In this section, we present two algorithms for obtaining the extremal positive definite solutions of Equation (1.1). The main idea of the algorithms is to avoid computing the inverses of matrices.
Algorithm 4.1
(INVERSE-FREE Algorithm)
Consider the iterative algorithm
$$ \begin{aligned} &X_{k + 1} = \Biggl(I - \sum _{i = 1}^{m} A_{i}^{*} Y_{k}^{\frac{\delta_{i}}{\delta}} A_{i}\Biggr)^{\frac{1}{r}}, \\ &Y_{k + 1} = Y_{k}\bigl[2I - X_{k}Y_{k}^{ - \frac{1}{\delta}} \bigr],\quad k = 0,1,2,\ldots, i = 1,2,\ldots,m, \\ &X_{0} = \alpha I,\qquad Y_{0} = \alpha^{\delta} I,\quad \alpha > 1, \end{aligned} $$
(4.1)
where δ is a negative integer such that \(\frac{\vert \delta \vert }{r} < 1\).
Theorem 4.2
Suppose that Equation (1.1) has a positive definite solution. Then the iterative Algorithm
4.1
generates subsequences
\(\{ X_{2k} \}\)
and
\(\{ X_{2k + 1} \}\)
that are decreasing and converge to the maximal solution
\(X_{L}\).
Proof
Suppose that Equation (1.1) has a solution. We first prove that the subsequences \(\{ X_{2k} \}\) and \(\{ X_{2k + 1} \}\) are decreasing and the subsequences \(\{ Y_{2k} \}\) and \(\{ Y_{2k + 1} \}\) are increasing. Consider the sequence of matrices generated by (4.1).
For \(k = 0\), we have
$$X_{1} = \Biggl(I - \sum_{i = 1}^{m} A_{i}^{*} Y_{0}^{\frac{\delta_{i}}{\delta}} A_{i} \Biggr)^{\frac{1}{r}} = \Biggl(I - \sum_{i = 1}^{m} \alpha^{\delta_{i}}A_{i}^{*}A_{i} \Biggr)^{\frac{1}{r}}. $$
Since \((\alpha^{r} - 1)I + \sum_{i = 1}^{m} \alpha^{\delta_{i}}A_{i}^{*}A_{i} > 0\), we have \(I - \sum_{i = 1}^{m} \alpha^{\delta_{i}}A_{i}^{*}A_{i} < \alpha^{r}I\). Then
$$ \Biggl(I - \sum_{i = 1}^{m} \alpha^{\delta_{i}}A_{i}^{*}A_{i} \Biggr)^{\frac{1}{r}} < \alpha I = X_{0}. $$
(4.2)
So we get \(X_{1} < X_{0}\). Also, \(Y_{1} = Y_{0}[2I - X_{0}Y_{0}^{ - \frac{1}{\delta}} ] = \alpha^{\delta} [2I - \alpha \alpha^{ - 1}I] = \alpha^{\delta} I = Y_{0}\).
For \(k = 1\), we have
$$X_{2} = \Biggl(I - \sum_{i = 1}^{m} A_{i}^{*} Y_{1}^{\frac{\delta_{i}}{\delta}} A_{i} \Biggr)^{\frac{1}{r}} = \Biggl(I - \sum_{i = 1}^{m} \alpha^{\delta_{i}}A_{i}^{*}A_{i} \Biggr)^{\frac{1}{r}} = X_{1}. $$
So we have \(X_{2} < X_{0}\). Also,
$$\begin{aligned}& Y_{2} = Y_{1}\bigl[2I - X_{1}Y_{1}^{ - \frac{1}{\delta}} \bigr] = \alpha^{\delta} \Biggl[2I - \Biggl(I - \sum _{i = 1}^{m} \alpha^{\delta_{i}}A_{i}^{*}A_{i} \Biggr)^{\frac{1}{r}}\alpha^{ - 1}\Biggr] \\& \quad = 2\alpha^{\delta} I - \alpha^{\delta - 1}\Biggl(I - \sum _{i = 1}^{m} \alpha^{\delta_{i}}A_{i}^{*}A_{i} \Biggr)^{\frac{1}{r}}. \end{aligned}$$
From (4.2) we get \(2\alpha^{\delta} I - \alpha^{\delta - 1}(I - \sum_{i = 1}^{m} \alpha^{\delta_{i}}A_{i}^{*}A_{i} )^{\frac{1}{r}} > \alpha^{\delta} I\). Thus, \(Y_{2} > Y_{0}\).
For \(k = 2\), we have
$$X_{3} = \Biggl(I - \sum_{i = 1}^{m} A_{i}^{*} Y_{2}^{\frac{\delta_{i}}{\delta}} A_{i} \Biggr)^{\frac{1}{r}}. $$
Since \(Y_{2} > Y_{0}\), we have \(Y_{2}^{\frac{\delta_{i}}{\delta}} > Y_{0}^{\frac{\delta_{i}}{\delta}}\), so we get \((I - \sum_{i = 1}^{m} A_{i}^{*} Y_{2}^{\frac{\delta_{i}}{\delta}} A_{i})^{\frac{1}{r}} < (I - \sum_{i = 1}^{m} A_{i}^{*}Y_{0}^{\frac{\delta_{i}}{\delta}} A_{i})^{\frac{1}{r}}\).
Thus, \(X_{3} < X_{1}\). Also, \(Y_{3} = Y_{2}[2I - X_{2}Y_{2}^{ - \frac{1}{\delta}} ]\). Since \(Y_{2} > Y_{0}\) and \(X_{2} < X_{0}\), we have \(Y_{2}^{ - \frac{1}{\delta}} > Y_{0}^{ - \frac{1}{\delta}}\) and \(- X_{2} > - X_{0}\), so that \(- X_{2}Y_{2}^{ - \frac{1}{\delta}} > - X_{0}Y_{0}^{ - \frac{1}{\delta}}\) and \(2I - X_{2}Y_{2}^{ - \frac{1}{\delta}} > 2I - X_{0}Y_{0}^{ - \frac{1}{\delta}}\), and we get \(Y_{2}[2I - X_{2}Y_{2}^{ - \frac{1}{\delta}} ] > Y_{0}[2I - X_{0}Y_{0}^{ - \frac{1}{\delta}} ]\). Thus, \(Y_{3} > Y_{1}\).
Similarly, we can prove:
$$\begin{aligned}& X_{0} > X_{2} > X_{4} >\cdots \quad \mbox{and} \quad X_{1} > X_{3} > X_{5} >\cdots, \\& Y_{0} < Y_{2} < Y_{4} < \cdots \quad \mbox{and} \quad Y_{1} < Y_{3} < Y_{5} < \cdots\,. \end{aligned}$$
Hence, \(\{ X_{2k} \}\) and \(\{ X_{2k + 1} \}\), \(k = 0,1,2,\ldots\) , are decreasing, whereas \(\{ Y_{2k} \}\) and \(\{ Y_{2k + 1} \}\), \(k = 0,1,2,\ldots\) , are increasing.
Now, we show that \(\{ X_{2k} \}\) and \(\{ X_{2k + 1} \}\) are bounded from below by \(X_{L}\) (\(X_{k} > X_{L}\)) and that \(\{ Y_{2k} \}\) and \(\{ Y_{2k + 1} \}\) are bounded from above by \(X_{L}^{\delta}\).
By induction on k we get
$$X_{0} - X_{L} = \alpha I - \Biggl(I - \sum _{i = 1}^{m} A_{i}^{*}X_{L}^{\delta_{i}}A_{i} \Biggr)^{\frac{1}{r}}. $$
Since \(X_{L}\) is a solution of (1.1), we have that \((I - \sum_{i = 1}^{m} A_{i}^{*}X_{L}^{\delta_{i}}A_{i} )^{\frac{1}{r}} < I\) and \(\alpha I - (I - \sum_{i = 1}^{m} A_{i}^{*}X_{L}^{\delta_{i}}A_{i} )^{\frac{1}{r}} > \alpha I - I = (\alpha - 1)I > 0\), \(\alpha > 1\), that is, \(X_{0} > X_{L}\). Also, \(X_{1} - X_{L} = (I - \sum_{i = 1}^{m} A_{i}^{*}X_{0}^{\delta_{i}}A_{i} )^{\frac{1}{r}} - (I - \sum_{i = 1}^{m} A_{i}^{*}X_{L}^{\delta_{i}}A_{i})^{\frac{1}{r}}\).
Since \(X_{0} > X_{L}\), we have \(X_{0}^{\delta_{i}} < X_{L}^{\delta_{i}}\), and therefore \((I - \sum_{i = 1}^{m} A_{i}^{*}X_{0}^{\delta_{i}}A_{i} )^{\frac{1}{r}} > (I - \sum_{i = 1}^{m} A_{i}^{*}X_{L}^{\delta_{i}}A_{i})^{\frac{1}{r}}\). So we get \(X_{1} > X_{L}\). Also, \(X_{L}^{\delta} - Y_{0} = (I - \sum_{i - 1}^{m} A_{i}^{*}X_{L}^{\delta_{i}}A_{i} )^{\frac{\delta}{r}} - \alpha^{\delta} I\).
Since \(X_{0} > X_{L}\), we have \(\alpha I > X_{L}\) and \(\alpha^{\delta_{i}}I < X_{L}^{\delta_{i}}\)
\(\forall i = 1,2,\ldots, m\), so we have
$$ \begin{aligned} &I - \sum_{i = 1}^{m} \alpha^{\delta_{i}} A_{i}^{*}A_{i} > I - \sum _{i = 1}^{m} A_{i}^{*}X_{L}^{\delta_{i}}A_{i} \quad \text{and} \\ & \Biggl(I - \sum_{i = 1}^{m} \alpha^{\delta_{i}} A_{i}^{*}A_{i} \Biggr)^{\frac{\delta}{r}} < \Biggl(I - \sum_{i = 1}^{m} A_{i}^{*}X_{L}^{\delta_{i}}A_{i} \Biggr)^{\frac{\delta}{r}}. \end{aligned} $$
(4.3)
From (4.2) we get
$$ \Biggl(I - \sum_{i = 1}^{m} \alpha^{\delta_{i}} A_{i}^{*}A_{i} \Biggr)^{\frac{\delta}{r}} > \alpha^{\delta} I. $$
(4.4)
From (4.3) and (4.4) we get \((I - \sum_{i = 1}^{m} A_{i}^{*}X_{L}^{\delta_{i}}A_{i} )^{\frac{\delta}{r}} > \alpha^{\delta} I\), so we have \(X_{L}^{\delta} > Y_{0}\) and \(X_{L}^{\delta} > Y_{1}\).
Assume that \(X_{2k} > X_{L}\), \(X_{2k + 1} > X_{L}\) at \(k = t\) that is, \(X_{2t} > X_{L}\), \(X_{2t + 1} > X_{L}\). Also, \(Y_{2t} < X_{L}^{\delta}\) and \(Y_{2t + 1} < X_{L}^{\delta}\).
Now, for \(k = t + 1\),
$$X_{2t + 2} - X_{L} = \Biggl(I - \sum _{i = 1}^{m} A_{i}^{*} Y_{2t + 1}^{\frac{\delta_{i}}{\delta}} A_{i}\Biggr)^{\frac{1}{r}} - \Biggl(I - \sum_{i = 1}^{m} A_{i}^{*}X_{L}^{\delta_{i}} A_{i}\Biggr)^{\frac{1}{r}}. $$
Since \(Y_{2t + 1} < X_{L}^{\delta}\), we have \(Y_{2t + 1}^{\frac{\delta_{i}}{\delta}} < X_{L}^{\delta_{i}}\) and thus \(\sum_{i = 1}^{m} A_{i}^{*} Y_{2t + 1}^{\frac{\delta_{i}}{\delta}} A_{i} < \sum_{i = 1}^{m} A_{i}^{*}X_{L}^{\delta_{i}} A_{i}\). Therefore, \((I - \sum_{i = 1}^{m} A_{i}^{*} Y_{2t + 1}^{\frac{\delta_{i}}{\delta}} A_{i})^{\frac{1}{r}} > (I - \sum_{i = 1}^{m} A_{i}^{*}X_{L}^{\delta_{i}} A_{i})^{\frac{1}{r}}\).
Hence, \(X_{2t + 2} - X_{L} > 0\), that is, \(X_{2t + 2} > X_{L}\). Also, \(X_{L}^{\delta} - Y_{2t + 2} = X_{L}^{\delta} - Y_{2t + 1}[2I - X_{2t + 1}Y_{2t + 1}^{ - \frac{1}{\delta}} ]\).
Since \(Y_{2t + 1} < X_{L}^{\delta}\) and \(X_{2t + 1} > X_{L}\), we have \(Y_{2t + 1}^{ - \frac{1}{\delta}} < X_{L}^{ - 1}\) and \(- X_{2t + 1} < - X_{L}\),
$$- X_{2t + 1}Y_{2t + 1}^{ - \frac{1}{\delta_{i}}} < - X_{L}^{ - 1}X_{L} = - I,\quad \mbox{and}\quad 2I - X_{2t + 1}Y_{2t + 1}^{ - \frac{1}{\delta_{i}}} < 2I - I = I. $$
Then \(Y_{2t + 1}[2I - X_{2t + 1}Y_{2t + 1}^{ - \frac{1}{\delta}} ] < X_{L}^{\delta}\). Hence, \(Y_{2t + 2} < X_{L}^{\delta}\), and thus
$$X_{2t + 3} - X_{L} = \Biggl(I - \sum _{i = 1}^{m} A_{i}^{*} Y_{2t + 2}^{\frac{\delta_{i}}{\delta}} A_{i}\Biggr)^{\frac{1}{r}} - \Biggl(I - \sum_{i = 1}^{m} A_{i}^{*}X_{L}^{\delta_{i}} A_{i}\Biggr)^{\frac{1}{r}}. $$
Since, \(Y_{2t + 2} < X_{L}^{\delta}\), we have \(Y_{2t + 2}^{\frac{\delta_{i}}{\delta}} < X_{L}^{\delta_{i}}\) and \(\sum_{i = 1}^{m} A_{i}^{*} Y_{2t + 2}^{\frac{\delta_{i}}{\delta}} A_{i} < \sum_{i = 1}^{m} A_{i}^{*}X_{L}^{\delta_{i}} A_{i}\). Therefore, \((I - \sum_{i = 1}^{m} A_{i}^{*} Y_{2t + 2}^{\frac{\delta_{i}}{\delta}} A_{i})^{\frac{1}{r}} > (I - \sum_{i = 1}^{m} A_{i}^{*}X_{L}^{\delta_{i}} A_{i})^{\frac{1}{r}}\).
Hence, \(X_{2t + 3} - X_{L} > 0\), that is, \(X_{2t + 3} > X_{L}\), and thus \(X_{L}^{\delta} - Y_{2t + 3} = X_{L}^{\delta} - Y_{2t + 2}[2I - X_{2t + 2}Y_{2t + 2}^{ - \frac{1}{\delta}} ]\).
Since \(Y_{2t + 2} < X_{L}^{\delta}\) and \(X_{2t + 2} > X_{L}\), we have \(Y_{2t + 2}^{ - \frac{1}{\delta}} < X_{L}^{ - 1}\) and \(- X_{2t + 2} < - X_{L}\), and thus
$$- X_{2t + 2}Y_{2t + 2}^{ - \frac{1}{\delta}} < - X_{L}^{ - 1}X_{L} = - I\quad \mbox{and} \quad 2I - X_{2t + 2}Y_{2t + 2}^{ - \frac{1}{\delta}} < 2I - I = I. $$
Then \(Y_{2t + 2}[2I - X_{2t + 2}Y_{2t + 2}^{ - \frac{1}{\delta}} ] < X_{L}^{\delta}\). Hence, \(Y_{2t + 3} < X_{L}^{\delta}\).
Since \(\{ X_{2k} \}\) and \(\{ X_{2k + 1} \}\) are decreasing and bounded from below by \(X_{L}\) and \(\{ Y_{2k} \}\) and \(\{ Y_{2k + 1} \}\) are increasing and bounded from above by \(X_{L}^{\delta}\), it follows that \(\lim_{k \to \infty} X_{k} = X\) and \(\lim_{k \to \infty} Y_{k} = Y\) exist.
Taking limits in (4.1) gives \(Y = X^{\delta}\) and \(X = (I - \sum_{i = 1}^{m} A_{i}^{*}X^{\delta_{i}} A_{i})^{\frac{1}{r}}\), that is, X is a solution. Hence, \(X = X_{L}\). □
Remark
We have proved that the maximal solution is unique; see the Appendix.
Now, we consider the case \(0 < \alpha < 1\).
Algorithm 4.3
(INVERSE-FREE Algorithm)
Consider the iterative (simultaneous) algorithm
$$ \begin{aligned} &X_{k + 1} = \Biggl(I - \sum _{i = 1}^{m} A_{i}^{*} Y_{k}^{\frac{\delta_{i}}{\delta}} A_{i}\Biggr)^{\frac{1}{r}}, \\ &Y_{k + 1} = Y_{k}\bigl[2I - X_{k}Y_{k}^{ - \frac{1}{\delta}} \bigr],\quad k = 0,1,2,\ldots, i = 1,2,\ldots,m, \\ &X_{0} = \alpha I,\qquad Y_{0} = \alpha^{\delta} I, \quad 0 < \alpha < 1, \end{aligned} $$
(4.5)
where δ is a negative integer such that \(\frac{\vert \delta \vert }{r} < 1\).
Theorem 4.4
Suppose that Equation (1.1) has a positive definite solution such that
\(\sum_{i = 1}^{m} \alpha^{\delta_{i}}A_{i}^{*}A_{i} < (1 - \alpha^{r})I\). Then the iterative Algorithm
4.3
generates the subsequences
\(\{ X_{2k} \}\)
and
\(\{ X_{2k + 1} \}\)
that are increasing and converge to the minimal solution
\(X_{S}\).
Proof
Suppose that Equation (1.1) has a solution. We first prove that the subsequences \(\{ X_{2k} \}\) and \(\{ X_{2k + 1} \}\) are increasing and the subsequences \(\{ Y_{2k} \}\) and \(\{ Y_{2k + 1} \}\) are decreasing. Consider the sequence of matrices generated by (4.5).
For \(k = 0\), we have
$$X_{1} = \Biggl(I - \sum_{i = 1}^{m} A_{i}^{*} Y_{0}^{\frac{\delta_{i}}{\delta}} A_{i} \Biggr)^{\frac{1}{r}} = \Biggl(I - \sum_{i = 1}^{m} \alpha^{\delta_{i}}A_{i}^{*}A_{i} \Biggr)^{\frac{1}{r}}. $$
From the condition of the theorem we have
$$ I - \sum_{i = 1}^{m} \alpha^{\delta_{i}}A_{i}^{*}A_{i} > \alpha^{r}I. $$
(4.6)
Then,
$$ \Biggl(I - \sum_{i = 1}^{m} \alpha^{\delta_{i}}A_{i}^{*}A_{i} \Biggr)^{\frac{1}{r}} > \alpha I = X_{0}. $$
(4.7)
So we get \(X_{1} > X_{0}\). Also, \(Y_{1} = Y_{0}[2I - X_{0}Y_{0}^{ - \frac{1}{\delta}} ] = \alpha^{\delta} [2I - \alpha \alpha^{ - 1}I] = \alpha^{\delta} I = Y_{0}\).
For \(k = 1\), we have
$$X_{2} = \Biggl(I - \sum_{i = 1}^{m} A_{i}^{*} Y_{1}^{\frac{\delta_{i}}{\delta}} A_{i} \Biggr)^{\frac{1}{r}} = \Biggl(I - \sum_{i = 1}^{m} \alpha^{\delta_{i}}A_{i}^{*}A_{i} \Biggr)^{\frac{1}{r}} = X_{1}. $$
So we have \(X_{2} > X_{0}\). Also,
$$\begin{aligned}& Y_{2} = Y_{1}\bigl[2I - X_{1}Y_{1}^{ - \frac{1}{\delta}} \bigr] = \alpha^{\delta} \Biggl[2I - \Biggl(I - \sum _{i = 1}^{m} \alpha^{\delta_{i}}A_{i}^{*}A_{i} \Biggr)^{\frac{1}{r}}\alpha^{ - 1}\Biggr] \\ & \hphantom{Y_{2}} = 2\alpha^{\delta} I - \alpha^{\delta - 1}\Biggl(I - \sum_{i = 1}^{m} \alpha^{\delta_{i}}A_{i}^{*}A_{i} \Biggr)^{\frac{1}{r}}. \end{aligned}$$
From (4.7) we get \(2\alpha^{\delta} I - \alpha^{\delta - 1}(I - \sum_{i = 1}^{m} \alpha^{\delta_{i}}A_{i}^{*}A_{i} )^{\frac{1}{r}} < \alpha^{\delta} I\). Thus, \(Y_{2} < Y_{0}\).
For \(k = 2\), we have
$$X_{3} = \Biggl(I - \sum_{i = 1}^{m} A_{i}^{*} Y_{2}^{\frac{\delta_{i}}{\delta}} A_{i} \Biggr)^{\frac{1}{r}}. $$
Since \(Y_{2} < Y_{0}\), we have \(Y_{2}^{\frac{\delta_{i}}{\delta}} < Y_{0}^{\frac{\delta_{i}}{\delta}}\), so we get \((I - \sum_{i = 1}^{m} A_{i}^{*} Y_{2}^{\frac{\delta_{i}}{\delta}} A_{i})^{\frac{1}{r}} > (I - \sum_{i = 1}^{m} A_{i}^{*}Y_{0}^{\frac{\delta_{i}}{\delta}} A_{i})^{\frac{1}{r}}\). Thus, \(X_{3} > X_{1}\). Also, \(Y_{3} = Y_{2}[2I - X_{2}Y_{2}^{ - \frac{1}{\delta}} ]\).
Since \(Y_{2} < Y_{0}\) and \(X_{2} > X_{0}\), we have \(Y_{2}^{ - \frac{1}{\delta}} < Y_{0}^{ - \frac{1}{\delta}}\) and \(- X_{2} < - X_{0}\), so that \(- X_{2}Y_{2}^{ - \frac{1}{\delta}} < - X_{0}Y_{0}^{ - \frac{1}{\delta}}\) and \(2I - X_{2}Y_{2}^{ - \frac{1}{\delta}} < 2I - X_{0}Y_{0}^{ - \frac{1}{\delta}}\), and we get \(Y_{2}[2I - X_{2}Y_{2}^{ - \frac{1}{\delta}} ] < Y_{0}[2I - X_{0}Y_{0}^{ - \frac{1}{\delta}} ]\). Thus, \(Y_{3} < Y_{1}\).
Similarly, we can prove that
$$\begin{aligned}& X_{0} < X_{2} < X_{4} < \cdots \quad \mbox{and} \quad X_{1} < X_{3} < X_{5} < \cdots, \\ & Y_{0} > Y_{2} > Y_{4} >\cdots \quad \mbox{and} \quad Y_{1} > Y_{3} > Y_{5} >\cdots\,. \end{aligned}$$
Hence \(\{ X_{2k} \}\) and \(\{ X_{2k + 1} \}\), \(k = 0,1,2,\ldots\) , are increasing, whereas \(\{ Y_{2k} \}\) and \(\{ Y_{2k + 1} \}\), \(k = 0,1,2,\ldots\) , are decreasing.
Now, we show that \(\{ X_{2k} \}\) and \(\{ X_{2k + 1} \}\) are bounded from above by \(X_{S}\) (\(X_{S} > X_{k}\)), and \(\{ Y_{2k} \}\) and \(\{ Y_{2k + 1} \}\) are bounded from below by \(X_{S}^{\delta}\).
By induction on k we obtain
$$X_{S} - X_{0} = \Biggl(I - \sum _{i = 1}^{m} A_{i}^{*}X_{S}^{\delta_{i}}A_{i} \Biggr)^{\frac{1}{r}} - \alpha I > 0, $$
that is, \(X_{S} > X_{0}\). Also,
$$X_{S} - X_{1} = \Biggl(I - \sum _{i = 1}^{m} A_{i}^{*}X_{S}^{\delta_{i}}A_{i} \Biggr)^{\frac{1}{r}} - \Biggl(I - \sum_{i = 1}^{m} A_{i}^{*}X_{0}^{\delta_{i}}A_{i} \Biggr)^{\frac{1}{r}}. $$
Since \(X_{S} > X_{0}\), we have \(X_{S}^{\delta_{i}} < X_{0}^{\delta_{i}}\), and therefore \((I - \sum_{i = 1}^{m} A_{i}^{*}X_{S}^{\delta_{i}}A_{i} )^{\frac{1}{r}} > (I - \sum_{i = 1}^{m} A_{i}^{*}X_{0}^{\delta_{i}}A_{i})^{\frac{1}{r}}\). So we get \(X_{S} > X_{1}\). Also, \(Y_{0} - X_{S}^{\delta} = \alpha^{\delta} I - (I - \sum_{i - 1}^{m} A_{i}^{*}X_{S}^{\delta_{i}}A_{i} )^{\frac{\delta}{r}}\).
Since \(X_{S} > X_{0}\), we have \(X_{S} > \alpha I\) and \(X_{S}^{\delta_{i}} < \alpha^{\delta_{i}}I\)
\(\forall i = 1,2,\ldots, m\), and we get
$$ \begin{aligned} &I - \sum_{i = 1}^{m} \alpha^{\delta_{i}} A_{i}^{*}A_{i} < I - \sum _{i = 1}^{m} A_{i}^{*}X_{S}^{\delta_{i}}A_{i} \quad \text{and} \\ &\Biggl(I - \sum_{i = 1}^{m} \alpha^{\delta_{i}} A_{i}^{*}A_{i} \Biggr)^{\frac{\delta}{r}} > \Biggl(I - \sum_{i = 1}^{m} A_{i}^{*}X_{S}^{\delta_{i}}A_{i} \Biggr)^{\frac{\delta}{r}}. \end{aligned} $$
(4.8)
From (4.6) we get
$$ \Biggl(I - \sum_{i = 1}^{m} \alpha^{\delta_{i}} A_{i}^{*}A_{i} \Biggr)^{\frac{\delta}{r}} < \alpha^{\delta} I. $$
(4.9)
From (4.8) and (4.9) we get \(\alpha^{\delta} I > (I - \sum_{i = 1}^{m} A_{i}^{*}X_{S}^{\delta_{i}}A_{i} )^{\frac{\delta}{r}}\), and thus \(Y_{0} > X_{S}^{\delta}\) and \(Y_{1} > X_{S}^{\delta}\).
Assume that \(X_{2k} < X_{S}\) and \(X_{2k + 1} < X_{S}\) at \(k = t\), that is, \(X_{2t} < X_{S}\) and \(X_{2t + 1} < X_{S}\). Also, \(Y_{2t} > X_{S}^{\delta}\) and \(Y_{2t + 1} > X_{S}^{\delta}\).
Now, for \(k = t + 1\), we have
$$X_{S} - X_{2t + 2} = \Biggl(I - \sum _{i = 1}^{m} A_{i}^{*}X_{S}^{\delta_{i}}A_{i} \Biggr)^{\frac{1}{r}} - \Biggl(I - \sum_{i = 1}^{m} A_{i}^{*} Y_{2t + 1}^{\frac{\delta_{i}}{\delta}} A_{i} \Biggr)^{\frac{1}{r}}. $$
Since \(Y_{2t + 1} > X_{S}^{\delta}\), we have \(Y_{2t + 1}^{\frac{\delta_{i}}{\delta}} > X_{S}^{\delta_{i}}\) and \(\sum_{i = 1}^{m} A_{i}^{*} Y_{2t + 1}^{\frac{\delta_{i}}{\delta}} A_{i} > \sum_{i = 1}^{m} A_{i}^{*}X_{S}^{\delta_{i}} A_{i}\). Therefore, \((I - \sum_{i = 1}^{m} A_{i}^{*} Y_{2t + 1}^{\frac{\delta_{i}}{\delta}} A_{i})^{\frac{1}{r}} < (I - \sum_{i = 1}^{m} A_{i}^{*}X_{S}^{\delta_{i}} A_{i})^{\frac{1}{r}}\).
Hence, \(X_{S} - X_{2t + 2} > 0\), that is, \(X_{S} > X_{2t + 2}\). Also,
$$Y_{2t + 2} - X_{S}^{\delta} = Y_{2t + 1}\bigl[2I - X_{2t + 1}Y_{2t + 1}^{ - \frac{1}{\delta}} \bigr] - X_{S}^{\delta}. $$
Since \(Y_{2t + 1} > X_{S}^{\delta}\) and \(X_{2t + 1} < X_{S}\), we have \(Y_{2t + 1}^{ - \frac{1}{\delta}} > X_{S}^{ - 1}\) and \(- X_{2t + 1} > - X_{S}\), and then
$$- X_{2t + 1}Y_{2t + 1}^{ - \frac{1}{\delta_{i}}} > - X_{S}X_{S}^{ - 1} = - I\quad \mbox{and}\quad 2I - X_{2t + 1}Y_{2t + 1}^{ - \frac{1}{\delta_{i}}} > 2I - I = I. $$
Then \(Y_{2t + 1}[2I - X_{2t + 1}Y_{2t + 1}^{ - \frac{1}{\delta}} ] > X_{S}^{\delta}\). Hence, \(Y_{2t + 2} > X_{S}^{\delta}\). Consider
$$X_{S} - X_{2t + 3} = \Biggl(I - \sum _{i = 1}^{m} A_{i}^{*}X_{S}^{\delta_{i}}A_{i} \Biggr)^{\frac{1}{r}} - \Biggl(I - \sum_{i = 1}^{m} A_{i}^{*} Y_{2t + 2}^{\frac{\delta_{i}}{\delta}} A_{i} \Biggr)^{\frac{1}{r}}. $$
Since \(Y_{2t + 2} > X_{S}^{\delta}\), we have \(Y_{2t + 2}^{\frac{\delta_{i}}{\delta}} > X_{S}^{\delta_{i}}\) and \(\sum_{i = 1}^{m} A_{i}^{*} Y_{2t + 2}^{\frac{\delta_{i}}{\delta}} A_{i} > \sum_{i = 1}^{m} A_{i}^{*}X_{S}^{\delta_{i}} A_{i}\). Therefore, \((I - \sum_{i = 1}^{m} A_{i}^{*} Y_{2t + 2}^{\frac{\delta_{i}}{\delta}} A_{i})^{\frac{1}{r}} < (I - \sum_{i = 1}^{m} A_{i}^{*}X_{S}^{\delta_{i}} A_{i})^{\frac{1}{r}}\).
Hence, \(X_{S} - X_{2t + 3} > 0\), that is, \(X_{S} > X_{2t + 3}\). Now consider
$$Y_{2t + 3} - X_{S}^{\delta} = Y_{2t + 2}\bigl[2I - X_{2t + 2}Y_{2t + 2}^{ - \frac{1}{\delta}} \bigr] - X_{S}^{\delta}. $$
Since \(Y_{2t + 2} > X_{S}^{\delta}\) and \(X_{2t + 2} < X_{S}\), we have \(Y_{2t + 2}^{ - \frac{1}{\delta}} > X_{S}^{ - 1}\) and \(- X_{2t + 2} > - X_{S}\), and thus
$$- X_{2t + 2}Y_{2t + 2}^{ - \frac{1}{\delta}} > - X_{S}X_{S}^{ - 1} = - I\quad \mbox{and} \quad 2I - X_{2t + 2}Y_{2t + 2}^{ - \frac{1}{\delta}} > 2I - I = I. $$
Then \(Y_{2t + 2}[2I - X_{2t + 2}Y_{2t + 2}^{ - \frac{1}{\delta}} ] > X_{S}^{\delta}\), and hence \(Y_{2t + 3} > X_{S}^{\delta}\).
Since \(\{ X_{2k} \}\) and \(\{ X_{2k + 1} \}\) are increasing and bounded from above by \(X_{S}\) and \(\{ Y_{2k} \}\) and \(\{ Y_{2k + 1} \}\) are decreasing and bounded from below by \(X_{S}^{\delta}\), it follows that \(\lim_{k \to \infty} X_{k} = X\) and \(\lim_{k \to \infty} Y_{k} = Y\) exist. Taking the limits in (4.5) gives \(Y = X^{\delta}\) and \(X = (I - \sum_{i = 1}^{m} A_{i}^{*}X^{\delta_{i}} A_{i})^{\frac{1}{r}}\), that is, X is a solution of Equation (1.1). Hence, \(X = X_{S}\). □
Remark
We have proved that the minimal solution is unique; see the Appendix.