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.