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<title>A note on the unique solvability condition for generalized absolute value matrix equation: A note on the unique solvability condition for generalized absolute value matrix equation</title>
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<h1>A note on the unique solvability condition for generalized absolute value matrix equation</h1>
<p class="authors">
<span class="author">Shubham Kumar\(^\ast \) Deepmala\(^\ast \)</span>
</p>
<p class="date">received: June 15, 2022; accepted: July 25, 2022; published online: August 25, 2022.</p>
</div>
<div class="abstract"><p> We present the spectral radius condition \(\rho \big(\vert A^{-1} \vert \cdot \vert B \vert \big){\lt}1\) for the unique solvability of the <i class="it">generalized absolute value matrix equation</i> (GAVME) \(AX + B \vert X \vert = D\). For some instances, our condition is superior to the earlier published singular values conditions \(\sigma _{\max }(\vert B \vert )\) \({\lt}\) \(\sigma _{\min }(A)\) <span class="cite">
	[
	<a href="#M Dehghan 2020 GAVME" >3</a>
	]
</span> and \(\sigma _{\max }(B)\) \({\lt}\) \(\sigma _{\min }(A)\) <span class="cite">
	[
	<a href="#Kai Xie 2021 GAVME" >12</a>
	]
</span>. For the validity of our condition, we provided some examples. </p>
<p><b class="bf">MSC.</b> 15A06, 15A18, 90C30. </p>
<p><b class="bf">Keywords.</b> Generalized absolute value matrix equation, unique solution, singular value, spectral radius. </p>
</div>
<p>\(^\ast \)Mathematics Discipline, PDPM-Indian Institute of Information Technology, Design and Manufacturing, Jabalpur, M.P., India, e-mails: <span class="tt">shub.srma@gmail.com, dmrai23@gmail.com</span>. <br />The work of the first author was supported by the Ministry of Education (Government of India) through GATE fellowship registration No. MA19S43033021. </p>
<h1 id="a0000000002">1 Introduction</h1>
<p> In this note, we consider the following GAVME </p>
<div class="equation" id="Equ GAVME">
<p>
  <div class="equation_content">
    \begin{equation}  \label{Equ GAVME} AX + B\vert X \vert =D, \end{equation}
  </div>
  <span class="equation_label">1</span>
</p>
</div>
<p> where \(A,B,D \in \mathbb {R}^{n\times n}\) and \(X \in \mathbb {R}^{n\times n}\) is unknown. <br /></p>
<p>The GAVME is a generalization form of the following <i class="it">generalized absolute value equation</i> (GAVE) </p>
<div class="equation" id="Equ555">
<p>
  <div class="equation_content">
    \begin{equation}  \label{Equ GAVE} Ax + B\vert x \vert =d, \end{equation}
  </div>
  <span class="equation_label">2</span>
</p>
</div>
<p> where \(A,B \in \mathbb {R}^{n\times n}\), \(d \in \mathbb {R}^{n}\) are known and \(x\in \mathbb {R}^{n}\) is unknown. </p>
<p>The GAVE was first introduced by Rohn <span class="cite">
	[
	<a href="#Jiri Rohn 2004" >7</a>
	]
</span> and studied more detail in <span class="cite">
	[
	<a href="#M Achache 2019" >1</a>
	, 
	<a href="#N Anane 2020" >2</a>
	, 
	<a href="#Mangasarian 2006" >5</a>
	, 
	<a href="#Mangasarian 2007i" >6</a>
	]
</span>, where authors provided its unique solvability conditions and discussed its numerical solution. The importance of <i class="it">absolute value equations</i> (AVEs) is due to their broad applications in many mathematics and applied sciences domains. For instance, the linear complementarity problem, bimatrix games, mixed-integer programming, system of linear interval matrix and convex quadratic optimization can be formulated as AVEs. Because of that reason, AVEs attract the attention of researchers in this field. The GAVME is a generalization of the GAVE. There are many different types of conditions for the unique solution of GAVE (<a href="#Equ555">2</a>) studied (see <span class="cite">
	[
	<a href="#Shi Liang Wu 2016" >8</a>
	, 
	<a href="#Shi Liang Wu 2018" >9</a>
	, 
	<a href="#Shi-Liang Wu 2019" >10</a>
	, 
	<a href="#Shi Liang Wu 2021" >11</a>
	]
</span> and references therein), but GAVME (<a href="#Equ GAVME">1</a>) has few conditions for detecting the unique solution. The study of GAVE and GAVME is challenging and exciting because both the equations contain the non-differentiable term. </p>
<p>The GAVME was first considered by Dehghan <i class="it">et al.</i> <span class="cite">
	[
	<a href="#M Dehghan 2020 GAVME" >3</a>
	]
</span> and provided a multi-splitting Picard-iterative method for its solution. Kai <span class="cite">
	[
	<a href="#Kai Xie 2021 GAVME" >12</a>
	]
</span> extended the unique solvability result and improved the convergence condition of Dehghan <i class="it">et al.</i> <span class="cite">
	[
	<a href="#M Dehghan 2020 GAVME" >3</a>
	]
</span>. For more details about the GAVME one can see (<span class="cite">
	[
	<a href="#M Dehghan 2020 GAVME" >3</a>
	, 
	<a href="#Kai Xie 2021 GAVME" >12</a>
	]
</span> and references therein). </p>
<p>We will prove that if the following spectral radius condition </p>
<div class="displaymath" id="a0000000003">
  \begin{equation*}  \label{Equ555} \rho \big(\vert A^{-1} \vert \cdot \vert B \vert \big){\lt}1, \end{equation*}
</div>
<p> satisfy for nonsingular matrix A, then the GAVME (<a href="#Equ GAVME">1</a>) has a unique solution for any matrix D, where \(\rho (A)\) denote the spectral radius of the matrix \(A \in \mathbb {R}^{n \times n}\). For some instances, our result is superior to the following two singular value conditions </p>
<div class="equation" id="Equ3">
<p>
  <div class="equation_content">
    \begin{equation}  \label{Equ3} \sigma _{\max }\big(\vert B \vert \big) < \sigma _{\min }(A), \end{equation}
  </div>
  <span class="equation_label">3</span>
</p>
</div>
<div class="equation" id="Equ4">
<p>
  <div class="equation_content">
    \begin{equation}  \label{Equ4} \sigma _{\max }(B) < \sigma _{\min }(A), \end{equation}
  </div>
  <span class="equation_label">4</span>
</p>
</div>
<p> where \(\sigma _{\max }(A)\) denote the maximum and \(\sigma _{\min }(A)\) denote the minimum singular values of matrix \(A \in \mathbb {R}^{n \times n}\). In papers by Dehghan <i class="it">et al</i>. <span class="cite">
	[
	<a href="#M Dehghan 2020 GAVME" >3</a>
	]
</span> and Kai <span class="cite">
	[
	<a href="#Kai Xie 2021 GAVME" >12</a>
	]
</span>, the conditions (<a href="#Equ3">3</a>) and (<a href="#Equ4">4</a>) are provided, respectively. These conditions (<a href="#Equ3">3</a>) and (<a href="#Equ4">4</a>) are also used to determine whether the GAVME is uniquely solvable, but occasionally they are insufficient to determine whether the GAVME (<a href="#Equ GAVME">1</a>) is uniquely solvable. For example, see <a href="#Ex1">example 1</a> and <a href="#Ex2">example 2</a>, where our condition is sufficient to determine the unique solution of the GAVME (<a href="#Equ GAVME">1</a>), while conditions (<a href="#Equ3">3</a>) and (<a href="#Equ4">4</a>) are invalid to judge the unique solution of GAVME (<a href="#Equ GAVME">1</a>). </p>
<h1 id="a0000000004">2 Main Results</h1>
<p> In this section, for our main result, the following Lemma is required. </p>
<p><div class="lemma_thmwrapper " id="Equ5i">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">1</span>
    <span class="lemma_thmtitle"><span class="cite">
	[
	<a href="#Lotfi 2013" >4</a>
	]
</span></span>
  </div>
  <div class="lemma_thmcontent">
  <p>  If nonsingular matrix A and matrix B is satisfy the conditions </p>
<div class="displaymath" id="a0000000005">
  \begin{equation*}  \label{Equ5i} \rho \big(\vert A^{-1} \vert \cdot \vert B \vert \big){\lt}1, \end{equation*}
</div>
<p> then GAVE <a href="#Equ555" class="eqref">2</a> has a unique solution for any d. </p>

  </div>
</div> </p>
<p>Under the same unique solvability condition of the GAVE in the previous <a href="#Equ5i">lemma 1</a>, we prove that GAVME (<a href="#Equ GAVME">1</a>) also has a unique solution. So based on <a href="#Equ5i">lemma 1</a>, we give our main result. </p>
<p><div class="theorem_thmwrapper " id="Main Thoerem">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">1</span>
  </div>
  <div class="theorem_thmcontent">
  <p> If matrices A and B satisfy the condition </p>
<div class="equation" id="Equ5">
<p>
  <div class="equation_content">
    \begin{equation} \label{Equ5} \rho \big(\vert A^{-1} \vert \cdot \vert B \vert \big)<1, \end{equation}
  </div>
  <span class="equation_label">5</span>
</p>
</div>
<p> then GAVME <a href="#Equ GAVME" class="eqref">1</a> has a unique solution for any matrix \(D\), where \(A\) is invertible. </p>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000006">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> Suppose \(X=(x_{1},x_{2},...,x_{n})\) and \(D=(d_{1},d_{2},...,d_{n}),\) where \(x_{k}\) and \(d_{k}\) are the \(k\)th column of the matrices X and D, respectively. By \(\vert X \vert =(\vert x_{1}\vert ,\vert x_{2}\vert ,...,\vert x_{n}\vert )\), the GAVME (<a href="#Equ GAVME">1</a>) can be rewritten as \(A(x_{1},x_{2},...,x_{n}) + B(\vert x_{1}\vert ,\vert x_{2}\vert ,...,\vert x_{n}\vert )\) = \((d_{1},d_{2},...,d_{n}),\) or equivalently, </p>
<div class="equation" id="Equ GAVME2">
<p>
  <div class="equation_content">
    \begin{equation}  \label{Equ GAVME2} Ax_{k} + B \vert x_{k} \vert =d_{k}, \end{equation}
  </div>
  <span class="equation_label">6</span>
</p>
</div>
<p> where \(k=1,2,3,...,n.\) Then with the aid of <a href="#Equ5i">lemma 1</a>, GAVE (<a href="#Equ GAVME2">6</a>) has a unique solution for any k when the condition \(\rho \big(\vert A^{-1} \vert \cdot \vert B \vert \big){\lt}1\) holds. From here we can calculate all \(x_{k}\) separately. This completes the proof. <div class="proof_wrapper" id="a0000000007">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>
<p>In <a href="#Main Thoerem">theorem 1</a>, we have a new sufficient condition for the GAVME (<a href="#Equ GAVME">1</a>). To check our condition’s validity we are considering the following examples, see <a href="#Ex1">example 1</a> and <a href="#Ex2">example 2</a>, in which our condition (<a href="#Equ5">5</a>) is holds, while both the conditions (<a href="#Equ3">3</a>) and (<a href="#Equ4">4</a>) are not holds. </p>
<p><div class="example_thmwrapper " id="Ex1">
  <div class="example_thmheading">
    <span class="example_thmcaption">
    Example
    </span>
    <span class="example_thmlabel">1</span>
  </div>
  <div class="example_thmcontent">
  <p> Consider the matrices A and B given below </p>
<div class="displaymath" id="a0000000008">
  \begin{equation*}  A= \begin{bmatrix}  1 

&  0

\\ 0 

&  1 

\end{bmatrix} ~  ~  and ~  ~  B= \begin{bmatrix}  -0.4 

&  -0.9 

\\ 0.48 

&  0.04 

\end{bmatrix}\end{equation*}
</div>
<p>Here \(\sigma _{\min }(A)=1,\) \(\sigma _{\max }(B)= \sigma _{\max }(\vert B \vert )=1.0172,\) and \(\rho \big(\vert A^{-1} \vert \cdot \vert B \vert \big)=0.9015 {\lt}1\). Clearly, both the conditions <a href="#Equ3" class="eqref">3</a> and <a href="#Equ4" class="eqref">4</a> are not satisfying, while our condition <a href="#Equ5" class="eqref">5</a> is satisfying. Moreover, GAVME <a href="#Equ GAVME" class="eqref">1</a> has a unique solution according to <a href="#Main Thoerem">theorem 1</a>. </p>

  </div>
</div> </p>
<p><div class="example_thmwrapper " id="Ex2">
  <div class="example_thmheading">
    <span class="example_thmcaption">
    Example
    </span>
    <span class="example_thmlabel">2</span>
  </div>
  <div class="example_thmcontent">
  <p> Consider the following GAVME <a href="#Equ GAVME" class="eqref">1</a> </p>
<div class="displaymath" id="a0000000009">
  \begin{align*} & \begin{bmatrix}  2 

&  -4 

&  0 

\\ 0 

&  1.2 

&  1.1 

\\ -2 

&  0.8 

&  0 

\end{bmatrix}\begin{bmatrix}  x_1 

&  x_2 

&  x_3 

\\ x_4 

&  x_5 

&  x_6 

\\ x_7 

&  x_8 

&  x_9 

\end{bmatrix} + \begin{bmatrix}  1 

&  -1 

&  0 

\\ 0 

&  1 

&  1 

\\ -1 

&  0 

&  0 

\end{bmatrix}\begin{bmatrix}  |x_1| 

&  |x_2| 

&  |x_3| 

\\ |x_4| 

&  |x_5| 

&  |x_6| 

\\ |x_7| 

&  |x_8| 

&  |x_9| 

\end{bmatrix} = \\ & =\begin{bmatrix}  -9 

&  -17 

&  -45 

\\ 8.7 

&  8.3 

&  21.8 

\\ -3.6 

&  0.2 

&  11.4 

\end{bmatrix}\end{align*}
</div>
<p>Clearly, \(\rho \big(\vert A^{-1} \vert \cdot \vert B \vert \big)=0.9091 {\lt}1.\) Further, \(\sigma _{\max }\big(\vert B \vert \big)\)= 1.8019 \(\nless \) \(\sigma _{\min }(A)=0.9038\) and \(\sigma _{\max }(B)= 1.8019\) \(\nless \) \(\sigma _{\min }(A)=0.9038.\) This GAVME has a unique solution </p>
<div class="displaymath" id="a0000000010">
  \begin{equation*}  X =\begin{bmatrix}  2 

&  1 

&  -5 

\\ 3 

&  4 

&  8 

\\ 1 

&  -5 

&  2 

\end{bmatrix} . \end{equation*}
</div>

  </div>
</div> </p>
<p><div class="example_thmwrapper " id="Ex3">
  <div class="example_thmheading">
    <span class="example_thmcaption">
    Example
    </span>
    <span class="example_thmlabel">3</span>
  </div>
  <div class="example_thmcontent">
  <p> Consider the matrix A= \(I_{n \times n}\) and matrix </p>
<div class="displaymath" id="a0000000011">
  \begin{equation*}  B=b_{ij}= \begin{cases}  0.97 &  \text{if } i=j\\ 0 &  \text{if } i{\gt}j\\ 0.8 &  \text{if } i{\lt}j \end{cases}\end{equation*}
</div>
<p>We take n=25, to satisfy the condition of <a href="#Main Thoerem">theorem 1</a>. We get \(\rho \big(\vert A^{-1} \vert \cdot \vert B \vert \big)\)= 0.97 \({\lt}\) 1, so the GAVME <a href="#Equ GAVME" class="eqref">1</a> has a unique solution. </p>

  </div>
</div> </p>
<p>From <a href="#Main Thoerem">theorem 1</a>, we get the Corollary that follows. <div class="corollary_thmwrapper " id="Main Corollary">
  <div class="corollary_thmheading">
    <span class="corollary_thmcaption">
    Corollary
    </span>
    <span class="corollary_thmlabel">1</span>
  </div>
  <div class="corollary_thmcontent">
  <p> If A is invertible matrix and satisfy the following spectral radius condition </p>
<div class="equation" id="a0000000012">
<p>
  <div class="equation_content">
    \begin{equation}  \rho \big(\vert A^{-1} \vert \big)<1, \end{equation}
  </div>
  <span class="equation_label">10</span>
</p>
</div>
<p> then absolute value matrix equation \(AX + \vert X \vert =D\) has a unique solution for any matrix \(D\). </p>

  </div>
</div> </p>
<h1 id="a0000000013">3 Conclusions</h1>
<p> In this note, a new condition is stated for the unique solvability of the generalized absolute value matrix equation. Since the generalized absolute value equation and generalized absolute value matrix equation both have the same known terms on the left-hand side, namely matrices A and B. So in the future, we can also use the unique solvability conditions of the GAVE to detect the unique solvability of GAVME. This needs further investigation. The numerical results for the GAVME are also an interesting topic in the future. </p>
<p><small class="footnotesize">  </small></p>
<div class="bibliography">
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</dd>
  <dt><a name="Shi Liang Wu 2021">11</a></dt>
  <dd><p><a href =" https://doi.org/10.1007/s11590-020-01672-2"> <i class="sc">S.L. Wu, S. Shen</i>, <i class="it">On the unique solution of the generalized absolute value equation</i>, Optim. Lett., <b class="bf">15</b> (2021), 2017–-2024. <a href=" https://doi.org/10.1007/s11590-020-01672-2"> https://doi.org/10.1007/s11590-020-01672-2</a> <img src="img-0001.png" alt="\includegraphics[scale=0.1]{ext-link.png}" style="width:12.0px; height:10.700000000000001px" />
</a> </p>
</dd>
  <dt><a name="Kai Xie 2021 GAVME">12</a></dt>
  <dd><p><a href ="https://doi.org/10.11648/j.ajam.20210904.12"> <i class="sc">K. Xie</i>, <i class="it">On the Unique Solvability of the Generalized Absolute Value Matrix Equation</i>, American Journal of Applied Mathematics, <b class="bf">9</b> (2021) no. 4, p. 104. <a href="https://doi.org/10.11648/j.ajam.20210904.12">https://doi.org/10.11648/j.ajam.20210904.12</a> <img src="img-0001.png" alt="\includegraphics[scale=0.1]{ext-link.png}" style="width:12.0px; height:10.700000000000001px" />
</a> </p>
</dd>
</dl>


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