There representatons are based on the full-rank decomposition of A^{l}, where l ≥ ind(A). Such an approach in representation of the Drazin inverse is not employed before even for complex matrices.

2. Results

Firstly, we investigate general representations of {1, 2}-inverses, {1, 2, 3}, {1, 2, 4}-inverses, the Moore-Penrose and the weighted Moore-Penrose for operators in arbitrary Hilbert spaces.

We shall frequently use the following observation. If S ∈ B(H_{1}, H_{2}) is onto, then SS^{∗} is invertible and S^{†} is the right inverse of S. Analogously, if T ∈ B(H_{1}, H_{2}) is one-to-one with closed range, then T^{∗}T is invertible and T^{†} is the left inverse of T.

In the beginning, we state an analogy of the well-known result from [10, pp. 20, 28].

Lemma 2.1. If A = P Q is the full-rank decomposition of A ∈ B(H_{1}, H_{2}) according to (1.1), then:

(a) Any right inverse of Q can be represented in the following form: Q_{r}^{−1} = W_{1}(QW_{1})^{−1}, for an arbitrary operator W_{1} ∈ B(H_{3}, H_{1}) such that QW_{1} is invertible.

(b) Any left inverse of P can be represented in the following form: P_{l}^{−1} = (W_{2}P)^{−1}W_{2}, for an arbitrary operator W_{2} ∈ B(H_{2}, H_{3}) such that W_{2}P is invertible.

(c) Any reflexive generalized inverse X of A has the form X = Q_{r}^{−1}P_{l}^{−1} for an arbitrary right inverse Q_{r}^{−1} of Q and an arbitrary left inverse P_{l}^{−1} of P.

In the literature there are known general representations for various classes of generalized inverses, for the set of complex matrices. The general representation of {1, 2} inverses for matrices is investigated in [9] and [10, pp. 20, 28]. The general representations of {1, 2, 3} and {1, 2, 4} inverses for matrices are investigated in [9]. In [6] there is given a general representation and conditions for the existence of the group inverse for a given complex matrix. The general representation of the Moore-Penrose inverse is given in [2[, for arbitrary Hilbert spaces.

In the following theorem we give general representations of {1, 2}, {1, 2, 3} and {1, 2, 4} inverses for an arbitrary bounded operator on Hilbert spaces. As a consequence we obtain the known representation of the Moore-Penrose inverse from [2].

Theorem 2.1. Let A = P Q be a full-rank decomposition of A ∈ B(H_{1}, H_{2}) according to (1.1). Then:

(a) X ∈ A{1, 2} if and only if there exist operators W_{1} ∈ B(H_{3}, H_{1}) and W_{2} ∈ B(H_{2}, H_{3}), such that QW_{1} and W_{2}P are invertible in B(H_{3}). In such a case, X possesses the following general representation

X = W_{1}(QW_{1})^{−1}(W_{2}P)^{−1}W_{2}  (2.1)

(b) X ∈ A{1, 2, 3} if and only if there exists an operator W_{1} ∈ B(H_{3}, H_{1}), such that QW_{1} is invertible in B(H_{3}). In the case when it exists, a general representation for X is as follows:

X = W_{1}(QW_{1})^{−1}(P^{∗}P)^{−1}P^{∗}.  (2.2)

(c) X ∈ A{1, 2, 4} if and only if there exists an operator W_{2} ∈ B(H_{2}, H_{3}), such that W_{2}P is invertible in B(H_{3}). In this case

X = Q^{∗}(QQ^{∗})^{−1}(W_{2}P)^{−1}W_{2}.

(d) A^{†} = Q^{†}P^{†} = Q^{∗}(QQ^{∗})^{−1}(P^{∗}P)^{−1}P^{∗} = Q^{∗}(P^{∗}AQ^{∗})^{−1}P^{∗}.

Proof. (a) Follows from Lemma 2.1.

(b) If X has the form (2.2), then it is easy to verify X ∈ A{1, 2, 3}. We need to prove that the form (2.2) holds for all {1, 2, 3} inverses of A. Indeed, if X ∈ A{1, 2, 3}, then X = Q_{r}^{−1}P_{l}^{−1}, and from the equation (3) it follows that (P P_{l}^{−1})^{∗} = P P_{l}^{−1}. Thus P^{∗}P P_{l}^{−1} = P^{∗}. Operator P^{∗}P is invertible, so that P_{l}^{−1} = (P^{∗}P)^{−1}P^{∗}. The right inverse of Q retains the general form Q_{r}^{−1} = W_{1}(QW_{1})^{−1} from Lemma 2.1. Consequently,

X = W_{1}(QW_{1})^{−1}(P^{∗}P)^{−1}P^{∗}.

The proof of the statement (c) is similar as the proof of (b). Also, (d) follows from (b) and (c). For the part (d) see also [2].

Now, we shall consider the weighted Moore-Penrose inverse under the various hypothesis. The weighted Moore-Penrose inverse is investigated in [1], [7] and [10] for the set of complex matrices and in [8] for matrices over an integral domain. If M and N are positive, then A_{M,N}^{†} and A_{M′,N′}^{†} always exist [1], [10], and A_{M,N}^{†} = A_{M′,N′}^{†}. In [8] and [7] it is derived a representation and conditions for the existence of A_{M,N}^{†} and A_{M′,N′}^{†}, respectively, under the more general assumptions that the matrices M and N are invertible (not necessary positive).

Theorem 2.2. Let A = P Q be a full-rank decomposition of A according to (1.1). Then:

(a) If M ∈ B(H_{2}) and N ∈ B(H_{1}) are invertible operators, then A_{M,N}^{†} exists if and only if P^{∗}MP and QN^{−1}Q^{∗} are invertible selfadjoint operators. In that case

\begin{aligned}A_{M,N}^{†} & = N^{−1}Q^{∗}(QN^{−1}Q^{∗})^{−1}(P^{∗}MP)^{−1}P^{∗}M^{∗} \\ & = N^{−1}Q^{∗}(Q(QN^{−1})^{∗})^{−1}((MP)^{∗}P)^{−1}(MP)^{∗}. \end{aligned}  (2.3)

(b) Let M ∈ B(H_{2}) and N ∈ B(H_{1}) be invertible operators, such that QN^{−1}Q^{∗} is left invertible and P^{∗}MP right invertible. Then A_{M′,N′}^{†} exists if and only if QN^{−1}Q^{∗} and P^{∗}MP are invertible and

\begin{aligned}E & = N^{−1}Q^{∗}(QN^{−1}Q^{∗})^{−1} = (QN^{−1})^{∗}(Q(QN^{−1})^{∗})^{−1}, \\ F & = (P^{∗}MP)^{−1}P^{∗}M = ((MP)^{∗}P)^{−1}(MP)^{∗}. \end{aligned}  (2.4)

In this case is A_{M′,N′}^{†} = Q_{r}^{−1}P_{l}^{−1}, where Q_{r}^{−1} = E and P_{l}^{−1} = F.

(c) If M ∈ B(H_{2}) and N ∈ B(H_{1}) are positive and invertible operators, then

\begin{aligned}A_{M,N}^{†} = A_{M′,N′}^{†} & = (QN^{−1})^{∗}(Q(QN^{−1})^{∗})^{−1}((MP)^{∗}P)^{−1}(MP)^{∗} \\ & = N^{−1}Q^{∗}(QN^{−1}Q^{∗})^{−1}(P^{∗}MP)^{−1}P^{∗}M. \end{aligned}  (2.5)

Proof. (a) If M and N are invertible operators and A_{M,N}^{†} exists, using the principles from [8], from (3M) and (4N) we get

P^{∗}MP P_{l}^{−1} = P^{∗}M^{∗},  Q_{r}^{−1}Q(N^{−1})^{∗}Q^{∗} = N^{−1}Q^{∗},  (2.6)
P^{∗}MP = P^{∗}M^{∗}P,  QN^{−1}Q^{∗} = Q(N^{−1})^{∗}Q^{∗}.  (2.7)

From (2.7) we conclude that P^{∗}MP and QN^{−1}Q^{∗} are selfadjoint operators. We now prove that P^{∗}MP = P^{∗}M^{∗}P and QN^{−1}Q^{∗} = Q(N^{−1})^{∗}Q^{∗} are invertible. Indeed, from (1) and (3M) we get the following equation (see [8]):

(QXM^{−1}X^{∗}Q^{∗})(P^{∗}M^{∗}P) = I.

This means that P^{∗}M^{∗}P is left invertible. Also, since P^{∗}M^{∗}P is selfadjoint, we conclude that P^{∗}M^{∗}P is invertible. Similarly, (1) and (4N) imply the following

(QN^{−1}Q^{∗})(P^{∗}X^{∗}N^{∗}XP) = I,

which means that QN^{−1}Q^{∗} is right invertible, so it is also invertible.

Using invertibility of QN^{−1}Q^{∗} and P^{∗}M^{∗}P, from (2.6) it follows

P_{l}^{−1} = (P^{∗}M^{∗}P)^{−1}(MP)^{∗}, Q_{r}^{−1} = N^{−1}Q^{∗}(Q(N^{−1})^{∗}Q^{∗})^{−1}.  (2.8)

Now, the representations (2.3) follows from A_{M,N}^{†} = Q_{r}^{−1}P_{l}^{−1}, (2.8) and (2.7).

(b) Suppose that M ∈ B(H_{2}) and N ∈ B(H_{1}) are invertible operators, such that QN^{−1}Q^{∗} is left invertible and P^{∗}MP right invertible and A_{M′,N′}^{†} exists. From the equations (1) and (3M′) in the same way as in [7] we get (QXM^{−1}X^{∗}Q^{∗})(P^{∗}MP) = I, which means that P^{∗}MP is left invertible. Similarly, from (1) and (4N′) we obtain (QN^{−1}Q^{∗})(P^{∗}X^{∗}NXP) = I, which implies the right invertibility of QN^{−1}Q^{∗}. According to the assumptions, we conclude that P^{∗}MP and QN^{−1}Q^{∗} are invertible. The identities (2.4) can be proved using the method from [7].

On the other hand, if P^{∗}MP and QN^{−1}Q^{∗} are invertible and (2.4) holds, one can verify that Q_{r}^{−1}P_{l}^{−1} (where Q_{r}^{−1} = E and

P_{l}^{−1} = F) satisfies the equations which define A_{M′,N′}^{†}.

(c) Firstly, we prove that Q(QN^{−1})^{∗} and (MP)^{∗}P are positive and invertible in B(H_{3}). If x ∈ H_{3} and ∥x∥ = 1, then

(Q(QN^{−1})^{∗}x, x) = (N^{−1}Q^{∗}x, Q^{∗}x) > 0.

Suppose that inf_{∥x∥=1}(Q(QN^{−1})^{∗}x, x) = 0. Then there exists a sequence of unit

vectors (x_{n})_{n} in H_{3}, such that lim_{n}(N^{−1}Q^{∗}x_{n}, Q^{∗}x_{n}) = 0. Since N^{−1} is positive and invertible, it follows that there exists a subsequence (x_{n_{k}})_{k} of (x_{n})_{n}, such that lim_{k} Q^{∗}x_{n_{k}} = 0. Now, it follows that Q^{∗} is not one-to-one with closed range, so

Q is not onto. We get the contradiction, so Q(QN^{−1})^{∗} is positive and invertible in B(H_{3}). Analogously, we can prove that (MP)^{∗}P is positive and invertible in B(H_{3}).

The rest of the proof follows from parts (a) and (b).

Now, we consider {2}-generalized inverses with prescribed range and kernel. Fundamental results for matrices can be found in [1] and [5].

Let A ∈ B(H_{1}, H_{2}) and X ∈ B(H_{2}, H_{1}) be a {2}-inverse of A, such that R(X) = T is a closed subspace of H_{1} and N (X) = S is a closed subspace of H_{2}. Then we write X = A_{T,S}^{(2)}. For given closed subspaces T of H_{1} and S of H_{2}, it is

a natural question when A_{T,S}^{(2)} exists? The answer in the case of arbitrary Hilbert spaces is given in the following theorem.

We state the following elementary result.

Lemma 2.2. Let A ∈ B(H_{1}, H_{2}), T and S be closed subspaces of H_{1} and H_{2} respectively. Then the following statements are equivalent:

(a) A has a {2}-inverse X ∈ B(H_{2}, H_{1}) such that R(X) = T and N (X) = S;

(b) A: T → A(T) is invertible and A(T) ⊕ S = H_{2}.

In the case when (a) or (b) holds, X is unique and is denoted by A_{T,S}^{(2)}.

Now, we generalize the result from [5].

Theorem 2.3. Suppose that A, T and S satisfy the condition (a) or (b) from Lemma 2.2 and let Y ∈ B(H_{2}, H_{1}) be such that R(Y) = T and N (Y) = S. If there exists a Hilbert space H_{3} and a left invertible operator E ∈ B(H_{3}, H_{1}) such that R(E) = T, then

W = E^{∗}Y AE ∈ B(H_{3})

is invertible in B(H_{3}) and

A_{T,S}^{(2)} = EW^{−1}E^{∗}Y.

Proof. Notice that E: H_{3} → T is invertible, A: T → AT is inverible and Y : AT → T is invertible. Since N (E^{∗})^{⊥} = R(E) = T, it follows that E^{∗}: T → H_{3} is invertible, so W is invertible. Now, it is easy to verify that EW^{−1}E^{∗}Y is a {2}-inverse of A. Also, N (Y) = S, W^{−1}E^{∗}Y : AT → H_{3} is invertible and R(EW^{−1}E^{∗}Y) = R(E) = T, N (EW^{−1}E^{∗}Y) = S, so

A_{T,S}^{(2)} = EW^{−1}E^{∗}Y.

Theorem 2.4. Let A: H_{1} → H_{2} and X: H_{2} → H_{1}. Then X ∈ A{2} if and only if there exist Hilbert spaces H_{3}, H_{4}, H_{5} and operators

C ∈ B(H_{4}, H_{1}), D ∈ B(H_{2}, H_{3}), W_{1} ∈ B(H_{5}, H_{4}), W_{2} ∈ B(H_{3}, H_{5}),

such that DAC is g-invertible and W_{2}DACW_{1} is invertible. In this case:

X = CW_{1}(W_{2}DACW_{1})^{−1}W_{2}D.  (2.9)

Proof. If X possesses the form (2.9), it is not difficult to verify X ∈ A{2}. On the other hand, using the method from [10], it is easy to verify that X ∈ A{2} if and only if there exist operators C and D, such that

X = C(DAC)^{(1,2)}D, C ∈ B(H_{4}, H_{1}), D ∈ B(H_{2}, H_{3}).

According to part (a) of Theorem 2.1, X ∈ A{2} if and only if there exist operators W_{1} and W_{2}, such that W_{2}DACW_{1} is invertible, and X possesses the form (2.9).

We introduce a general representation of the Drazin inverse based on an arbitrary full-rank factorization of A^{l}, l ≥ k = asc(A) = des(A). The following theorem is a natural generalization of a Cline’s result from [6], introduced for complex matrices. We shall assume that A is not a nilpotent operator, i.e. A^{D} ≠ 0.

Theorem 2.5. Let X be a Banach space. If A ∈ B(X), l ≥ k = asc(A) = des(A) < ∞ and A^{l} = P_{A^{l}}Q_{A^{l}} is the full-rank decomposition of A^{l}, then

A^{D} = P_{A^{l}}(Q_{A^{l}}AP_{A^{l}})^{−1}Q_{A^{l}}.

Proof. If asc(A) = des(A) = k < ∞, then it is well-known that N (A^{l}) = N (A^{k}) and R(A^{l}) = R(A^{k}) for all l ≥ k,

X = X_{1} ⊕ X_{2},  (2.10)

where X_{1} = N (A^{l}) and X_{2} = R(A^{l}), A(X_{i}) ⊂ X_{i} for i = 1, 2, A_{1} = A|_{X_{1}} is nilpotent and A_{2} = A|_{X_{2}} is invertible (A is not nilpotent) [3], [4]. We can write

A = [ \begin{matrix}A_{1} & 0 \\ 0 & A_{2} \end{matrix} ] , A^{D} = [ \begin{matrix}0 & 0 \\ 0 & A_{2}^{−1} \end{matrix} ]

with respect to the decomposition (2.10) ([3], [4]). Since N (A^{l}) and R(A^{l}) are complementary and closed subspaces of X, it follows that A^{l} is g-invertible, so there exists the full-rank decomposition A^{l} = P_{A^{l}}Q_{A^{l}}, where P_{A^{l}} ∈ B(Z, X) is left invertible and Q_{A^{l}} ∈ B(X , Z) is right invertible, for some Banach space Z. By the isomorphism theorem [3], we can take that Z = X_{2}. We conclude that P_{A^{l}} and Q_{A^{l}} have the following representations with respect to (2.10):

P_{A^{l}} = [ M ; P̃ ]  and  Q_{A^{l}} = [ N Q̃] ,

where P̃, Q̃ ∈ B(X_{2}), M ∈ B(X_{2}, X_{1}), N ∈ B(X_{1}, X_{2}). Now, P_{A^{l}} is left invertible and Q_{A^{l}} is right invertible, so P_{A^{l}} and Q_{A^{l}} are g-invertible operators, N (P_{A^{l}}) = {0} and R(Q_{A^{l}}) = X_{2}. It follows that R(P_{A^{l}}) = R(A^{l}) = X_{2} and N (Q_{A^{l}}) = N (A^{l}) = X_{1}, so M = 0, N = 0 and

P_{A^{l}} = [ 0 ; P̃ ]  and  Q_{A^{l}} = [ 0 Q̃] .

It is easy to verify that P̃ is left invertible and Q̃ is right invertible in B(X_{2}). But

[ \begin{matrix}0 & 0 \\ 0 & A_{2}^{l} \end{matrix} ] = A^{l} = P_{A^{l}}Q_{A^{l}} = [ \begin{matrix}0 & 0 \\ 0 & P̃Q̃ \end{matrix} ] ,

so A_{2}^{l} = P̃Q̃. Since A_{2}^{l} is invertible, it follows that P̃ and Q̃ are invertible in B(X_{2}).

Now, Q_{A^{l}}AP_{A^{l}} = Q̃A_{2}P̃ is invertible in B(X_{2}), so

A^{D} = [ \begin{matrix}0 & 0 \\ 0 & A_{2}^{−1} \end{matrix} ] = [ \begin{matrix}0 & 0 \\ 0 & P̃(Q̃A_{2}P̃)^{−1}Q̃ \end{matrix} ] = P_{A^{l}}(Q_{A^{l}}AP_{A^{l}})^{−1}Q.

As a corollary, we get the following result.

Corollary 2.1. If X is a Banach space, A ∈ B(X) and asc(A) = des(A) = k < ∞ and A^{l} = P_{A^{l}}Q_{A^{l}} is an arbitrary full-rank decomposition of A^{l}, l ≥ k, then (a) (A^{D})^{l} = P_{A^{l}}(Q_{A^{l}}A^{l}P_{A^{l}})^{−1}Q_{A^{l}} = P_{A^{l}}(Q_{A^{l}}P_{A^{l}})^{−2}P_{A^{l}};

(b) AA^{D} = P_{A^{l}}(Q_{A^{l}}P_{A^{l}})^{−1}Q_{A^{l}};

(c) If X is a Hilbert space, then (A^{D})^{†} = (Q_{A^{l}})^{†}Q_{A^{l}}AP_{A^{l}}(P_{A^{l}})^{†}.

Proof. (a) Follows from (A^{D})^{l} = (A^{l})^{#} and Theorem 2.5.

(b) According to Theorem 2.5 it follows that Q_{A^{l}}P_{A^{l}} = Q̃P̃, so an easy computation shows that

P_{A^{l}}(Q_{A^{l}}P_{A^{l}})^{−1}Q_{A^{l}} = [ \begin{matrix}0 & 0 \\ 0 & I \end{matrix} ] = AA^{D}.

(c) Follows from Theorem 2.1 (d) and Theorem 2.5.

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(received 03.09.1998.)

University of Niš, Faculty of Philosophy, Department of Mathematics, Ćirila i Metodija 2, 18000 Niš, YUGOSLAVIA E-mail: dragan@archimed.filfak.ni.ac.yu dragan@filfak.filfak.ni.ac.yu

pecko@archimed.filfak.ni.ac.yu pecko@filfak.filfak.ni.ac.yu