1 Introduction

Throughout this paper, we denote the real \(m \times n\) matrix space by \({{\mathbf {R}}}^{m \times n}\) and denote the transpose, the Moore–Penrose generalized inverse, the range space and the null space of a real matrix \(A \in {{\mathbf {R}}}^{m \times n}\) by \(A^\top \), \(A^+,\)\({\mathcal {R}}(A)\) and \({\mathcal {N}}(A)\), respectively. \(I_n\) represents the identity matrix of size \(n\). \(P_{{\mathcal {L}}}\) stands for the orthogonal projector on the subspace \({\mathcal {L}}\subset {\mathbf {R}}^n.\) We write \(A \ge 0\) (\(A > 0\)) if \(A\) is a symmetric positive semi-definite matrix (symmetric positive definite matrix). Furthermore, for a matrix \(A \in {{\mathbf {R}}}^{m \times n},\) let \(E_A\) and \(F_A\) stand for two orthogonal projectors: \(E_A=I_m-AA^+\), \(F_A=I_n-A^+A.\)

Recently, Bai and Bai [1] have established some necessary and sufficient conditions for the nonsingularity of the block two-by-two matrix by making use of the singular value decompositions and the Moore–Penrose generalized inverses of the matrix blocks. This paper is a special case of [1], which considers the solution of block \( 2\times 2\) linear systems of the form

$$\begin{aligned} \left[ \begin{array}{ll} A &{}\quad B^\top \\ B &{}\quad -\,D \\ \end{array} \right] \left[ \begin{array}{c} x \\ y \\ \end{array} \right] =\left[ \begin{array}{c} f \\ g \\ \end{array} \right] , \end{aligned}$$
(1)

where \(A \in {\mathbf {R}}^{n\times n}, D \in {\mathbf {R}}^{p\times p}\) are symmetric and positive semi-definite and \(B \in {\mathbf {R}}^{p \times n}.\) Linear systems of the form (1) arise in a number of applications including mixed finite element solution of the Navier–Stokes, the Maxwell equations and constraint optimization [2,3,4,5]. There have been many methods for solving the (generalized) saddle point problems (1) [6,7,8,9,10,11,12]. However, before to solve (1), a significant problem is under which conditions the equation of (1) is solvable ? In this paper, some simple criteria for the linear systems (1) to have solutions and the unique solution are provided, and the solvability conditions are expressed by \(A, B, D, f\) and \(g.\)

2 Main Results

To begin with, we introduce some lemmata.

Lemma 1

[13] If \(L \in \mathbf{R}^{m \times q}, \ b \in \mathbf{R}^{m},\) then \(Ly=b\) has a solution \(y \in \mathbf{R}^{q}\) if and only if one of the following conditions is satisfied:

  1. (i)

    \(LL^+b=b;\)

  2. (ii)

    \( b\in {\mathcal {R}}(L);\)

  3. (iii)

    \(N(L^\top )\subseteq N(b^\top ),\) that is, \( L^\top x=0\Rightarrow b^\top x=0\) for some \(x\in \mathbf{R}^{m}.\)

In this case, the general solution of the equation can be described as \( y=L^+ b+F_Lz\), where \(z \in \mathbf{R}^q\) is an arbitrary vector.

Lemma 2

[13] If \(A \in {{\mathbf {R}}}^{m \times n}, B \in {{\mathbf {R}}}^{n \times p},\) then

$$\begin{aligned} \mathrm{(i)}&{\mathcal {R}}(AB)\subseteq {\mathcal {R}}(A), \ {\mathcal {N}}(AB)\supseteq {\mathcal {N}}(B); \\ \mathrm{(ii)}&{\mathcal {R}}(I_m-AA^+)= {\mathcal {N}}(AA^+), \ {\mathcal {R}}(I_n-A^+A)= {\mathcal {N}}(A^+A); \\ \mathrm{(iii)}&{\mathcal {N}}(I_m-AA^+)= {\mathcal {R}}(AA^+), \ {\mathcal {N}}(I_n-A^+A)= {\mathcal {R}}(A^+A); \\ \mathrm{(iv)}&{\mathcal {R}}(AA^+)={\mathcal {R}}(A), \ {\mathcal {N}}(AA^+)= {\mathcal {N}}(A^+)={\mathcal {N}}(A^\top ); \\ \mathrm{(v)}&{\mathcal {R}}(A^+A)={\mathcal {R}}(A^+)={\mathcal {R}}(A^\top ), \ {\mathcal {N}}(A^+A)= {\mathcal {N}}(A); \\ \mathrm{(vi)}&{\mathcal {N}}(A^\top )={\mathcal {R}}(A)^\bot . \end{aligned}$$

Lemma 3

[13] If \(A \in {{\mathbf {R}}}^{m \times n} \) and \(P_{{\mathcal {L}}}\) is an orthogonal projector on the subspace \({\mathcal {L}}\), then

$$\begin{aligned} \mathrm{(i)}&AA^+=P_{{\mathcal {R}}(A)}, \ \ A^+A=P_{{\mathcal {R}}(A^\top )}; \\ \mathrm{(ii)}&E_A=I_m -AA^+= P_{{\mathcal {N}}(A^\top )}, \ \ F_A=I_n -A^+A= P_{{\mathcal {N}}(A)};\\ \mathrm{(iii)}&P_{{\mathcal {L}}}A=A \Leftrightarrow {\mathcal {R}}(A)\subseteq {\mathcal {L}}, \ \ A P_{{\mathcal {L}}}=A \Leftrightarrow {\mathcal {R}}(A^\top )\subseteq {\mathcal {L}}. \end{aligned}$$

Lemma 4

[13] Let \(P_{{\mathcal {L}}}\) and \(P_{{\mathcal {M}}}\) be the orthogonal projectors on the subspaces \({\mathcal {L}}\) and \({\mathcal {M}},\) respectively. Then \(P_{\mathcal {L}}P_{\mathcal {M}}\) is an orthogonal projector if and only if \(P_{\mathcal {L}}P_{\mathcal {M}}=P_{\mathcal {M}}P_{\mathcal {L}}.\) In this case, \(P_{\mathcal {L}}P_{\mathcal {M}}=P_{{\mathcal {L}}\cap {\mathcal {M}}}.\)

Lemma 5

[14] Assume that \(A \in {{\mathbf {R}}}^{m \times n}\) and \({\mathcal {T}}\) is a subspace of \({{\mathbf {R}}}^{n}\). Let \(\tilde{{\mathcal {T}}}={\mathcal {R}}(P_{\mathcal {T}}A^\top )=P_{\mathcal {T}} {\mathcal {R}}(A^\top ),\) then

$$\begin{aligned} \tilde{{\mathcal {T}}}={\mathcal {T}} \cap ({\mathcal {T}}\cap {\mathcal {N}}(A))^\bot , \quad \ \tilde{{\mathcal {T}}}^\bot ={\mathcal {T}}^\bot {\mathop {\oplus }\limits ^{\bot }} ({\mathcal {T}}\cap {\mathcal {N}}(A)). \end{aligned}$$

Lemma 6

Suppose that a matrix \(M\) is partitioned as

$$\begin{aligned} M=\left[ \begin{array}{ll} L_{11} &{}\quad L_{12} \\ L_{21} &{}\quad L_{22} \\ \end{array} \right] , \end{aligned}$$

where \(L_{11}\) and \(L_{22}\) are square. If \(M\) and \(L_{11}\) are nonsingular, then

$$\begin{aligned} M^{-1}=\left[ \begin{array}{ll} L_{11}^{-1}+ L_{11}^{-1}L_{12} K^{-1} L_{21} L_{11}^{-1} &{}\quad -\,L_{11}^{-1}L_{12} K^{-1} \\ - K^{-1} L_{21} L_{11}^{-1} &{}\quad K^{-1} \\ \end{array} \right] , \end{aligned}$$

where \(K=L_{22}- L_{21}L_{11}^{-1}L_{12}.\)

We first consider a special case: \(D=0.\) In this case, the equation of (1) can be equivalently written as

$$\begin{aligned} \left\{ \begin{array}{ll} Ax+B^\top \!{y}=f, \\ Bx=g. \end{array} \right. \end{aligned}$$
(2)

According to Lemma 1, the second equation of (2) has a solution if and only if \(BB^+g=g.\) In which case, the general solution of \(Bx=g\) is

$$\begin{aligned} x=B^+g+F_Bz, \end{aligned}$$
(3)

where \(z\in {\mathbf {R}}^{n}\) is an arbitrary vector. Substituting (3) into the first equation of (2), we obtain

$$\begin{aligned} AF_Bz+B^\top y=f-AB^+g. \end{aligned}$$
(4)

It follows from Lemma 1 that the equation of (4) with unknown vector \(y\) has a solution if and only if

$$\begin{aligned} Qz=F_Bf-F_BAB^+g, \end{aligned}$$
(5)

and the general solution of (4) with respect to \(y\) is

$$\begin{aligned} y=(B^\top )^+(f-AB^+g)-(B^\top )^+AF_Bz+E_Bv, \end{aligned}$$
(6)

where \(Q=F_BAF_B\) and \(v\in {\mathbf {R}}^{p}\) is an arbitrary vector. Using Lemma 1 again, we know that the equation of (5) with respect to \(z\) has a solution if and only if

$$\begin{aligned} E_QF_Bf=E_QF_BAB^+g. \end{aligned}$$
(7)

In this case, the general solution of (5) is

$$\begin{aligned} z=Q^+(F_Bf-F_BAB^+g)+F_Qu, \end{aligned}$$
(8)

where \(u \in {\mathbf {R}}^{n}\) is an arbitrary vector. Substituting (8) into (3) and (6), we obtain

Theorem 1

Assume that \(A \in {\mathbf {R}}^{n\times n}, B \in {\mathbf {R}}^{p \times n}\) and \(D=0.\) Let \(Q=F_BAF_B;\) then the equation of (1) has a solution if and only if

$$\begin{aligned} BB^+g=g, \ \ E_QF_Bf=E_QF_BAB^+g. \end{aligned}$$
(9)

In which case, the general solution of (1) is

$$\begin{aligned} x= & {} B^+g+F_BQ^+F_B(f-AB^+g)+F_B F_Q u, \end{aligned}$$
(10)
$$\begin{aligned} y= & {} (B^\top )^+(f-AB^+g)-(B^\top )^+AF_BQ^+F_B(f-AB^+g)\nonumber \\&-(B^\top )^+A F_B F_Qu+E_Bv, \end{aligned}$$
(11)

where \(u \in {\mathbf {R}}^{n}\) and \(v\in {\mathbf {R}}^{p}\) are arbitrary vectors.

Corollary 1

Assume that \(A \in {\mathbf {R}}^{n\times n}, B \in {\mathbf {R}}^{p \times n}\) and \(A\ge 0, \ D=0.\) Let \(Q=F_BAF_B;\) then the equation of (1) has a solution if and only if

$$\begin{aligned} BB^+g=g, \ \ E_QF_Bf=0, \end{aligned}$$
(12)

or equivalently,

$$\begin{aligned} g\in {\mathcal {R}}(B), \ \ f \in {\mathcal {R}}([A, B^\top ]). \end{aligned}$$
(13)

In which case, the general solution of (1) is

$$\begin{aligned} x= & {} B^+g+F_BQ^+F_B(f-AB^+g)+F_B F_Q u, \end{aligned}$$
(14)
$$\begin{aligned} y= & {} (B^\top )^+(f-AB^+g)-(B^\top )^+AF_BQ^+F_B(f-AB^+g)+E_Bv, \end{aligned}$$
(15)

where \(u \in {\mathbf {R}}^{n}\) and \(v\in {\mathbf {R}}^{p}\) are arbitrary vectors.

Proof

If \(A\ge 0,\) it is easily seen that \({\mathcal {R}}(Q)= {\mathcal {R}}(F_BA),\) then we have

$$\begin{aligned} E_QF_BA=0, \quad A F_B F_Q=0. \end{aligned}$$
(16)

Thus, we can obtain (12) by (9). On the other hand, we notice that

$$\begin{aligned} F_BE_Q=F_B(I_n-F_BAF_B(F_BAF_B)^+)=F_B-QQ^+, \end{aligned}$$

which implies that \(F_BE_Q=E_QF_B.\) From Lemma 4, we know \(F_BE_Q\) is an orthogonal projector, that is,

$$\begin{aligned} F_BE_Q=P_{{\mathcal {R}}(F_B)\cap {\mathcal {R}}(E_Q)}. \end{aligned}$$
(17)

By Lemma 2, we have

$$\begin{aligned}&{\mathcal {R}}(F_B)={\mathcal {R}}(I_n-B^+B)={\mathcal {N}}(B^+B)={\mathcal {N}}(B),\\&{\mathcal {R}}(E_Q)={\mathcal {R}}(I_n-QQ^+)={\mathcal {N}}(QQ^+)={\mathcal {N}}(Q^\top )={\mathcal {R}}(Q)^\bot . \end{aligned}$$

By Lemma 5,

$$\begin{aligned} {\mathcal {R}}(Q)={\mathcal {R}}(F_BA)=F_B{\mathcal {R}}(A)=P_{{\mathcal {N}}(B)}{\mathcal {R}}(A)={\mathcal {N}}(B)\cap ({\mathcal {N}}(B)\cap {\mathcal {N}}(A))^\bot . \end{aligned}$$

Therefore,

$$\begin{aligned} F_BE_Q=P_{{\mathcal {N}}(B)\cap ({\mathcal {N}}(B)^\bot \oplus ({\mathcal {N}}(B)\cap {\mathcal {N}}(A)))}=P_{{\mathcal {N}}(B)\cap {\mathcal {N}}(A)}. \end{aligned}$$
(18)

Using Lemmas 2, 3 and (18), we have

$$\begin{aligned} E_QF_Bf=0\Leftrightarrow f \in ({\mathcal {N}}(B)\cap {\mathcal {N}}(A))^\bot ={\mathcal {R}}(B^\top )+ {\mathcal {R}}(A)={\mathcal {R}}([A,B^\top ]), \end{aligned}$$

which implies that (12) and (13) are equivalent. By (10), (11) and (16), we can get (14) and (15). \(\square \)

Corollary 2

Assume that \(A \in {\mathbf {R}}^{n\times n}, B \in {\mathbf {R}}^{p \times n}\) and \(A\ge 0, \ D=0.\) Let \(Q=F_BAF_B;\) then the solution of (1) is unique if and only if one of the following conditions is satisfied:

  1. (a)

    \(\left[ \begin{array}{cc} A &{}\quad B^\top \\ B &{}\quad 0 \\ \end{array} \right] \) is nonsingular;

  2. (b)

    \( E_B=0, \ E_QF_B=0 ;\)

  3. (c)

    \(\text{ rank }(B)=p\) (that is, \(B\) is of full row rank),   \(A+B^\top B\) is nonsingular.

In which case, the unique solution of (1) is

$$\begin{aligned}&x=B^+g+F_BQ^+F_B(f-AB^+g), \end{aligned}$$
(19)
$$\begin{aligned}&y=(B^\top )^+(f-AB^+g)-(B^\top )^+AF_BQ^+F_B(f-AB^+g). \end{aligned}$$
(20)

Proof

Observe that the coefficient matrix of (1) is square, and therefore, the equation of (1) has a unique solution if and only if the condition (a) holds. From (14) and (15), the solution is unique if and only if \( E_B=0, E_QF_B=0.\) By (18), we have

$$\begin{aligned}E_QF_B=0\Leftrightarrow {\mathcal {N}}(B)\cap {\mathcal {N}}(A)=\{0\}. \end{aligned}$$

It follows from \(A\ge 0\) that \({\mathcal {N}}(A+B^\top B)={\mathcal {N}}(A)\cap {\mathcal {N}}(B),\) which implies that (b) and (c) are equivalent. \(\square \)

Corollary 3

Assume that \(A \in {\mathbf {R}}^{n\times n}, B \in {\mathbf {R}}^{p \times n}\) and \(A > 0, \ D=0.\) Let \(Q=F_BAF_B;\) then the equation of (1) has a solution if and only if \(g\in {\mathcal {R}}(B).\) In which case, the general solution of (1) is

$$\begin{aligned}&x=B^+g+F_BQ^+F_B(f-AB^+g), \\&y=(B^\top )^+(f-AB^+g)-(B^\top )^+AF_BQ^+F_B(f-AB^+g)+E_Bv, \end{aligned}$$

where \(v\in {\mathbf {R}}^{p}\) are arbitrary vectors.

Now, if \(A\ge 0 \) and \( D\ge 0,\) can we achieve the similar result as that of Corollary 1? The answer is affirmative.

Theorem 2

Assume that \(A \in {\mathbf {R}}^{n\times n}, D \in {\mathbf {R}}^{p\times p}, B \in {\mathbf {R}}^{p \times n}\) and \(A\ge 0, \ D \ge 0.\) Then the equation of (1) has a solution if and only if

$$\begin{aligned} f \in {\mathcal {R}}([A, B^\top ]), \quad \ g \in {\mathcal {R}}([B, -D]). \end{aligned}$$
(21)

Proof

The necessary part is clear. We now prove the sufficient part. When the conditions (21) hold, there exist vectors \(a,b, c\) and \(d\) such that

$$\begin{aligned} f=Aa+B^\top b, \quad \ g=Bc-Dd. \end{aligned}$$
(22)

Let \([x_0^\top , y_0^\top ]^\top \) be a solution of the equation

$$\begin{aligned} \left[ \begin{array}{ll} A &{}\quad B^\top \\ B &{}\quad -\,D \\ \end{array} \right] \left[ \begin{array}{c} x \\ y \\ \end{array} \right] =\left[ \begin{array}{c} 0 \\ 0 \\ \end{array} \right] . \end{aligned}$$

That is,

$$\begin{aligned} Ax_0+B^\top y_0=0,\quad \ Bx_0-Dy_0=0. \end{aligned}$$
(23)

It follows from (23) that

$$\begin{aligned} x_0^\top Ax_0+x_0^\top B^\top y_0=0, \quad y_0^\top Bx_0-y_0^\top Dy_0=0. \end{aligned}$$
(24)

Thus, we have \(x_0^\top Ax_0+y_0^\top Dy_0=0.\) Notice that \(A\ge 0, \ D \ge 0,\) we can get

$$\begin{aligned} Ax_0=0, \quad Dy_0=0. \end{aligned}$$
(25)

Substituting (25) into (23), we obtain

$$\begin{aligned} B^\top y_0=0, \quad Bx_0=0. \end{aligned}$$
(26)

It follows from (22), (25) and (26) that

$$\begin{aligned}{}[f^\top , g^\top ]\left[ \begin{array}{c} x_0 \\ y_0 \\ \end{array} \right] =a^\top Ax_0 +b^\top Bx_0+ c^\top B^\top y_0 - d^\top Dy_0=0. \end{aligned}$$

Therefore, according to the third condition of Lemma 1, we know the equation of (1) is solvable.

Theorem 3

Assume that \(A \in {\mathbf {R}}^{n\times n}, D \in {\mathbf {R}}^{p\times p}, B \in {\mathbf {R}}^{p \times n}\) and \(A\ge 0, \ D \ge 0.\) Then the solution of (1) is unique if and only if one of the following conditions is satisfied:

  1. (c1)

    \(\left[ \begin{array}{ll} A &{} B^\top \\ B &{} -D \\ \end{array} \right] \) is nonsingular;

  2. (c2)

    \(A+B^\top B \ \text{ and } \ D+BB^\top \) are nonsingular;

  3. (c3)

    \({\mathcal {N}}(A)\cap {\mathcal {N}}(B)=\{0\}, \ {\mathcal {N}}(D)\cap {\mathcal {N}}(B^\top )=\{0\}.\)

In which case, the unique solution of (1) is

$$\begin{aligned}&x=(A+B^\top B)^{-1}(f+B^\top g)-(A+B^\top B)^{-1}(B^\top \nonumber \\&\qquad \quad -\, B^\top D)H^{-1}B(A+B^\top B)^{-1}(f+B^\top g) \nonumber \\&\qquad \quad +\, \ (A+B^\top B)^{-1}(B^\top - B^\top D)H^{-1}g, \end{aligned}$$
(27)
$$\begin{aligned}&y=H^{-1}B(A+B^\top B)^{-1}(f+B^\top g)- H^{-1}g, \end{aligned}$$
(28)

where \(H=D+B(A+B^\top B)^{-1}(B^\top - B^\top D).\)

Proof

“\((c1) \Rightarrow (c2)\)” If \(A+B^\top B\) or \(D+BB^\top \) is singular, without loss of generality, we assume that \(A+B^\top B\) is singular, then there exists a nonzero vector \({\tilde{x}}\) such that \({\tilde{x}}\in {\mathcal {N}}(A+B^\top B).\) Notice that \({\mathcal {N}}(A+B^\top B)= {\mathcal {N}}(A)\cap {\mathcal {N}}(B),\) thus, \(A{\tilde{x}}=0, B{\tilde{x}}=0,\) that is,

$$\begin{aligned} \left[ \begin{array}{ll} A &{}\quad B^\top \\ B &{} \quad -\,D \\ \end{array} \right] \left[ \begin{array}{c} {\tilde{x}} \\ 0 \\ \end{array} \right] =\left[ \begin{array}{c} 0 \\ 0 \\ \end{array} \right] , \end{aligned}$$

which implies that \(\left[ \begin{array}{ll} A &{}\quad B^\top \\ B &{}\quad -\,D \\ \end{array} \right] \) is singular, this is in contradiction with the condition (c1). “\((c2) \Rightarrow (c1)\)” Let \([u^\top , v^\top ]^\top \) be a solution of the equation

$$\begin{aligned} \left[ \begin{array}{ll} A &{}\quad B^\top \\ B &{}\quad -\,D \\ \end{array} \right] \left[ \begin{array}{c} x \\ y \\ \end{array} \right] =\left[ \begin{array}{c} 0 \\ 0 \\ \end{array} \right] . \end{aligned}$$

That is,

$$\begin{aligned} Au+B^\top v=0, \quad Bu-Dv=0. \end{aligned}$$

From the proof of Theorem 2, we can get

$$\begin{aligned} Au=0, \quad B^\top v=0,\quad Bu=0, \quad Dv=0. \end{aligned}$$

Hence,

$$\begin{aligned} (A+B^\top B)u=0, \quad (D+BB^\top )v=0. \end{aligned}$$
(29)

It follows from (29) and condition (c2) that \(u=0,v=0,\) which implies that \(\left[ \begin{array}{ll} A &{}\quad B^\top \\ B &{}\quad -\,D \\ \end{array} \right] \) is nonsingular. The equivalence of (c2) and (c3) is clear.

The equation of (1) can be equivalently written as

$$\begin{aligned} \left[ \begin{array}{ll} A+B^\top B &{}\quad B^\top - B^\top D \\ B &{}\quad -\,D \\ \end{array} \right] \left[ \begin{array}{c} x \\ y \\ \end{array} \right] =\left[ \begin{array}{ll} I_n &{}\quad B^\top \\ 0 &{}\quad I_p \\ \end{array} \right] \left[ \begin{array}{c} f \\ g \\ \end{array} \right] =\left[ \begin{array}{c} f +B^\top g\\ g \\ \end{array} \right] . \end{aligned}$$
(30)

Applying Lemma 6 to (30), we can easily obtain the expressions of (27) and (28). \(\square \)