The Inversion-Free Iterative Methods for Solving the Nonlinear Matrix Equation X + A H X − 1 A + B H X − 1 B =

and Applied Analysis 3 Lemma 6 (see [23]). If C and P are Hermitian matrices of the same order with P > 0, then 2C − CPC ≤ P. Lemma 7. If M,P ∈ C are both Hermitian positive semidefinite, thenMP + PM is a Hermitian positive semidefinite matrix as well. Proof. Obviously,MP+PM is aHermitianmatrix. If one ofM and P is positive definite, without loss of generality, suppose thatM > 0. Then by the assumption of the lemma, we know thatM, P are well defined andM > 0, P ≥ 0. Then we get MP + PM = M 1/2 M 1/2 P + PM 1/2 M 1/2 = M 1/2 (M 1/2 PM −(1/2) +M −(1/2) PM 1/2 )M 1/2 ≥ 0. (12) Assume that neither ofM and P is positive definite. Then for β > 0, we have M + βI > 0. Thus by the above analysis, we have (M + βI)P + P(M + βI) ≥ 0. Let β → 0; we get MP + PM ≥ 0. This completes the proof. Now we are in a position to prove that {Y k } generated by Algorithm 1 with the initial matrix Y 0 = I converges to the minimal positive definite solution of (3). Theorem 8. The nonlinear matrix equation (3) has a positive definite solution and the sequence {Y k } is generated by Algorithm 1with the initialmatrixY 0 = I. Let ?̃? be theminimal positive definite solution of (3). Then the sequence {Y k } is well defined, Y 0 ≤ Y 1 ≤ Y 2 ≤ ⋅ ⋅ ⋅ < Y k ≤ ⋅ ⋅ ⋅ ≤ ?̃?, and lim k→∞ Y k = ?̃?. (13) Proof. Let Y ∗ be any positive definite solution of (3). Firstly, we will prove 0 < Y k ≤ Y k+1 ≤ Y ∗ for all k ≥ 0 by mathematical induction. For k = 0, we have Y 0 = I > 0. By (6) we have Y 1 = 2I − I + A H A + B H B = I + A H A + B H B ≥ I = Y 0 . (14) On the other hand, by Lemma 5, we have I−AA−BB > 0. This together with Lemma 6 and (3) yields Y 1 = 2I − I (I − A H A − B H B) I ≤ (I − A H A − B H B) −1 ≤ (I − A H Y ∗ A − B H Y ∗ B) −1

Motivated and inspired by the works mentioned above, in this paper, we propose two new inversion-free iterative methods for obtaining the maximal positive definite solution of (1).We prove that the sequences generated by the two iterative schemes are monotonically increasing and bounded above.In addition, we also provide some numerical results to illustrate the effectiveness of the proposed algorithm.
Throughout this paper, we use the following notations: for  ∈ C × , we write   ,  −1 and ‖‖ to denote the conjugate transpose, the inverse, and the Frobenius norm of the matrix , respectively.For any  = (  ),  = (  ), we write  ≥  (or  ≤ ) if  −  is a Hermitian positive semidefinite matrix.And we write  >  (or  < ) if  −  is a Hermitian positive definite matrix.We use 0 to denote the zero matrix of size implied by context and  to denote the identity matrix of size implied by context.This paper is organized as follows.In Section 2, we present two new iterative methods to solve the nonlinear matrix equation (1).In Section 3, the convergence analysis of the proposed methods is given.Some numerical experiments are reported in Section 4. Finally, we conclude this paper in Section 5.

New Inversion-Free Iterative Methods
In this section, we present two new inversion free iterative algorithms for solving problem (1).Let  =  −1 ; then the nonlinear matrix equation ( 1) is equivalent to Obviously, if  * is a Hermitian positive definite solution of (3), we have 0 <  −1 * ≤ .That is,  * ≥ .Premultiplying and postmultiplying by  on (3) simultaneously, we get Adding  to both sides of the above equation, we have Obviously,  solves (5) if  is a solution of (3).Conversely, if  is a nonsingular solution of (5),  solves (3) as well.Thus, we just need to solve the matrix equation ( 5) if we want to get a Hermitian positive definite solution of (3).By this point, we present the following iterative scheme.
On the other hand, premultiplying and postmultiplying by  on (3), respectively, we have Adding the above two equations, we can get By some simple calculating, we obtain By the above analysis, we know that if  is a solution of (3),  is a solution of ( 9).Now we prove that a Hermitian positive definite solution of ( 9) is also a solution of (3).
Theorem 3. Let  be a Hermitian positive definite solution of the nonlinear matrix equation (9).Then  is also a positive definite solution of the nonlinear matrix equation (3).
Proof.By the nonlinear matrix equation ( 9), we have This implies that  −1 +    +    −  is a solution of the matrix equation  +  = 0, where  ∈ C × is unknown.Since  is positive definite,  and − have no common eigenvalue and the matrix equation  +  = 0 has a unique solution (see [31]).As 0 solves the equation  +  = 0, we must have Namely,  is a positive definite solution of (3).
Thus, we just need to solve the matrix equation ( 9) if we want to get a Hermitian positive definite solution of (3).By this point, we present the following iterative scheme.
Obviously, for all  ≥ 0, sequence {  } generated by Algorithm 4 with the initial matrix  0 =  are all Hermitian matrices.

Convergence Analysis
In this section, we will prove that the sequences {  } generated by Algorithms 1 and 4 with the initial matrix  0 =  converge to the minimal positive definite solution of (3).In the first place, we introduce the following lemmas.
Proof.Obviously, + is a Hermitian matrix.If one of  and  is positive definite, without loss of generality, suppose that  > 0. Then by the assumption of the lemma, we know that  1/2 ,  1/2 are well defined and  1/2 > 0,  1/2 ≥ 0. Then we get Assume that neither of  and  is positive definite.Then for  > 0, we have  +  > 0. Thus by the above analysis, we have ( + ) + ( + ) ≥ 0. Let  → 0 + ; we get  +  ≥ 0. This completes the proof.Now we are in a position to prove that {  } generated by Algorithm 1 with the initial matrix  0 =  converges to the minimal positive definite solution of (3).
By the principle of mathematical induction, 0 <   ≤  +1 ≤  * is true for all  ≥ 0. The sequence {  } is now well defined, monotonically increasing, and bounded above.Let lim  → ∞   = Ỹ.Then Ỹ is a positive definite solution of the nonlinear matrix equation ( 3) by (6).Since Ỹ ≤  * for any positive definite solution  * of (3), Ỹ is the minimal positive definite solution of (3).This completes the proof.Remark 9. Let X be the maximal positive definite solution of the nonlinear matrix equation (1).By the relationship between (1) and (3), we get that X−1 is the minimal positive definite solution of (3).So from Theorem 8, we know that the sequence {  } generated by Algorithm 1 with the initial matrix  0 =  converges to the inverse of the maximal positive definite solution of (1).Now we consider the convergence theorem of Algorithm 4.

Theorem 10. The nonlinear matrix equation (3) has a positive definite solution and the sequence {𝑌 𝑘 } is generated by
Algorithm 4 with the initial matrix  0 = .Let Ỹ be the minimal positive definite solution of (3).Then the sequence Proof.Let  * be any positive definite solution of (3).Firstly, we will prove 0 <   ≤  +1 ≤  * for all  ≥ 0 by induction.

Numerical Experiments
In this section, we will give some numerical examples to support our Algorithms 1 and 4. All experiments were run on a PC with Pentium(R) Dual-Core CPU E5800 @2.40 GHz.
All the programming is implemented in MATLAB R2011b (7.13).We report the number of required iterations (Iter.), the norm of the residual (Res.) when the process is stopped, the required CPU time (CPU), and the number of matrix-matrix (MM) products required.In our implementation, we stop all considered algorithms when ‖ +1 −   ‖  ≤ 10 −10 with  ≥ 1.
We compare our Algorithm 1 (A1) and Algorithm 4 (A2) with the following inverse-free methods for solving (1).
(i) In [28], for finding positive definite solution of (1), Long et al. proposed the following iteration: (ii) In [30], for solving positive definite solution of (1), Liu and Chen proposed the following iteration: ) . ( We could obtain the maximal solution  + (the first 4 digits) by the iterative schemes A1-A2 and B1-B3.
Our numerical results are reported in Table 1. ) . ( We will obtain the maximal solution  + (the first 4 digits) by the iterative schemes A1-A2 and B1-B3.The maximal solution is Our numerical results are reported in Table 2.

) Example 11 .
For the first experiment, we consider (1) when  and  are given as in Example 4.1 from[28]:

Table 1 :
Numerical results of Example 11.

Table 2 :
Numerical results of Example 12.

Table 3 :
Numerical results of Example 13.

Table 4 :
Numerical results of Example 14. Example 12.In this test, the matrices  and  are given as in Example 4.2 from [28]: