Generalized inverses by bordered matrices

R.B.Bapat

Indian Statistical Institute, New Delhi, 110016, India

Bing Zheng ^{†}

Department of Mathematics, Shanghai University, Shanghai 200436, People’s Republic of China

Abstract

Several authors have considered nonsingular borderings A = ( \begin{matrix}B & C \\ D & X \end{matrix} ) of B and investigated the properties of submatrices of A^{−1}. Under specific conditions on the bordering, one can recover any g-inverse of B as a submatrix of A^{−1}. We consider borderings A of B, where A might be singular, or even rectangular. If A is m × n and if B is an r × s submatrix of A, we study the consequences of the equality m + n − rank(A) = r + s − rank(B) with reference to the g-inverses of A. It is shown that under this condition many properties enjoyed by nonsingular borderings have analogs for singular (or rectangular) borderings as well. We also consider g-inverses of the bordered matrix when certain rank additivity conditions are satisfied. It is shown that any g-inverse of B can be realized as a submatrix of a suitable g-inverse of A, under certain conditions.

AMS subject classifications: 15A09, 15A15. Keywords: generalized inverse, Moore-Penrose inverse, bordered matrix, rank additivity

1 Introduction

Let A be an m × n matrix over the complex field and let A^{∗} denote the conjugate transpose of A. We recall that a generalized inverse G of A is an n × m matrix which satisfies the first of the four Penrose equations

†The current address: Department of Mathematics, Shanghai Normal University, Shanghai, 200234, P.R.China. The present work was done while this author was visiting Indian Statistical Institute during October 2001- September 2002. He would like to express his thanks for the warm hospitality he enjoyed during his visit. Supported by Chinese Scholarship Council.

(1) AXA = A (2) XAX = X (3) (AX)^{∗} = AX (4) (XA)^{∗} = XA.

For a subset {i, j, …} of the set {1, 2, 3, 4}, the set of n × m matrices satisfying the equations indexed by {i, j, …} is denoted by A{i, j, …}. A matrix in A{i, j, …} is called an {i, j, …}-inverse of A and is denoted by A^{(i,j,...)}. In particular, the matrix G is called a {1}-inverse or a g-inverse of A if it satisfies (1). As usual, a g-inverse of A is denoted by A^{−}. If G satisfies (1) and (2) then it is called a reflexive inverse or a {1, 2}-inverse of A. Similarly, G is called a {1, 2, 3}-inverse of A if it satisfies (1),(2) and (3). The Moore-Penrose inverse of A is the matrix G satisfying (1)-(4). Any matrix A admits a unique Moore-Penrose inverse, denoted A^{†}. If A is n × n then G is called the group inverse of A if it satisfies (1), (2) and AG = GA. The matrix A has group inverse, which is unique and denoted by A^{♯}, if and only if rank(A) = rank(A^{2}). We refer to [3, 5] for basic results on g-inverse.

Suppose

A = ( \begin{matrix}& q_{1} & q_{2} \\ p_{1} & B & C \\ p_{2} & D & X \end{matrix} )  (1)

is a partitioned matrix. We say that A is obtained by bordering B. We will generally partition a g-inverse A^{−} of A as

A^{−} = ( \begin{matrix}& p_{1} & p_{2} \\ q_{1} & E & F \\ q_{2} & G & Y \end{matrix} )  (2)

which is in conformity with A^{∗}.

We say that the g-inverses of A have “block independence property” if for any g-inverses

A_{i}^{−} = ( \begin{matrix}E_{i} & F_{i} \\ G_{i} & Y_{i} \end{matrix} ) , i = 1, 2

of A, ( \begin{matrix}E_{1} & F_{1} \\ G_{1} & Y_{2} \end{matrix} ) , ( \begin{matrix}E_{1} & F_{1} \\ G_{2} & Y_{1} \end{matrix} ) etc. are also g-inverses of A.

If A is an m × n matrix, then the following function will play an important role in this paper:

ψ(A) = m + n − rank(A).

An elementary result is given next. For completeness, we include a proof.

Lemma 1 If B is a submatrix of A, then ψ(B) ≤ ψ(A).

Proof. Let

A = ( \begin{matrix}& q_{1} & q_{2} \\ p_{1} & B & C \\ p_{2} & D & X \end{matrix} ).

Then

rank(A) ≤ rank ( B C) + rank ( D X)
≤ rank(B) + rank(C) + p_{2}
≤ rank(B) + q_{2} + p_{2}.

From this inequality, we get ψ(B) ≤ ψ(A).

Note that a matrix B with rank(B) = r can be completed to a nonsingular matrix A of order n if and only if ψ(B) ≤ n [9, Theorem 5]. As another example of a result concerning ψ, if

A = ( \begin{matrix}& q_{1} & q_{2} \\ p_{1} & B & C \\ p_{2} & D & O \end{matrix} )

is a nonsingular matrix of order n, n = p_{1} + p_{2} = q_{1} + q_{2}, Then A^{−1} is of the form

A^{−1} = ( \begin{matrix}& p_{1} & p_{2} \\ q_{1} & E & F \\ q_{2} & G & O \end{matrix} )

if and only if ψ(B) = ψ(A). This will follow from Theorem 8.

Several authors [3, 4, 7, 9, 10, 11] have considered nonsingular borderings A of B and investigated the properties of submatrices of A^{−1}. Under specific conditions on the bordering, one can recover a special g-inverse of B as a submatrix of A^{−1}. It turns out that in all such cases the condition ψ(B) = ψ(A) holds. The main theme of the present paper is to investigate borderings A of B, where A might be singular, or even rectangular. We show that if ψ(A) = ψ(B) is satisfied then many properties enjoyed by nonsingular borderings have analogs for singular (or rectangular) borderings as well. For example, any g-inverse of B can be obtained as a submatrix of A^{−} where A is a bordering of B with ψ(A) = ψ(B). This will be shown in Section 4. In Section 5 we show how to obtain the Moore-Penrose inverse and the group inverse by a general, not necessarily nonsingular, bordering. In the next two sections we consider general borderings A of B and obtain some results concerning A^{−}.

We say that rank additivity holds in the matrix equation A = A_{1} + ⋯ + A_{k} if rank(A) = rank(A_{1}) + ⋯ + rank(A_{k}). Let R(A) and N(A) denote the range space of A and the null space of A respectively. We will need the following well-known result.

Lemma 2 [1]. Let A, B be m × n matrices. Then the following conditions are equivalent:

(i) rank(B) = rank(A) + rank(B − A).

(ii) Every B^{−} is a g-inverse of A.

(iii) AB^{−}(B − A) = O, (B − A)B^{−}A = O for any B^{−}.

(iv) There exists a B^{−} that is a g-inverse of both A and B − A.

It follows from the proof of Lemma 1 that if ψ(B) = ψ(A) then rank additivity holds in

( \begin{matrix}B & C \\ D & X \end{matrix} ) = ( \begin{matrix}B & O \\ O & O \end{matrix} ) + ( \begin{matrix}O & C \\ O & O \end{matrix} ) + ( \begin{matrix}O & O \\ D & X \end{matrix} )  (3)

and in

( \begin{matrix}B & C \\ D & X \end{matrix} ) = ( \begin{matrix}B & O \\ O & O \end{matrix} ) + ( \begin{matrix}O & O \\ D & O \end{matrix} ) + ( \begin{matrix}O & C \\ O & X \end{matrix} ) .  (4)

In Section 2 we discuss necessary and sufficient conditions for the block matrix ( \begin{matrix}E & F \\ G & Y \end{matrix} ) to be a g-inverse of ( \begin{matrix}B & C \\ D & X \end{matrix} ) under the assumption of rank additivity in (3)

and (4). In section 3, necessary and sufficient conditions for the block matrix ( \begin{matrix}E & F \\ G & Y \end{matrix} )

to be a g-inverse of ( \begin{matrix}B & C \\ D & X \end{matrix} ) are considered under the assumption ψ(A) = ψ(B). Some related results are also proved. These results can be viewed as a generalization of the works in [4, 7, 10].

2 G-inverses of a bordered matrix

Let A = ( \begin{matrix}B & C \\ D & X \end{matrix} ) be a block matrix which is a bordering of B. In this sec-

tion we will study some necessary and sufficient conditions for a partitioned matrix

( \begin{matrix}E & F \\ G & Y \end{matrix} ) , conformal with A^{∗}, to be a g-inverse of A.

Theorem 3 Let A = ( \begin{matrix}B & C \\ D & X \end{matrix} ). Then rank additivity holds in (3) and (4) and

H = ( \begin{matrix}E & F \\ G & Y \end{matrix} )is a g-inverse of A if and only if the following conditions hold.

(i). BEB = B, CGC = C, DF D = D, XGC = DF X = −DEC, X = XY X − DEC

(ii). CY D, BF X, CY X, XGB, XY D, BEC, DEB, CGB, BF D are null matrices Furthermore, if EBE = E, then X = XY X.

Proof. “Only if” part: Assume rank additivity in (3) and (4) and that H is a ginverse of A. Then by (ii) of Lemma 2, H is also a g-inverse of each summand matrix in (3) and (4). Using the definition of g-inverse, we easily get BEB = B, CGC = C, DF D = D, XY D = O, CY X = O, and

DF X + XY X = X, XGC + XY X = X.  (5)

On the other hand, by (iii) of Lemma 2, we have

( \begin{matrix}B & O \\ O & O \end{matrix} ) ( \begin{matrix}E & F \\ G & Y \end{matrix} ) ( \begin{matrix}O & C \\ O & O \end{matrix} ) = ( \begin{matrix}O & O \\ O & O \end{matrix} ) ⇒ BEC = O,

( \begin{matrix}O & C \\ O & O \end{matrix} ) ( \begin{matrix}E & F \\ G & Y \end{matrix} ) ( \begin{matrix}B & O \\ O & O \end{matrix} ) = ( \begin{matrix}O & O \\ O & O \end{matrix} ) ⇒ CGB = O,

( \begin{matrix}B & O \\ O & O \end{matrix} ) ( \begin{matrix}E & F \\ G & Y \end{matrix} ) ( \begin{matrix}O & O \\ D & X \end{matrix} ) = ( \begin{matrix}O & O \\ O & O \end{matrix} ) ⇒ BF D = O, BF X = O,

( \begin{matrix}O & C \\ O & O \end{matrix} ) ( \begin{matrix}E & F \\ G & Y \end{matrix} ) ( \begin{matrix}O & O \\ D & X \end{matrix} ) = ( \begin{matrix}O & O \\ O & O \end{matrix} ) ⇒ CY D = O, CY X = O,

( \begin{matrix}O & C \\ O & X \end{matrix} ) ( \begin{matrix}E & F \\ G & Y \end{matrix} ) ( \begin{matrix}O & O \\ D & O \end{matrix} ) = ( \begin{matrix}O & O \\ O & O \end{matrix} ) ⇒ CY D = O, XY D = O.

\begin{matrix}( \begin{matrix}O & O \\ D & X \end{matrix} ) ( \begin{matrix}E & F \\ G & Y \end{matrix} ) ( \begin{matrix}B & O \\ O & O \end{matrix} ) = ( \begin{matrix}O & O \\ O & O \end{matrix} ) \\ ( \begin{matrix}O & C \\ O & X \end{matrix} ) ( \begin{matrix}E & F \\ G & Y \end{matrix} ) ( \begin{matrix}B & O \\ O & O \end{matrix} ) = ( \begin{matrix}O & O \\ O & O \end{matrix} ) \end{matrix} \} ⇒ XGB = O, DEB = O,

\begin{matrix}( \begin{matrix}O & O \\ D & X \end{matrix} ) ( \begin{matrix}E & F \\ G & Y \end{matrix} ) ( \begin{matrix}O & C \\ O & O \end{matrix} ) = ( \begin{matrix}O & O \\ O & O \end{matrix} ) \\ ( \begin{matrix}O & O \\ D & O \end{matrix} ) ( \begin{matrix}E & F \\ G & Y \end{matrix} ) ( \begin{matrix}O & C \\ O & X \end{matrix} ) = ( \begin{matrix}O & O \\ O & O \end{matrix} ) \end{matrix} \} ⇒ XGC = DF X = −DEC. (6)

Also, (5) and (6) imply X = XY X − DEC.

“If” part: If the conditions (i) and (ii) hold, then it is easy to verify that H is a g-inverse of each summand matrix in (3) and (4). By (iv) in Lemma 2, rank additivity holds in (3) and (4). It is also easily verified that H is a g-inverse of A.

If EBE = E, then DEC = O and so X = XY X.

We note certain consequences of Theorem 3.

Corollary 4 Let A = ( \begin{matrix}B & C \\ D & X \end{matrix} ). Then rank additivity holds in (3) and (4) and the

matrix H = ( \begin{matrix}E & F \\ G & O \end{matrix} ) is a g-inverse of A if and only if the following conditions hold.

(i). BEB = B, CGC = C, DF D = D, DEC = −X.

(iii). BEC, DEB, CGB, BF D are null matrices.

Furthermore if EBE = E, then X = O.

Corollary 5 Let A = ( \begin{matrix}B & C \\ D & O \end{matrix} ). Then R(B) ∩ R(C) = \{0\}, R(B^{∗}) ∩ R(D^{∗}) = \{0\}

and H = ( \begin{matrix}E & F \\ G & Y \end{matrix} ) is a g-inverse of A if and only if the following conditions hold.

(i). BEB = B, CGC = C, DF D = D.

(ii). CY D, DEC, BEC, DEB, CGB, BF D are null matrices.

In this case, the g-inverses of A have block independence property.

Remark: As the conditions R(B) ∩ R(C) = {0}, R(B^{∗}) ∩ R(D^{∗}) = {0} together with X = O imply rank additivity in (3) and (4), Corollary 5 is a direct consequence of Theorem 3. In particular, conditions (i) and (ii) indicate that the block matrices

in ( \begin{matrix}E & F \\ G & Y \end{matrix} ) can be independently chosen if it is a g-inverse of A. In other words,

the g-inverses of A = ( \begin{matrix}B & C \\ D & O \end{matrix} ) have block independence property. Thus Corollary

5 complements the known result (see Theorem 3.1 in [13] and Lemma 5(1.2e) in [6]) that the g-inverses of A have block independence property if and only if

rank(A) = rank ( \begin{matrix}B \\ D \end{matrix} ) + rank(C)
= rank ( B C) + rank(D).

The next result can also be viewed as a generalization of Corollary 5. This type of rank additivity has been considered, for example, in [12].

Theorem 6 Let A = ( \begin{matrix}B & C \\ D & X \end{matrix} ) and suppose rank(A) = rank(B) + rank(C) +

rank(D) + rank(X). Then H = ( \begin{matrix}E & F \\ G & Y \end{matrix} ) is a g-inverse of A if and only if the

following conditions hold.

(i) BEB = B, CGC = C, DF D = D, XY X = X.

(ii) BF X, CY D, CY X, DF X, XGB, XGC, XY D, BEC, BF D, CGB, DEB, DEC are null matrices.

Proof. Note that the condition rank(A) = rank(B) + rank(C) + rank(D) + rank(X) implies rank additivity in

A = ( \begin{matrix}B & O \\ O & O \end{matrix} ) + ( \begin{matrix}O & C \\ O & O \end{matrix} ) + ( \begin{matrix}O & O \\ D & O \end{matrix} ) + ( \begin{matrix}O & O \\ O & X \end{matrix} ) ,

Now the proof is similar to that of Theorem 3.

A generalization of Theorem 6 is stated next; the proof is omitted.

Theorem 7 Let A = ( A_{i,j} ) , i = 1, 2, ⋯, m, j = 1, 2, ⋯, n be an m×n block matrix.

If rank(A) = ∑_{i=1}^{m} ∑_{j=1}^{n} rank(A_{i,j}), then G = ( G_{l,s}) , l = 1, 2, ⋯, n, s = 1, 2, ⋯, m is a

g-inverse of A if and only if the following equations hold.

A_{i,j}G_{j,l}A_{l,s} = \{ \begin{matrix}A_{i,j} & (i, j) = (l, s) \\ O & (i, j) ≠ (l, s) \end{matrix} .

3 G-inverses of block matrix A with ψ(A) = ψ(B)

Let A and H be matrices of order m × n and n × m respectively, partitioned as follows:

A = ( \begin{matrix}& q_{1} & q_{2} \\ p_{1} & B & C \\ p_{2} & D & X \end{matrix} )and H = ( \begin{matrix}& p_{1} & p_{2} \\ q_{1} & E & F \\ q_{2} & G & Y \end{matrix} ),  (7)

where p_{1} + p_{2} = m and q_{1} + q_{2} = n. By η(A) we denote the row nullity of A, which by definition is the number of rows minus the rank of A. If m = n, A is nonsingular, H = A^{−1} and if A and H are partitioned as in (7) then it was proved by Fiedler and Markham [9], and independently by Gustafson [8], that

η(B) = η(Y).  (8)

The following result, proved in [2], will be used in the sequel. We include an alternative simple proof for completeness.

Lemma 8 Let A and H be matrices of order m×n and n×m respectively, partitioned as in (7). Assume rank(A) = r and rank(H) = k. Then the following assertions are true.

(i) If AHA = A, then

−(m − r) ≤ η(Y) − η(B) ≤ n − r.

(ii) If HAH = H, then

−(n − k) ≤ η(B) − η(Y) ≤ m − k.

Proof. (i) According to a result on bordered matrices and g-inverse([10], Theorem 1) there exist matrices P, Q and Z of order m×(m−r), (n−r)×n and (n−r)×(m−r) respectively, such that the matrix

S = ( \begin{matrix}A & P \\ Q & Z \end{matrix} )

is nonsingular and the submatrix formed by the first n rows and the first m columns of T = S^{−1} is W. Thus we may write

S =  ( \begin{matrix}& q_{1} & q_{2} & m − r \\ p_{1} & B & C & P_{1} \\ p_{2} & D & X & P_{2} \\ n − r & Q_{1} & Q_{2} & Z \end{matrix} )and T =  ( \begin{matrix}& p_{1} & p_{2} & n − r \\ q_{1} & E & F & U_{1} \\ q_{2} & G & Y & U_{2} \\ m − r & V_{1} & V_{2} & W \end{matrix} ).

Since S is nonsingular, we have, using (8),

η(B) = η(( \begin{matrix}Y & U_{2} \\ V_{2} & W \end{matrix} )) = q_{2} + m − r − rank ( \begin{matrix}Y & U_{2} \\ V_{2} & W \end{matrix} ) .

Now by Lemma 1

rank(Y) ≤ rank ( \begin{matrix}Y & U_{2} \\ V_{2} & W \end{matrix} ) ≤ rank(Y) + m + n − 2r,

and hence

−(m − r) ≤ η(Y) − η(B) ≤ n − r.

The result (ii) follows from (i).

The following result, proved using Lemma 8, will be used in the sequel.

Theorem 9 Let A = ( \begin{matrix}B & C \\ D & X \end{matrix} ) with ψ(A) = ψ(B). Then for any g-inverse A^{−} =

( \begin{matrix}E & F \\ G & Y \end{matrix} ) of A, Y = O.

Proof. Assume the sizes of the block matrices in A to be as in (7). By Lemma 8 we have

−(m − r) ≤ η(Y) − η(B) ≤ n − r.

It follows that

−m + r ≤ q_{2} − rank(Y) − p_{1} + rank(B).

Using ψ(A) = ψ(B) and the inequality above we get rank(Y) = 0 and hence Y = O.

Theorem 10 Let A = ( \begin{matrix}B & C \\ D & X \end{matrix} ) . Then ψ(A) = ψ(B) and H = ( \begin{matrix}E & F \\ G & Y \end{matrix} ) is a g-inverse of A if and only if the following equations hold.

(i). Y = O, BEB = B, GC = I, DF = I.

(ii). DEC = −X.

(iii). BEC = O, DEB = O, BF = O, GB = O.

Furthermore, if EBE = E, then X = O.

Proof. If H is a g-inverse of A with ψ(A) = ψ(B), then by Theorem 9, we know Y = O. From the proof of Lemma 1, the condition ψ(A) = ψ(B) also indicates rank additivity in (3) and (4). Note that C and D are also of full column rank and of full row rank respectively under the condition ψ(A) = ψ(B). Then the proof of the theorem is similar to that of Theorem 3.

The proof of the following result is also similar and is omitted.

Theorem 11 Let A = ( \begin{matrix}B & C \\ D & X \end{matrix} ) , H = ( \begin{matrix}E & F \\ G & Y \end{matrix} ) and consider the statements (i) Y = O, BEB = B, GC = I, DF = I, BF = O, GB = O (ii) EB + F D is hermitian (iii) BE + CG is hermitian (iv) EBE + F DE = E (v) EBE + ECG = E. Then

(a) ψ(A) = ψ(B) and H ∈ A{1, 2, 3} if and only if (i), (ii), (iv) hold, DEC = −X, EC = F DEC and DEB = O.

(b) ψ(A) = ψ(B) and H ∈ A{1, 2, 4} if and only if (i), (iii), (v) hold, DEC = −X, DE = DECG and BEC = O.

(c) ψ(A) = ψ(B) and H = A^{†} if and only if (i)-(v) hold, DE + XG = O and EC + F X = O.

The two previous results will be used in the proof of the next result.

Theorem 12 let A = ( \begin{matrix}B & C \\ D & X \end{matrix} ) . Then the following conditions are equivalent:

(1) ψ(A) = ψ(B) and ( \begin{matrix}B & C \\ D & X \end{matrix} )^{†} = ( \begin{matrix}B^{†} & D^{†} \\ C^{†} & X^{†} \end{matrix} )

(2) ψ(A) = ψ(B) and ( \begin{matrix}B^{†} & D^{†} \\ C^{†} & X^{†} \end{matrix} ) is a g-inverse of A.

(3) X = O, C^{†}C = I, DD^{†} = I, BD^{†} = O, C^{†}B = O.

(4) X = O, C^{†}C = I, DD^{†} = I, BD^{∗} = O, C^{∗}B = O.

(5) ψ(A) = ψ(B) and ( \begin{matrix}E & D^{†} \\ C^{†} & X^{†} \end{matrix} ) is a g-inverse of A for some E ∈ B^{\{1,2\}}.

(6) ψ(A) = ψ(B) and ( \begin{matrix}E & D^{†} \\ C^{†} & X^{†} \end{matrix} ) is a g-inverse of A for some E.

(7) ψ(A) = ψ(B) and ( \begin{matrix}B & C \\ D & X \end{matrix} )^{†} = ( \begin{matrix}B^{†} & F \\ G & Y \end{matrix} ) for some matrices F, G, Y.

(8) ψ(A) = ψ(B) and ( \begin{matrix}B^{†} & F \\ C^{†} & Y \end{matrix} ) is a \{1, 2, 3\}-inverse of A for some F, Y.

(9) ψ(A) = ψ(B) and ( \begin{matrix}B^{†} & D^{†} \\ G & Y \end{matrix} ) is a \{1, 2, 4\}-inverse of A for some G, Y.

Proof. Clearly, (1) ⇒ (2). (2 ⇒ (3): This follows from Theorem 10.

(3) ⇔ (4) : Since BD^{†} = O and C^{†}B = O are equivalent to BD^{∗} = O and C^{∗}B = O respectively, we have this implication.

(3) ⇒ (1) : Note that BD^{†} = O and C^{†}B = O imply DB^{†} = O and B^{†}C = O. Then it is easy to verify that ( \begin{matrix}B^{†} & D^{†} \\ C^{†} & X^{†} \end{matrix} ) is A^{†} thus (1) holds.

Clearly, (1) ⇒ (5) ⇒ (6).

(6) ⇒ (3) : By Theorem 10, if ( \begin{matrix}E & D^{†} \\ C^{†} & X^{†} \end{matrix} ) is a g-inverse of A for some matrix E, then we have X^{†} = O, C^{†}C = I, DD^{†} = I, BD^{†} = O and C^{†}B = O. Note that X^{†} = O ⇔ X = O, thus (3) holds.

(6) ⇒ (1) : This follows from (6) ⇒ (3) and (3) ⇒ (1). Obviously, (1) ⇒ (7), (1) ⇒ (8) and (1) ⇒ (9).

(7) ⇒ (1) : By Theorem 10, we have X = O, Y = O, GC = I, DF = I, BF = O and GB = O. Clearly, G ∈ C{1, 2, 4} and F ∈ D{1, 2, 3}. Using the hermitian property

of the matrices ( \begin{matrix}B & C \\ D & X \end{matrix} ) , ( \begin{matrix}B^{†} & F \\ G & Y \end{matrix} ), ( \begin{matrix}B^{†} & F \\ G & Y \end{matrix} ) ( \begin{matrix}B & C \\ D & X \end{matrix} ) , BB^{†} and B^{†}B, it is easy

to conclude that CG and F D are also hermitian. Thus F = D^{†} and G = C^{†}. Note that Y = X^{†} = O and (1) is proved.

Similarly, using Theorem 11 we can show (8) ⇒ (1) and (9) ⇒ (1) and the proof is complete.

4 Obtaining any g-inverse by bordering

By Theorem 10 if A = ( \begin{matrix}B & C \\ D & X \end{matrix} ) with ψ(A) = ψ(B) and if H = ( \begin{matrix}E & F \\ G & O \end{matrix} ) is a g-inverse of A, then E is a g-inverse of B which also satisfies DEC = −X, BEC = O and DEB = O. Such E, hereafter, will be denoted by E_{(C,D,X)}. Note that E_{(C,D,X)} is not uniquely determined by C, D, X, since A^{−} is not unique. In this section we will investigate the converse problem, that is: for a given g-inverse E of B, how to construct C, D and X so that H = ( \begin{matrix}E & F \\ G & O \end{matrix} ) is a g-inverse of A = ( \begin{matrix}B & C \\ D & X \end{matrix} ) with ψ(A) = ψ(B) for some matrices of proper sizes. We first state some well-known lemmas to be used later, see, for example, [3 5].

Lemma 13 The following three statements are equivalent: (i) E is a g-inverse of B (ii) BE is an idempotent matrix and rank(BE) = rank(B) and (iii) EB is an idempotent matrix and rank(EB) = rank(B).

Lemma 14 E is a {1, 2}-inverse of B if and only if E is a g-inverse of B and rank(E) = rank(B).

Lemma 15 Let H = UV be a rank factorization of a square matrix. Then the following three statements are equivalent: (i) H is an idempotent matrix, (ii) I − H is an idempotent matrix, and (iii) V U = I.

Theorem 16 (i) Let E be a g-inverse of the p_{1} × q_{1} matrix B with rank(C) = r. Then there exist C, D, and X such that E = E_{(C,D,X)}, where rank(C) ≤ p_{1} − r and rank(D) ≤ q_{1} − r.

(ii) If E = E_{(C,D,X)}, then there exist matrices U, V, Ū and V̄ such that

I − BE = ( C U ) ( \begin{matrix}G \\ V \end{matrix} ) and I − EB = ( F Ū ) ( D ; V̄ )

are the rank factorizations of I − BE and I − EB respectively.

(iii) rank(E_{(C,D,X)}) = rank(B) + rank(R), where

R = ( \begin{matrix}−X & DEU \\ V̄ EC & V̄ EU \end{matrix} )  (9)

for some matrices U and V̄ as in (ii).

Proof. For a given g-inverse E of B, we use rank factorizations of I − BE and I − EB, by which there exist C, D, X, F, G, U, Ū, V, and V̄ satisfying the following identities

I − BE = ( C U ) ( \begin{matrix}G \\ V \end{matrix} ) ,  (10)

I − EB = ( F Ū ) ( D ; V̄ ) ,  (11)

X = −DEC.

To prove (i), we only need to show that these C, D, X, F and G along with Y = O satisfy the conditions (i) and (ii) in Corollary 4. In fact, from (10) and (11), we have, in view of Lemma 15, that( \begin{matrix}G \\ V \end{matrix} ) ( C U ) = I and ( D ; V̄ ) ( F Ū ) = I, implying

GC = I and DF = I.

Again from (10) and (11), we have, by (I − BE)B = O and B(I − EB) = O,

( \begin{matrix}G \\ V \end{matrix} ) B = O and B ( F Ū ) = O,  (12)

and by BE(I − BE) = O and (I − EB)EB = O,

BE ( C U ) = O and ( D ; V̄ ) EB = O.  (13)

Now by (12), GB = O and BF = O, and by (13), BEC = O and DEB = O.

(ii) Let E = E_{(C,D,X)}. By Theorem 10, BEC = O, which means R(C) ⊆ N(BE) = R(I −BE). Note that C is of full column rank under the condition ψ(A) = ψ(B). Thus there exists a matrix U so that R(( C U )) = R(I − BE) and the matrix ( C U ) is of full column rank. Hence, there exists a matrix of full row rank which can be partitioned as ( \begin{matrix}G \\ V \end{matrix} ) such that

I − BE = ( C U ) ( \begin{matrix}G \\ V \end{matrix} ) .

On the other hand, DEB = O implies N(I − EB) = R(EB) ⊆ N(D). So there exists a matrix V̄ such that ( D ; V̄ ) is of full row rank and

N(I − EB) = N(( D ; V̄ )).

From this we conclude that there exists a matrix of full column rank which can be partitioned as ( F Ū ) such that

I − EB = ( F Ū ) ( D ; V̄ ) .

Now we prove (iii). If E = E_{(C,D,X)}, then from the proof of (ii) there exist matrices U, V, Ū and V̄ such that (10) and (11) hold. Hence BE ( C U ) = O and ( D ; V̄ ) EB = O. Therefore we have

( B ; D ; V̄ ) E ( B C U ) = ( \begin{matrix}B & O & O \\ O & DEC & DEU \\ O & V̄ EC & V̄ EU \end{matrix} )
= ( \begin{matrix}B & O \\ O & R \end{matrix} ) ,

where R = ( D ; V̄ ) E ( C U ) .

On the other hand,

( E F Ū ) ( \begin{matrix}B & O \\ O & R \end{matrix} ) ( \begin{matrix}E \\ G \\ V \end{matrix} ) = EBE + ( F Ū ) R ( \begin{matrix}G \\ V \end{matrix} )
= EBE + (I − EB)E(I − BE)
= E.

Thus we have rank(E_{(C,D,R)}) = rank(B) + rank(R).

Theorem 16(i) and its proof show that for a given matrix B and its g-inverse E we can find matrices C of full column rank with R(C) ⊆ N(BE) and D of full row rank with R(EB) ⊆ N(D), as well as X = −DEC, F and G such that matrix ( \begin{matrix}E & F \\ G & O \end{matrix} ) is

a g-inverse of A = ( \begin{matrix}B & C \\ D & O \end{matrix} ) with ψ(A) = ψ(B). Furthermore, we have the following.

Corollary 17 Let B and its g-inverse E be given. Then the matrix A = ( \begin{matrix}B & C \\ D & X \end{matrix} )

satisfies ψ(A) = ψ(B) and has a g-inverse of the form ( \begin{matrix}E & F \\ G & O \end{matrix} ) if and only if C is of full column rank with R(C) ⊆ N(BE) and D of full row rank with R(EB) ⊆ N(D). In this case, X = −DEC, F ∈ D{1, 3}, G ∈ C{1, 4}, BF = O and GB = O.

Proof. Necessity: This follows from Theorem 10.

Sufficiency: The proof of sufficiency is similar to that of Theorem 16(i), (ii).

As a special case we recover the following known result.

Corollary 18 [10,Theorem 1] Let E be a g-inverse of B. Then for any matrix C of full column rank with R(C) = N(BE) and any matrix D of full row rank with N(D) = R(EB), the matrix

A = ( \begin{matrix}B & C \\ D & −DEC \end{matrix} )

is nonsingular and

A^{−1} = ( \begin{matrix}E & F \\ G & O \end{matrix} ) ,

where F ∈ D{1, 3}, BF = O, G ∈ C{1, 4} and GB = O.

5 Moore-Penrose inverse and group inverse by bordering

For a given g-inverse E of B, Corollary 17 shows that C and D can be chosen with the conditions R(C) ⊆ N(BE) and R(D^{∗}) ⊆ N((EB)^{∗}) so that A = ( \begin{matrix}B & C \\ D & −DEC \end{matrix} )

satisfies ψ(A) = ψ(B) and has a g-inverse of the form ( \begin{matrix}E & F \\ G & O \end{matrix} ) . Further, Corollary 18 provides an approach to border the matrix B into a nonsingular matrix such that in its inverse, the block matrix on the upper left corner is E. We now show how to border the matrix if E is the Moore-Penrose inverse or the group inverse of B.

Theorem 19 Let B be given. Then the matrix A = ( \begin{matrix}B & C \\ D & X \end{matrix} ) satisfies ψ(A) = ψ(B)

and has a g-inverse of the form ( \begin{matrix}B^{†} & F \\ G & O \end{matrix} ) if and only if C has full column rank with R(C) ⊆ N(B^{∗}) and D has full row rank with R(D^{∗}) ⊆ N(B). In this case, X = −DB^{†}C = O and

A^{†} = ( \begin{matrix}B^{†} & D^{†} \\ C^{†} & O \end{matrix} ) .

Proof. Note that N(BB^{†}) = N(B^{∗}) and N((EB)^{∗}) = N(B^{†}) = N(B), and the necessity and sufficiency follow from Corollary 17.

It is easy to verify that ( \begin{matrix}B^{†} & D^{†} \\ C^{†} & O \end{matrix} ) is a g-inverse of A. Thus by Corollary 12,(2), we have

A^{†} = ( \begin{matrix}B^{†} & D^{†} \\ C^{†} & O \end{matrix} ) ,

where X = −DB^{†}C = O.

Combining Corollary 18 with Theorem 19, we have

Corollary 20 [4] Let B be a p_{1} × q_{1} matrix with rank(B) = r. Suppose the columns of C ∈ C_{p_{1}−r}^{p_{1}×(p_{1}−r)} are a basis of N(B^{∗}) and the columns of D^{∗} ∈ C_{q_{1}−r}^{q_{1}×(q_{1}−r)} are a basis for N(B). Then the matrix

A = ( \begin{matrix}B & C \\ D & O \end{matrix} )

is nonsingular and its inverse is

A^{−1} = ( \begin{matrix}B^{†} & D^{†} \\ C^{†} & O \end{matrix} ) .

If B is square and has group inverse, we can get a bordering ( \begin{matrix}B & ∗ \\ ∗ & O \end{matrix} ) of B such

that it has a g-inverse in the form ( \begin{matrix}B^{♯} & ∗ \\ ∗ & O \end{matrix} ) . Part (ii) of the following result is known. We generalize it to any bordering, not necessarily nonsingular, in part (i).

Theorem 21 Let B be n × n and with index 1. Then

(i) there exist matrices C of full column rank with R(C) ⊆ N(B) and D of full row rank with R(B) ⊆ N(D) which satisfy DC = I such that ( \begin{matrix}B^{♯} & C \\ D & O \end{matrix} ) is a g-inverse of

( \begin{matrix}B & C \\ D & O \end{matrix} ) with ψ(A) = ψ(B).

(ii)[7] for any matrix C of full column rank with R(C) = N(B) and any matrix D of full row rank with R(B) = N(D), the matrix

A = ( \begin{matrix}B & C \\ D & O \end{matrix} )

is nonsingular and

A^{−1} = ( \begin{matrix}B^{♯} & C(DC)^{−1} \\ (DC)^{−1}D & O \end{matrix} ) .

Proof. (i): Consider the rank factorization of I − BB^{♯} given by

I − BB^{♯} = ( C U ) ( \begin{matrix}D \\ V \end{matrix} ) .

Note that BB^{♯} = B^{♯}B, and we have

I − B^{♯}B = ( C U ) ( \begin{matrix}D \\ V \end{matrix} ) .

Obviously R(C) ⊆ N(A) and R(B) ⊆ N(D). As in the proof of Theorem 16(i), we conclude that ( \begin{matrix}B^{♯} & C \\ D & O \end{matrix} ) is a g-inverse of ( \begin{matrix}B & C \\ D & O \end{matrix} ) with ψ(A) = ψ(B), since X = −DB^{♯}C = O.

(ii): By Corollary 18, the nonsingularity of the matrix ( \begin{matrix}B^{♯} & C \\ D & O \end{matrix} ) under the conditions R(C) = N(A) and R(B) = N(D) can be easily seen. We now prove that for any matrix C of full column rank with R(C) = N(B) and any matrix D of full row rank with R(B) = N(D), DC is nonsingular.

In fact, if DCx = O, then Cx ∈ R(C) and Cx ∈ N(D). Since R(C) = N(B), N(D) = R(B) and R(B) ∩ N(B) = {0}, we have Cx = O and therefore x = 0. Thus DC is nonsingular.

By Lemma 15, C(DC)^{−1}D is an idempotent matrix and

I − BB^{♯} = I − B^{♯}B = C(DC)^{−1}D

is a rank factorization. From Corollary 18, we know that

( \begin{matrix}B & C(DC)^{−1} \\ D & O \end{matrix} ) and ( \begin{matrix}B & C \\ (DC)^{−1}D & O \end{matrix} )

are nonsingular and in fact

( \begin{matrix}B & C(DC)^{−1} \\ D & O \end{matrix} )^{−1} = ( \begin{matrix}B^{♯} & C(DC)^{−1} \\ D & O \end{matrix} ) .

Note that

( \begin{matrix}B & C(DC)^{−1} \\ D & O \end{matrix} ) = ( \begin{matrix}B & C \\ D & O \end{matrix} ) ( \begin{matrix}I & O \\ O & (DC)^{−1} \end{matrix} ) .

The result immediately follows from the two preceding above.

Remark: Theorem 21(ii) can be used to compute the group inverse of the matrix (I − T)^{♯} which plays an important role in the theory of Markov chains, where T is the transition matrix of a finite Markov chain. For an n-state ergodic chain, it is wellknown that the transition matrix T is irreducible and rank(I −T) = n−1 [5, Theorem 8.2.1]. Hence by Theorem 21(ii) we can compute the group inverse (I − T)^{♯} of I − T by a bordered matrix.

Let c be a right eigenvector of T and d a left eigenvector, that is c and d satisfy T c = c and d^{∗}T = d^{∗}, respectively. Then the bordered matrix ( \begin{matrix}I − T & c \\ d^{∗} & 0 \end{matrix} ) is nonsingular and

( \begin{matrix}I − T & c \\ d^{∗} & 0 \end{matrix} )^{−1} = ( \begin{matrix}(I − T)^{♯} & \frac{c}{d^{∗}c} \\ \frac{d}{d^{∗}c} & o \end{matrix} ) .

Thus the group inverse (I − T)^{♯} can be obtained by computing the inverse of a nonsingular matrix.

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