# Latest Inversion-Free Iterative Scheme for Solving a Pair of Nonlinear Matrix Equations

Authors: Sourav Shil; Hemant Kumar Nashine
Journal: Journal of Mathematics
 (2021)

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## Abstract

In this work, the following system of nonlinear matrix equations is considered,X1+A∗X1−1A+B∗X2−1B=IandX2+C∗X2−1C+D∗X1−1D=I, where A,B,C,andD are arbitrary n×n matrices and I is the identity matrix of order n. Some conditions for the existence of a positive-definite solution as well as the convergence analysis of the newly developed algorithm for finding the maximal positive-definite solution and its convergence rate are discussed. Four examples are also provided herein to support our results.


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## Body

## 1. Introduction and Preliminaries

Consider the system of matrix equations of the form(1)Ω1X+A∗P1XA+B∗P2XB=Q,(2)Ω2X+C∗Q1XC+D∗Q2XD=Q,where Q is an n×n Hermitian positive definite matrix (HPD, for short), A,B,C,andD are complex matrices of order n×n, Ω1X, Ω2X, P1X, P2X, Q1X, and Q2X are mappings from the set of positive definite matrices to itself, and A∗ is the conjugate transpose of A.We can see that the above equations incorporate a few linear as well as nonlinear matrix equations (NMEs, in short). Over the last few decades, many researchers studied systems (1) and (2) with different types of Pi,Qi, and Ωi, i=1,2.For (1), Ω1X=Xs, P1X=X−p1, and P2X=X−p2, p1,p2∈0,1 and s∈I+ [1]For (1), Ω1X=X, P1X=X−1, P2X=X−1, and Q=I [2, 3]For (1), Ω1X=X, P1X=X−1, and P2X=0 [4]For (1), Ω1X=X, P1X=X−1, and P2X=X−1, s,t∈ℕ [5, 6]For (1), Ω1X=Xs, P1X=P−p1, and P2X=X−p2, s,p1,p2≥1 [7]For (1) and (2), Ω1X=Ω2X=X, Q1X=X−p1, P1X=X−p2, and Q2X=Q2X=O, Q=I and p1,p2∈I+ [8]For (1) and (2), Ω1X=Ω2X=X, Q1X=±X−p1, P1X=±X−p2, and Q2X=Q2X=O, Q=I and p1,p2∈I+ [9, 10]For (1) and (2) Ω1X=Ω2X=X, P1X=X−p1, Q1X=X−p2, and Q2X=Q2X=O, Q=I and p1,p2∈0,1 [11]For different types of applications of the Riccati equation, one can check [4, 12, 13].Throughout the article,ℝ denotes the set of real numbers, ℕ denotes the set of natural numbers, ℝ+=0,+∞, I+ denotes the set of positive numbers, and ℕ∗=ℕ∪0.Hn (resp. Kn and Pn) denotes the set of all n×n Hermitian (resp. positive semidefinite and positive definite) matrices over ℂ and Mn stands for the set of all n×n matrices over ℂ.For a matrixD∈Hn, s1D≥s2D≥⋯≥snD denotes its singular values and D denotes the sum of these values, i.e., the trace norm of D. The Frobenius norm will be denoted by DF=∑i,j=1nDij21/2. For P,Q∈Hn, P≥Q (resp. P>Q) indicates that P−Q is positive semidefinite (resp. positive definite). O and I stand for the zero and unit matrix in Hn, respectively.


## 2. Conditions in Support of the Existence of a Positive Definite Solution

We begin with two useful lemmas.Lemma 1 (see [14]).
IfP and Q>0 are Hermitian matrices having the same order, then 2P−PQP≤Q−1.Lemma 2 (see [15]).
IfU≥V>O, then Up≥Vp>O, for all p∈0,1.Now, let us consider the system of NMEs of the form(3)X1+A∗X1−1A+B∗X2−1B=I,X2+C∗X2−1C+D∗X1−1D=I.TakingY=X1−1 and Z=X2−1, the system of nonlinear matrix equation (3) is equivalent to(4)Y−1+A∗YA+B∗ZB=I,Z−1+C∗ZC+D∗YD=I.Theorem 1.
LetY,Z be a positive-definite solution (PDS, for short) of (4). Then, I≤Y and I≤Z.Proof.
Suppose that system (4) has a PDS Y,Z>O,O. Then, from Lemma 2,(5)A∗YA+B∗ZB≥O,C∗ZC+D∗YD≥O,which gives(6)Y−1=I−A∗YA−B∗ZB≤I,Z−1=I−C∗ZC−D∗YD≤I.
This, in turn, implies thatI≤Y and I≤Z.Lemma 3 (see Theorem 2.1 in [16]).
LetU and V be positive operators acting on a Hilbert space H in such a way that P1I≥U≥p1I>O, P2I≥V≥p2I>O, and V≥U>O. Then,(7)Us≤P1p1s−1Vs,Us≤P2p2s−1Vs,hold for any s≥1.Remark 1.
In the case thatV>U>O in Lemma 3, we have Us<P1/p1s−1Vt and Us<P2/p2s−1Vs.Theorem 2.
Suppose that system (4) has a PDS Y,Z. Then,(i)
A∗A+B∗B<I and C∗C+D∗D<I(ii)
Y−1>AA∗, Y−1>DD∗, Z−1>CC∗, and Z−1>BB∗(iii)
Y<1/snA2I if A is nonsingular, Y<1/snD2I if D is nonsingular, Z<1/snC2I if C is nonsingular, and Z<1/snB2I if B is nonsingularProof.
To prove (i), sinceY,Z is a PDS of the NMEs (4), we have Y≥I and Z≥I. Together with (4), this gives(8)A∗A+B∗B≤A∗YA+B∗ZB=I−Y−1<I,and(9)C∗C+D∗D≤C∗ZC+D∗YD=I−Z−1<I.
To prove (ii), from (6), we have(10)A∗YA<I,A∗Y1/2Y1/2A<I,Y1/2AA∗Y1/2<I,AA∗<Y−1,B∗ZB<I,BB∗<Z−1.
Similarly, by (9), we can obtain CC∗<Z−1 and DD∗<Y−1.
To prove (iii), ifA is nonsingular, then, by AA∗<Y−1, we obtain(11)AA∗−1>Y.
We have(12)1s12AI≤AA∗−1≤1sn2AI.
Sincesn2AI≤AA∗≤s12AI, where s1A and snA are the maximal and the minimal singular value of the matrix A, respectively, by (11) and (12) and Remark 1, we obtain Y<AA∗−1<1/snA2I.
For the case of nonsingular matricesD,C, and B, we can obtain similarly.Theorem 3.
System (4) attains a PDS Y,Z iff A,B,C,D admit the subsequent factorizations:(13)A=W∗W−1/2K1,B=U∗U−1/2K2,C=U∗U−1/2L1,D=W∗W−1/2L2,where the matrices WandU are nonsingular and the columns W−∗K1K2 and U−∗L1L2 are orthonormal. In this case, Y,Z=W∗W,U∗U is a solution of (4).Proof.
If the system of NMEs (4) has a PDS Y,Z, then Y,Z=W∗W,U∗U for some nonsingular matrices W,U. Rewrite the equation (4) as(14)W∗W−1+A∗W∗WA+B∗U∗UB=I,U∗U−1+C∗U∗UC+D∗W∗WD=I,or(15)W−∗∗W−∗+W∗W1/2A∗W∗W1/2A+U∗U1/2B∗U∗U1/2B=I,U−∗∗U−∗+U∗U1/2C∗U∗U1/2C+W∗W1/2D∗W∗W1/2D=I,or equivalently(16)W−∗W∗W1/2AU∗U1/2B∗W−∗W∗W1/2AU∗U1/2B=I,U−∗U∗U1/2CW∗W1/2D∗U−∗U∗U1/2CW∗W1/2D=I.
Let(17)W∗W1/2A=K1,U∗U1/2B=K2,U∗U1/2C=L1,W∗W1/2D=L2.
Then,(18)A=W∗W−1/2K1,B=U∗U−1/2K2,C=U∗U−1/2L1,D=W∗W−1/2L2.
Now, (16) implies that W−∗K1K2 and U−∗L1L2 are orthonormal. Conversely, assume that A,B,C,andD have decomposition (13). Set Y,Z=W∗W,U∗U. Then,(19)Y−1+A∗YA+B∗ZB=W∗W−1+W∗W−1/2K1∗W∗WW∗W−1/2K1+U∗U−1/2K2∗U∗UU∗U−1/2K2=W−1W−∗+K1∗K1+K2∗K2=I,Z−1+C∗ZC+D∗YD=W∗W−1+U∗U−1/2L1∗U∗UU∗U−1/2L1+W∗W−1/2L2∗W∗WW∗W−1/2L2=U−1U−∗+L1∗L1+L2∗L2=I,which implies Y,Z=W∗W,U∗U is a PDS of (4).Example 1.
We can findW,U,K1,K2,L1,andL2 (see Theorem 7.2.7, page 440, in [17]) for the following numerical experiment (2) which satisfies all the conditions of Theorem 3:(20)W=1.03550.03410.03070.03411.04480.03590.03070.03591.0335,U=1.0314−0.00040.0020−0.00041.02760.00030.00200.00031.0489,K1=0.04460.10120.07500.06670.14940.08090.16140.14600.1117,K2=0.10870.01820.08880.07700.13760.15560.10890.06860.0387,L1=0.24420.00110.0017−0.00590.2283−0.02070.01220.02240.2800,L2=−0.01040.0009−0.0024−0.0014−0.0171−0.01430.00430.0110−0.1093.
Also, it is easy to derive(21)s1A=0.2981,snA=0.0146,s1B=0.2691,snB=0.048,s1C=0.2692,snC=0.2227,s1D=0.1065,snD=0.0101.
It meets all of the requirements of Theorem2.


## 3. Construction of Iteration Schemes

This section contains a new iteration scheme for the NMEs (4) in the context of [2, 11].By pre- and postmultiplying the first and the second equation of (4), respectively, by Y and Z, we obtain(22)Y−YI−A∗YA−B∗Z2BY=O,Z−ZI−C∗ZC−D∗YDZ=O.After some simple calculation, we have(23)Y=2Y−YI−A∗YA−B∗ZBY,Z=2Z−ZI−C∗ZC−D∗YDZ.Apparently,Y,Z is a solution of (23) if Y,Z solves equation (4). Conversely, if Y,Z is a nonsingular solution of equation (23), Y,Z solves equation (4), too.Thus, to attain a HPDS of (4), we need to solve (23). From (23), the iterative scheme is as follows.(24)Y^n=I−A∗YnA−B∗ZnBYn+1=2I−YnY^nYn,Z^n=I−C∗ZnC−D∗Yn+1DZn+1=2I−ZnZ^nZn.


## 4. Convergence Analysis

This section contains the proof that the sequencesYn,Zn generated by Algorithm 1, where Y0=I and Z0=I are initial conditions and converge to the minimal positive-definite solution of equation (4).Algorithm 1: An inversion-free iterative algorithm for equation (4).
Step 1: inputA,B,C,D∈ℂn×n.TakeY0=I and Z0=I for initialization and tolerance error ϵ≥0. Set n:=0.Step 2: find outYn+1 and Zn+1 by the following iterative scheme:Step 3: stop ifYn+1−YnF+Zn+1−ZnF≤ϵ. Otherwise, n:=n+1, and go to Step 2.Theorem 4.
Suppose that the system of NMEs (4) has a PDS and the sequence Yn,Zn is generated by Algorithm 1 with initial values Y0=I and Z0=I. Let the pair Y^,Z^ be the minimal PDS of (4). Then, the sequence Yn,Zn is well defined:(25)Y0≤Y1≤Y2≤⋯≤Yn≤⋯≤Y^,Z0≤Z1≤Z2≤⋯<Zn≤⋯≤Z^,limn⟶∞Yn=Y^,limn⟶∞Zn=Z^.Proof.
LetY,Z be any PDS of the system of NMEs (4). We will prove that O<Yn≤Yn+1≤Y and O<Zn≤Zn+1≤Z for all n∈ℕ∗=ℕ∪0 using mathematical induction.
Forn=0, we have Y0=I>0 and Z0=I>O. Then, by (24), we have(26)Y1=2I−I+A∗A+B∗B=I+A∗A+B∗B≥I=Y0,Z1=2I−I+C∗C+D∗Y1D=I+C∗C+D∗Y1D≥I=Z0.
Following Theorem4, I−A∗A−B∗B>0. Using this condition in Lemma 1 together with NMEs (4),(27)Y1=2I−II−A∗A−B∗BI≤I−A∗A−B∗B−1≤I−A∗YA−B∗ZB−1=Y,since Y,Z≥I,I. That is, Y1≤Y. Using Lemma 2, this implies that I−C∗C−D∗Y1D≥I−C∗ZC−D∗YD=Z−1>0. Using this fact in Lemma 1 together with NMEs (4), we obtain(28)Z1=2I−II−C∗C−D∗Y1DI≤I−C∗C−D∗Y1D−1≤I−C∗ZC−D∗YD−1=Z.
Hence,O<Yn≤Yn+1≤Y and O<Zn≤Zn+1≤Z holds true for n=0.
Assume thatO<Yn≤Yn+1≤Y and O<Zn≤Zn+1≤Z hold for n=k≥0. Since I−A∗Yk+1A−B∗Zk+1B≥I−A∗YA−B∗ZB=Y−1>O, by equation (24), Lemma 1, and the fact Yk+1>O, we obtain(29)Yk+2=2Yk+1−Yk+1I−A∗Yk+1A−B∗Zk+1BYk+1≤I−A∗Yk+1A−B∗Zk+1B−1≤I−A∗YA+B∗ZB−1=Y,that is, Yk+2≤Y. Then, by this condition and Lemma 2, we have I−C∗Zk+1C+D∗Yk+2D≥I−C∗ZC−D∗YD=Z−1>0. Using this fact in Lemma 1 together with the NMEs (22) and the fact Zi+1>O, we have(30)Zk+2=2Zk+1−Zk+1I−C∗Zk+1C−D∗Yk+2DZk+1≤I−C∗Zk+1C−D∗Yk+2D−1≤Z.
Also, from the NMEs (24), we get that(31)Yk+2−Yk+1=Yk+1−Yk+1I−A∗Yk+1A−B∗Zk+1BYk+1=Yk+1Yk+1−1−I−A∗Yk+1A−B∗Zk+1BYk+1.
Next, by using the factI−A∗YkA−B∗ZkB≥I−A∗Yk+1A−B∗Zk+1B>O together with the NMEs (24), Lemma 1, and the fact Yk>0, it follows that(32)Yk+1=2Yk−YkI−A∗YkA−B∗ZkBYk≤I−A∗YkA−B∗ZkB−1≤I−A∗Yk+1A−B∗Zk+1B−1,that is, Yk+1≤I−A∗Yk+1A−B∗Zk+1B−1 and Yk+1−1≥I−A∗Yk+1A−B∗Zk+1B. Using this fact with Yk+1>O, in equation (31), we have Yk+2≥Yk+1. Using this fact with Lemma 2, we have(33)I−C∗ZkC−D∗Yk+1D≥I−C∗Zk+1C−D∗Yk+2D>O.
This fact, together with the NMEs (24), Lemma 1, and the fact Zk>O, yields(34)Zk+1=2Zk−ZkI−C∗ZkC−D∗Yk+1DZk≤I−C∗ZkC−D∗Yk+1D−1≤I−C∗Zk+1C−D∗Yk+2D−1,that is, Zk+1≤I−C∗Zk+1C−D∗Yk+2D−1 or Zk+1−1≥I−C∗Zk+1C−D∗Yk+2D. Thereby,(35)Zk+2−Zk+1=Zk+1−Zk+1I−C∗Zk+1C−D∗Yk+2DZk+1=Zk+1Zk+1−1−I−C∗Zk+1C−D∗Yk+2DZk+1≥O.
From the above,O<Yn≤Yn+1≤Y and O<Zn≤Zn+1≤Z hold for n=k+1.
Thus, by using the principle of induction, we have that the relationsO<Yn≤Yn+1≤Y and O<Zn≤Zn+1≤Z hold for n∈ℕ∗. So, the sequence Yn,Zn is well defined, increasing, and bounded above. Let(36)limn⟶∞Yn=Y^,limn⟶∞Zn=Z^.
Then,Y^,Z^ is a PDS of the NMEs (4) by Algorithm 1. Since Y^,Z^≤Y,Z for any PDS Y,Z of NMEs (4), Y^,Z^ is the minimal PSD of NMEs (4).Remark 2.
IfX^1,X^2 is the maximal PDS of the system of NMEs (3), then by the relationship between the NMEs (3) and (4), we get that X^1−1,X^2−1 is the minimal PDS of NMEs (4). Therefore, from Theorem 4, the sequence Yn,Zn generated by Algorithm 1 with the initialization Y0=I and Z0=I converges to the inverse of the maximal PDS of NMEs (3), respectively.


## 5. Rate of Convergence

Lemma 4 (see [18]).
If0<θ≤1 and U,V∈Pn with U,V≥cI>O, then, for every unitarily invariant norm, |Uθ−Vθ|≤θcθ−1|U−V| and |U−θ−V−θ|≤θc−θ+1|U−V|.Theorem 5.
IfY,Z is a PDS of the matrix equation (4) under Algorithm 1 and ϵ>0 is arbitrary, then we get the following.
(A)(37)Yn−Y≤AY+ε2Y−Yn+BY+ε2Z−Zn,and(38)Y^n−Y−1≤A2Yn−Y+B2Zn−Z.
(B)(39)Zn−Z≤CZ+ε2Zn−Z+DZ+ε2Yn+1−Y,and(40)Z^n−Z−1≤C2Zn−Z+D2Yn+1−Y,for all n large enough.Proof.
To prove (A), letY,Z be the PDS of equation (4). Then, from Algorithm 1,(41)Yn+1=2Yn−YnI−A∗YnA−B∗ZnBYn=2Yn−YnI−A∗Y+Yn−YA−B∗Z+Zn−ZBYn=2Yn−YnI−A∗YA−B∗ZB+A∗Y−YnA+B∗Z−ZnBYn=2Yn−YnY−1Yn−YnA∗Y−YnAYn−YnB∗Z−ZnBYn.
Thus, sinceY−2Yn+YnY−1Yn=Y−YnY−1Y−Yn, we have(42)Y−Yn+1=Y−YnY−1Y−Yn+YnA∗Y−YnAYn+YnB∗Z−ZnBYn.
Now, sincelimn⟶∞Yn=Y, using Lemma 4, we have(43)Y−Yn+1≤Y−YnY−1Y−Yn+AYn2Y−Yn+BYnZ−Zn≤Y−YnY−1+AYn2Y−Yn+BYn2Z−Zn≤AY+ε2Y−Yn+BY+ε2Z−Zn.which is (35).
Since(44)Y^n−Y−1=A∗Y−YnA+B∗Z−ZnB,using Lemma 4, we have relation (38).
In the same way, we can prove (B).


## 6. Numerical Examples

Two examples are presented in this section in support of Algorithm1. Take Residue =Yn+1−Yn+Zn+1−Zn, tolerance  = 1e − 10, and norm as Frobenius norm. For comparative analysis, a basic fixed point algorithm (BFP, for short) has been used:(45)X1k+1=I−A∗X1k−1A−B∗X2k−1B,X2k+1=I−C∗X2k−1C−D∗X1k−1D.These examples, together with tables and graphs showing different input data, solutions, iteration number of different schemes, error, CPU time, average computational time, etc., are illustrated here. For a better understanding, we have used line graphs, bar graphs, pie graphs, and surface plotting.Example 2.
Take(46)A=0.03670.08930.06700.05740.13550.07170.15300.13390.1036,B=0.10520.01750.08610.07490.13390.15140.10360.06540.0367,C=0.23670.00110.0011−0.00560.2221−0.02020.01120.02130.2670,D=−0.01010.00110.0011−0.0011−0.0168−0.01010.00450.0112−0.1055.
After applying Algorithm1 with the initial conditions Y0=I and Z0=I, we obtain X1=0.9379−0.0581−0.0526−0.05810.9221−0.0617−0.0526−0.06170.9418andX2=0.94010.0007−0.00350.00070.9470−0.0006−0.0035−0.00060.9090.
Table1 presents a comparison between our algorithm and the basic fixed point algorithm.
Figures1 and 2 represent CPU time graph and iteration vs. error graph.
Figures3 and 4 represent pie chart for average CPU time based on ten experiments and solution’s surface plot.Table 1
Comparison of Algorithm 1 with basic fixed point iteration.
MethodrNo. of iterationErrorAverage CPU timeAlgo13120.5908e − 100.0070BFP3120.2180e − 100.0157Figure 1
CPU time graph and iteration vs error graph.Figure 2
CPU time graph and iteration vs error graph.Figure 3
Pie chart for average CPU time.Figure 4
Solution surface plot.Example 3.
In the current example, a new, randomly chosen set of coefficientsA,B,C,D∈ℂn×n is used. The construction is followed by(47)Ati,j=ri+j−1,Bti,j=r1×Ati,j,Cti,j=r2i+j−1,Dti,j=1r−1×Ati,j,Q1i,j=Ii,j+At∗i,j×Ati,j,Q2i,j=Ii,j+Bt∗i,j×Bti,j,A1i,j=Q1i,j−1/2×Ati,j×Q1i,j−1/2,B1i,j=Q1i,j−1/2×Bti,j×Q1i,j−1/2,C1i,j=Q2i,j−1/2×Cti,j×Q2i,j−1/2,D1i,j=Q2i,j−1/2×Dti,j×Q2i,j−1/2,N1=normA1∗×A1+B1∗×B1,fro′,N2=normC1∗×C1+D1∗×D1,fro′,A=1r×A1N1+iB1N1,B=1r×B1N1+iA1N1,C=1r×C1N2+iD1N1,D=1r×C1N2+iD1N1.
Here, “r” and “fro” stand for dimension and Frobenius norm of the matrix, respectively. A different set of initial conditions has been chosen here as(48)Y0=12×A×conjugate‐transposeA,Z0=12×B×conjugate‐transposeB.
The result of this experiment using Algorithm1 is shown in Table 2.Table 2
CPU analysis for different dimension.
DimNo. of iterationErrorCPU time (seconds)383.0319e − 120.007770466.4760e − 120.006651552.2293e − 110.0083591042.0735e − 120.00744720037.9969e − 150.03039750021.1787e − 110.111352100027.3933e − 130.758877Example 4.
Here, we consider some randomly generated matrices:(49)A1=I+12×rrandr,B1=I+1r2randr,C1=I+1r3randr,D1=I+1r×3randr,A=A1A1,B=B1B1,C=C1C1,D=D1D1,where r = dimension of the matrices, rand(r) = random matrices of order r, and tol = 1e − 10. After applying Algorithm 1 (ALgo1) and basic fixed-point method under the initial conditions Y0=I and Z0=I, we get Table 3 that shows the various outputs for different r. The 5th column of Table 3 ensures positive definiteness of Example 4 w.r.t. different order matrices.
Associated graphs are (Figures5–25)(i)
dim = 3Fig5:CPU timeFig6:it. no. vs. errorFig7:Sols surface plot′(ii)
dim = 5Fig8:CPU timeFig9:it. no. vs. errorFig10:Sols surface plot′(iii)
dim = 8Fig11:CPU timeFig12:it. no. vs. errorFig13:Sols surface plot′(iv)
dim = 12Fig14:CPU timeFig15:it. no. vs. errorFig16:Sols surface plot′(v)
dim = 20Fig17:CPU timeFig18:it. no. vs. errorFig19:Sols surface plot′(vi)
dim = 32Fig20:CPU timeFig21:it. no. vs. errorFig22:Sols surface plot′(vii)
dim = 64Fig23:CPU timeFig24:it. no. vs. errorFig25:Sols surface plot′(viii)
Figure26 represents the average CPU time through bar graphs for different dimensionsTable 3
Comparison of Algorithm 1 with basic fixed point iteration for different dimension.
DimAlgo 1BFPMin(eig)X1,X23No. of iter.17190.7646, 0.7620Error (×10−10)0.68360.48240Av. CPU time0.00640.00985No. of iter.11130.8689, 0.8729Error (×10−10)0.16860.3111Av. CPU time0.00620.00968No. of iter.9100.9175, 0.9238Error (×10−10)0.81590.48613Av. CPU time0.00720.010112No. of iter.890.9450, 0.9497Error (×10−10)0.32850.0948Av. CPU time0.00730.010420No. of iter.770.9672, 0.9701Error (×10−10)0.23880.9457Av. CPU time0.00770.010932No. of iter.670.9796, 0.9814Error (×10−10)0.71790.0356Av. CPU time0.00780.011664No. of iter.660.9899, 0.9907Error (×10−10)0.13320.48082Av. CPU time0.01240.0168Figure 5
CPU time vs no. of iteration for dimension 3.Figure 6
Iteration no. vs error for dimension 3.Figure 7
Solution surface plot for dimension 3.Figure 8
CPU time vs no. of iteration for dimension 5.Figure 9
Iteration no. vs error for dimension 5.Figure 10
Solution surface plot for dimension 5.Figure 11
CPU time vs no. of iteration for dimension 8.Figure 12
Iteration no. vs error for dimension 8.Figure 13
Solution surface plot for dimension 8.Figure 14
CPU time vs no of iteration for dimension 12.Figure 15
Iteration no. vs error for dimension 12.Figure 16
Solution surface plot for dimension 12.Figure 17
CPU time vs no. of iteration for dimension 20.Figure 18
Iteration no. vs error for dimension 20.Figure 19
Solution surface plot for dimension 20.Figure 20
CPU time vs no. of iteration for dimension 32.Figure 21
Iteration no. vs error for dimension 32.Figure 22
Solution surface plot for dimension 32.Figure 23
CPU time vs no. of iteration for dimension 64.Figure 24
Iteration no. vs error for dimension 64.Figure 25
Solution surface plot for dimension 64.Figure 26
Average CPU time for different dimensions.Remark 3.
We can infer from the above discussions that our algorithm is less expensive in terms of computation after evaluating these examples w.r.t. different sets of parameters.Remark 4.
We can infer from the above discussions that the solutions are positive definite.


## 7. Conclusion

In this study, a new iterative algorithm has been developed. All numerical tests are in agreement with the theoretical findings of this research done. Finally, based on the numerical results, we have concluded that the new iterative approach is extremely powerful and efficient in finding numerical solutions for a wide range of nonlinear matrix equations including complex ones. It also produces very accurate results with less iterations and lower computational costs, compared with the basic fixed-point approach.



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