Existence and Uniqueness of the Positive Definite Solution for the Matrix Equation 𝑋=𝑄+𝐴^∗(𝑋−𝐶)⁻¹𝐴

Abstract

We consider the nonlinear matrix equation 𝑋=𝑄+𝐴^∗(𝑋−𝐶)⁻¹𝐴, where 𝑄 is positive definite, 𝐶 is positive semidefinite, and 𝑋 is the block diagonal matrix defined by 𝑋=diag(𝑋,𝑋,…,𝑋). We prove that the equation has a unique positive definite solution via variable replacement and fixed point theorem. The basic fixed point iteration for the equation is given.

1. Introduction

We consider the matrix equation 𝑋=𝑄+𝐴^∗𝑋−𝐶⁻¹𝐴, (1) where 𝑄 is an 𝑛×𝑛 positive definite matrix, 𝐶 is an 𝑚𝑛×𝑚𝑛 positive semidefinite matrix, 𝐴 is arbitrary 𝑚𝑛×𝑛 matrix, and 𝑋 is the block diagonal matrix defined by 𝑋=diag(𝑋,𝑋,…,𝑋) in which 𝑋 is an 𝑛×𝑛 matrix. This matrix equation is connected with certain interpolation problem (see [1]).

When 𝐶=0, we write 𝐴 as the 𝑚×1 block matrix 𝐴𝐴=¹𝐴₂⋮𝐴₄, (2) where 𝐴₁,…,𝐴_𝑚 are 𝑛×𝑛 matrices. Then (1) is equal to 𝑋=𝑄+∑^𝑚_𝑖=1𝐴^∗_𝑖𝑋⁻¹𝐴_𝑖, which has been investigated by many authors (see [2–9]). However, are few little theoretical results when 𝐶≠0. Ran and Reurings [1] have proved that (1) has a unique positive definite solution. Based on this, Sun [10] presented some perturbation results for the unique solution.

Fixed point theorem is often used to discuss the existence and uniqueness of the solution. However, for (1), it is difficult to prove the uniqueness of the solution directly by using fixed point theorems. In this paper, we turn (1) into its equivalent equation via variable replacement. Then we consider its equivalent equation by using fixed point theorem. This provides a new proof for the existence and uniqueness of the positive definite solution. And this method is shown to be much easier than the way of [1]. In addition, the basic fixed point iteration for the equation is given.

In this paper we use 𝒞(𝑛) to denote 𝑛×𝑛 complex matrices, 𝒫(𝑛) to denote 𝑛×𝑛 positive definite matrices, and 𝒫(𝑛) to denote 𝑛×𝑛 positive semidefinite matrices. For 𝑋,𝑌∈𝒫(𝑛), we write 𝑋≥𝑌(𝑋>𝑌) if 𝑋−𝑌 is positive semidefinite (definite). 𝐴^∗ denotes the conjugate transpose of a matrix 𝐴. Let 𝜆=𝜆_max(𝐴^∗𝐴), 𝜆=𝜆_min(𝐴^∗𝐴), 𝐶₁=𝜆_max(𝐶₁), and 𝐶₁=𝜆_min(𝐶₁).

2. The Case of 𝑄>𝐶

In this section, we discuss (1) with 𝑄>𝐶. Let 𝑆₁(𝑛)⊂𝑃(𝑛) be the set defined by 𝑆₁(𝑛)=𝑋∈𝑃(𝑛)∣𝑋>𝐶. (3) We are interested in positive definite solutions of (1) in this set. A matrix 𝑋 is a solution of (1) if and only if it is a fixed point of the map 𝐺₁ defined by 𝐺₁(𝑋)=𝑄+𝐴^∗𝑋−𝐶⁻¹𝐴. (4) Let 𝐶₁=𝑄−𝐶. For 𝑄>𝐶, we know that 𝐶₁ is a positive definite matrix. Then (1) turns into 𝑋−𝑄=𝐴^∗𝑋−𝑄+𝐶₁⁻¹𝐴. (5) Let 𝑌=𝑋−𝑄. Then (1) eventually becomes 𝑌=𝐴^∗𝑌+𝐶₁⁻¹𝐴. (6)

In the following, we consider the existence and uniqueness of the positive definite solution of (6). And then we can easily get corresponding conclusions about (1).

Theorem 1. Equation (6) has a positive semidefinite solution for any 𝐴∈𝒞(𝑛).

Proof. A matrix 𝑌 is a solution of (6) if and only if it is a fixed point of the map 𝐹 defined by 𝐹(𝑌)=𝐴^∗(𝑌+𝐶₁)⁻¹𝐴. Note that 𝐹 maps [0,𝐹(0)] into itself, because 𝐹 is order reversing. Hence it has a fixed point in the [0,𝐹(0)]. That is to says (6) has a positive semidefinite solution.

Theorem 2. If 𝑌 is positive semidefinite solution of (6), then 𝑌∈[0,𝐹(0)].

Proof. By 𝑌≥0, and 𝐶₁>0, we know that (𝑌+𝐶₁)⁻¹>0 and 𝐴^∗(𝑌+𝐶₁)⁻¹𝐴≥0; then 𝑌=𝐴^∗(𝑌+𝐶₁)⁻¹𝐴≥0. By 𝑌≥0, we know that 𝑌=𝐴^∗(𝑌+𝐶₁)⁻¹𝐴≤𝐴^∗𝐶₁⁻¹𝐴=𝐹(0). That is, 𝑌∈[0,𝐹(0)].

For proving the uniqueness, we first verify the following lemma.

Lemma 3. Let 𝜂(𝑡)=(((1−𝑡)𝐶₁)/(𝐶₁+𝐶₁⁻¹𝜆))>0. For any 𝑌∈[0,𝐹(0)] and 0<𝑡<1, one has 𝐹²(𝑡𝑌)≥𝑡1+𝜂(𝑡)𝐹²(𝑌). (7)

Proof. By Theorem 2, we know 𝐹(𝑌)∈[0,𝐹(0)] for any 𝑌∈[0,𝐹(0)]. Then 𝐹²(𝑌)∈[0,𝐹(0)] for any 𝑌∈[0,𝐹(0)]. Let 𝐹(𝑡𝑌)=𝐴^∗(𝑡𝑌+𝐶₁)⁻¹𝐴, and 𝐹₁(𝑡𝑌)=𝐴^∗(𝑌+𝑡⁻¹𝐶₁)⁻¹𝐴. For any 𝑌∈[0,𝐹(0)] and 0<𝑡<1, we have 𝐹²(𝑡𝑌)−𝑡1+𝜂(𝑡)𝐹²(𝑌)=𝐴^∗𝐹(𝑡𝑌)+𝐶₁⁻¹𝐴−𝑡𝐴^∗𝐹(𝑌)+𝐶₁⁻¹𝐴−𝑡𝜂(𝑡)𝐹²(𝑌)=𝑡𝐴^∗𝐹₁(𝑡𝑌)+𝑡𝐶₁⁻¹𝐴−𝑡𝐴^∗𝐹(𝑌)+𝐶₁⁻¹𝐴−𝑡𝜂(𝑡)𝐹²(𝑌)≥𝑡𝐴^∗𝐹(𝑌)+𝑡𝐶₁⁻¹𝐴−𝑡𝐴^∗𝐹(𝑌)+𝐶₁⁻¹𝐴−𝑡𝜂(𝑡)𝐹²(𝑌)=𝑡𝐴^∗𝐹(𝑌)+𝑡𝐶₁⁻¹−𝐹(𝑌)+𝐶₁⁻¹𝐴−𝑡𝜂(𝑡)𝐹²(𝑌)=𝑡𝐴^∗𝐹(𝑌)+𝑡C₁⁻¹(1−𝑡)𝐶₁𝐹(𝑌)+𝐶₁⁻¹𝐴−𝑡𝜂(𝑡)𝐹²(𝑌)≥𝑡𝐴^∗𝐹(𝑌)+𝐶₁⁻¹(1−𝑡)𝐶₁𝐹(𝑌)+𝐶₁⁻¹𝐴−𝑡𝜂(𝑡)𝐹²(𝑌)()𝐶≥𝑡1−𝑡₁𝐶₁+𝐶₁⁻¹𝜆𝐴^∗𝐹(𝑌)+𝐶₁⁻¹𝐴−𝑡𝜂(𝑡)𝐹²(𝑌)=𝑡𝜂(𝑡)𝐹²(𝑌)−𝑡𝜂(𝑡)𝐹²(𝑌)=0. (8)

Theorem 4. Equation (6) has a unique positive definite solution 𝑌∈[0,𝐹(0)], and for any 𝑌₀∈[0,𝐹(0)], the iteration 𝑌_𝑛+1=𝐴^∗𝑌_𝑛+𝐶₁⁻¹𝐴,𝑛=0,1,2,…, (9) converges to 𝑌; that is, lim_𝑛→∞𝑌_𝑛=𝑌.

Proof. Consider the matrix sequence (9). Let 𝐹(𝑌_𝑛)=𝐴^∗(𝑌_𝑛+𝐶₁)⁻¹𝐴. We first suppose that 𝑌₀=0. Then we get 0≤𝑌₁=𝐴^∗𝑌₀+𝐶₁⁻¹𝐴=𝐹(0),0≤𝑌₂=𝐴^∗𝑌₁+𝐶₁⁻¹𝐴=𝐹²(0)(0),𝐹≤𝐹₂(0)≤𝑌₃=𝐴^∗𝑌₂+𝐶₁⁻¹𝐴=𝐹³(0)(0),𝐹≤𝐹₂(0)≤𝑌₄=𝐴^∗𝑌₃+𝐶₁⁻¹𝐴=𝐹⁴(0)≤𝐹³(0). (10) Analogously one can prove that 0≤𝐹^2𝑘(0)≤𝐹^2𝑘+2(0)≤𝐹^2𝑘+1(0)≤𝐹^2𝑘−1(0)≤𝐹(0). (11) Hence, the sequence {𝐹^2𝑘(0),𝑘≥0} is monotone increasing and is bounded from above by 𝐹(0). Thus, the sequence {𝐹^2𝑘(0),𝑘≥0} has a finite positive definite limit. Moreover, the sequence {𝐹^2𝑘+1(0),𝑘≥0} is monotone decreasing and is bounded from below by 0. Thus, the sequence {𝐹^2𝑘+1(0),𝑘≥0} has a finite positive definite limit. Let lim_𝑘→∞𝐹^2𝑘(0)=𝑌⁽¹⁾,lim_𝑘→∞𝐹^2𝑘+1(0)=𝑌⁽²⁾; (12) then 0≤𝑌^(𝑖)≤𝐹(0) for 𝑖=1,2. Clearly, both 𝑌⁽¹⁾ and 𝑌⁽²⁾ are the positive fixed points of 𝐹²(𝑌). Then 𝑌⁽¹⁾=𝐹²𝑌⁽¹⁾=𝐴^∗𝐹(𝑌⁽¹⁾)+𝐶₁⁻¹𝐴≥𝐴^∗𝜆𝐶₁⁻²𝐶₁+𝐶₁⁻¹𝐴≥𝐴^∗𝜆𝐶₁⁻²+1⁻¹𝐶₁⁻¹𝐴=1𝜆𝐶₁⁻²𝐴+1^∗𝐶₁⁻¹𝐴≥1𝜆𝐶₁⁻²𝐹+1²𝑌⁽²⁾=1𝜆𝐶₁⁻²𝑌+1⁽²⁾. (13) Let 𝑡₀=sup{𝑡∣𝑌⁽¹⁾≥𝑡𝑌⁽²⁾}; then 0<𝑡₀<+∞. Now we prove that 𝑡₀≥1. Assume, on the contrary, that 0<𝑡₀<1; then 𝑌⁽¹⁾≥𝑡₀𝑌⁽²⁾. By Lemma 3 and monotonicity of 𝐹²(𝑌), we have 𝑌⁽¹⁾=𝐹²𝑌⁽¹⁾≥𝐹²𝑡₀𝑌⁽²⁾≥1+𝜂𝑡₀𝑡₀𝐹²𝑌⁽²⁾=1+𝜂𝑡₀𝑡₀𝑌₂. (14) Since (1+𝜂(𝑡₀))𝑡₀>𝑡₀, then it is a contradiction to the definition of 𝑡₀. Hence we have 𝑡₀≥1; therefore 𝑌⁽¹⁾≥𝑌⁽²⁾. Similarly, we can get 𝑌⁽¹⁾≤𝑌⁽²⁾. Therefore, 𝑌⁽¹⁾=𝑌⁽²⁾. Let Ω=[0,𝐹(0)]. By Theorem 1, we know 𝐹(Ω)⊆Ω. Hence 𝐹²(Ω)⊆Ω. That is, 𝐹² has only one positive fixed point in Ω. In other words, the equation 𝑌=𝐹²(𝑌) has only one positive definite solution. Hence 𝑌⁽¹⁾=𝑌⁽²⁾. It follows that lim_𝑛→∞𝐹^𝑛(0) is a fixed point of 𝐹². Since the positive definite solution of 𝐹(𝑌)=𝑌 solves 𝑌=𝐹²(𝑌), then 𝐹²(𝑌)=𝑌 has only one positive definite solution. Hence lim_𝑛→∞𝐹^(𝑛)(0) is the unique fixed point of 𝐹.
For any 𝑌₀>0, we get 𝑌₁≥0,0≤𝑌₂=𝐴^∗𝑌₁+𝐶₁⁻¹𝐴≤𝐴^∗𝐶₁⁻¹𝐴=𝐹(0). (15) By induction, we have for any 𝑘=1,2,…𝐹^2𝑘(0)≤𝑌_2𝑘+1≤𝐹^2𝑘−1(0),𝐹^2𝑘−2(0)≤𝑌_2𝑘≤𝐹^2𝑘−1(0). (16) Now letting 𝑘→∞ on both sides of the above inequalities, we can get lim_𝑘→∞𝑌_𝑘=lim_𝑛→∞𝐹^(𝑛)(0)=𝑌. (17)
Since (1) is equal to (6) when 𝑋=𝑌+𝑄, then we know that 𝑋 is a positive definite solution of (1) if and only if 𝑌 is a positive semidefinite solution of (6). And furthermore, 𝑌∈[0,𝐹(0)] if and only if 𝑋∈[𝑄,𝐺₁(𝑄)]. Thus, we can get the following conclusions about (1).

Theorem 5. Equation (1) with 𝑄>𝐶 has a positive definite solution for any 𝐴∈𝒞(𝑛).

Theorem 6. If 𝑋 is positive definite solution of (1) with 𝑄>𝐶, then 𝑋∈[𝑄,𝐺₁(𝑄)]⊂𝑆₁(𝑛).

Theorem 7. Let 𝑄∈𝒫(𝑛) and 𝐶∈𝒫(𝑚𝑛) such that 𝑄>𝐶. Then (1) has a unique solution 𝑋 in 𝑆₁(𝑛) and for any 𝑋₀∈[𝑄,𝐺₁(𝑄)], the iteration 𝑋_𝑛+1=𝑄+𝐴^∗𝑋_𝑛−𝐶⁻¹𝐴,𝑛=0,1,2,…, (18) converges to 𝑋; that is, lim_𝑛→∞𝑋_𝑛=𝑋.

3. The Case of 𝑄<𝐶

In this section, we discuss (1) with 𝑄<𝐶 and 𝐴^∗(𝐶−𝑄)⁻¹𝐴<𝑄. In fact, 𝐶∈𝒫(𝑛) must be a positive definite matrix in this case, because 𝑄<𝐶 and because 𝑄 is a positive definite matrix. We consider the matrix equation 𝑋=𝑄−𝐴^∗𝐶−𝑋⁻¹𝐴, (19) which is equal to (1). Let 𝑆₂(𝑛)⊂𝑃(𝑛) be the set defined by 𝑆₂(𝑛)=𝑋∈𝑃(𝑛)∣𝑋<𝐶. (20) We are interested in positive definite solutions of (1) in this set. A matrix 𝑋 is a solution of (1) if and only if it is a fixed point of the map 𝐺₂ defined by 𝐺₂(𝑋)=𝑄−𝐴^∗𝐶−𝑋⁻¹𝐴. (21) Let 𝐶₂=𝐶−𝑄; For 𝑄<𝐶, we have 𝐶₂ being a positive definite matrix. Then (1) turns into 𝑄−𝑋=𝐴^∗𝑄−𝑋+𝐶₂⁻¹𝐴. (22) Let 𝑌=𝑄−𝑋. Then (1) eventually becomes 𝑌=𝐴^∗𝑌+𝐶₂⁻¹𝐴. (23) We note that (23) should have the same results as (6). Then we directly give the following conclusions without proof.

Theorem 8. Equation (1) with 𝑄<𝐶 and 𝐴^∗(𝐶−𝑄)⁻¹𝐴<𝑄 has a positive definite solution for any 𝐴∈𝒞(𝑛).

Theorem 9. If 𝑋 is a positive definite solution of (1) with 𝑄<𝐶 and 𝐴^∗(𝐶−𝑄)⁻¹𝐴<𝑄, then 𝑋∈[𝐺₂(𝑄),𝑄]⊂𝑆₂(𝑛).

Theorem 10. Let 𝑄∈𝒫(𝑛) and 𝐶∈𝒫(𝑚𝑛) such that 𝑄<𝐶. Then (1) has a unique solution 𝑋 in 𝑆₂(𝑛), and for any 𝑋₀∈[𝐺₂(𝑄),𝑄], the iteration 𝑋_𝑛+1=𝑄−𝐴^∗𝐶−𝑋_𝑛⁻¹𝐴,𝑛=0,1,2,…, (24) converges to 𝑋; that is, lim_𝑛→∞𝑋_𝑛=𝑋.

4. Numerical Examples

We now use numerical examples to illustrate our results. All computations were performed using MATLAB, version7.01. We denote 𝜀(𝑋)=‖𝑋−𝐴^∗(𝑋−𝐶)⁻¹𝐴−𝑄‖_∞ and use the stopping criterion 𝜀(𝑋)<1.0×10⁻¹⁰.

Example 1. Consider (1) with 𝑛=2, 𝑚=3, and ,,𝐴=2−210−5362−3165,𝑄=20−1−113𝐶=61−205219−1−13−2−2−171−120−1160−253−10902−22−204 (25) where the matrices 𝑄 and 𝐶 satisfy 𝑄>𝐶. Considering the iterative method (18) with 𝑋₀=𝑄, after 23 iterations we get the following result: 𝑋≈𝑋₂₃=27.286668734882351.124933045738341.1249330457383416.21667821374470 (26) and 𝜀(𝑋₂₃)=‖𝑋₂₃−𝐴^∗(𝑋₂₃−𝐶)⁻¹𝐴−𝑄‖_∞=8.622080827080936×10⁻¹¹. It is not difficult to verify that 𝑋∈[𝑄,𝐺₁(𝑄)]⊂𝑆₁(𝑛).

Example 2. Consider (1) with 𝑛=2, 𝑚=3, and ,,𝐴=0−110−101221−11,𝑄=4−1−15𝐶=151−1201119−2−13−1−1−2102−222−12190−103−202301−12−1030 (27) where the matrices 𝑄 and 𝐶 satisfy 𝑄<𝐶 and 𝐴^∗(𝐶−𝑄)⁻¹𝐴<𝑄. Considering the iterative method (24) with 𝑋₀=𝑄, after 7 iterations one gets an approximation to the positive definite solution 𝑋. It is 𝑋≈𝑋₇=3.48707713254602−1.33049056598352−1.330490565983524.39117111733130 (28) and 𝜀(𝑋₇)=‖𝑋₂₃−𝐴^∗(𝑋₇−𝐶)⁻¹𝐴−𝑄‖_∞=6.447897771266753×10⁻¹¹. It is not difficult to verify that 𝑋∈[𝐺₂(𝑄),𝑄]⊂𝑆₂(𝑛).

Acknowledgments

The work was supported by the National Natural Science Foundation of China (11071141), the Natural Science Foundation of Shandong Province of China (ZR2011AL018), the Specialized Research Foundation for the Doctoral Program of Higher Education of China (20123705110001), the Program for Scientific Research Innovation Team in Colleges and Universities of Shandong Province, and the Project of Shandong Province Higher Educational Science and Technology Program (J11LA06 and J13LI02).