In this section, we first give the following notations, which will be useful in the following proofs.

*If A* = [*a*_{ij}] Î ℝ^{n×n} *is strictly row diagonally dominant, then, for all i, j* Î *N, j ≠ i, t* = 1, 2, … ,

(a) 1 > *d*_{j} ≥ *s*_{ji} ≥ *u*_{ji} ≥ *υ*_{ji}^{(0)} ≥ *p*_{ji}^{(1)} ≥ *υ*_{ji}^{(1)} ≥ *p*_{ji}^{(2)} ≥ *υ*_{ji}^{(2)} ≥ · · · ≥ *p*_{ji}^{(t)} ≥ *υ*_{ji}^{(t)} ≥ · · · ≥ 0;

(b) 1 ≥ *h*_{i} ≥ 0, 1 ≥ *h*_{i}^{(t)} ≥ 0.

*Proof.* Since *A* is a strictly row diagonally dominant matrix, that is, $\left|{a}_{jj}\right|\text{\hspace{0.5em}<\hspace{0.5em}}{\displaystyle \sum _{k\ne j}\left|{a}_{jk}\right|,}\text{\hspace{0.5em}}j\text{\hspace{0.5em}}\in \text{\hspace{0.5em}N,}$, obviously, 1 > *d*_{j} ≥ 0, *j* Î *N.* By the definitions of *d*_{j}, *s*_{ji}, we have 1 > *d*_{j} ≥ *s*_{ji} ≥ 0, *j, i* Î *N, j ≠ i.* And then, by the definitions of *s*_{ji}, *u*_{ji}, we have *s*_{ji} ≥ *u*_{ji}, *j,i* Î *N, j ≠ i.* Hence,
$$\frac{\left|{a}_{ji}\right|}{\left|{a}_{jj}\right|{s}_{ji}-{\displaystyle \sum _{k\ne j,i}\left|{a}_{jk}\right|{s}_{ki}}}=\frac{\left|{a}_{jj}\right|{s}_{ji}-{\displaystyle \sum _{k\ne j,i}\left|{a}_{jk}\right|{d}_{k}}}{\left|{a}_{jj}\right|{s}_{ji}-{\displaystyle \sum _{k\ne j,i}\left|{a}_{jk}\right|{s}_{ki}}}\le 1,$$
from the definition of *h*_{i}, we have 0 ≤ *h*_{i} ≤ 1, *i* Î *N.* Furthermore, by the definitions of *u*_{ji}, *υ*_{ji}^{(0)}, we have *u*_{ji} ≥ *υ*_{ji}^{(0)}, *j, i* Î *N, j ≠ i.*

Since ${h}_{i}=\underset{j\ne i}{\text{max}}\ufe5b\frac{\left|{a}_{ji}\right|}{\left|{a}_{jj}\right|{s}_{ji}-{\displaystyle \sum _{k\ne j,i}\left|{a}_{jk}\right|{s}_{ki}}}\ufe5c,i\in N,$, we have
$${h}_{i}\ge \frac{\left|{a}_{ji}\right|}{\left|{a}_{ji}\right|{s}_{ji}-{\displaystyle \sum _{k\ne j,i}\left|{a}_{jk}\right|{s}_{ki}}},\text{\hspace{0.5em}}i.e.,\text{\hspace{0.5em}}{s}_{ji}{h}_{i}\ge \frac{\left|{a}_{ji}\right|+{\displaystyle \sum _{k\ne j,i}\left|{a}_{jk}\right|{s}_{ki}{h}_{i}}}{\left|{a}_{jj}\right|}={\upsilon}_{ji}^{(0)},j,i\in N,j\ne i.$$
From the definitions of *υ*_{ji}^{(0)}, *p*_{ji}^{(1)}, we have *υ*_{ji}^{(0)} ≥ *p*_{ji}^{(1)} ≥ 0, *j, i* Î *N, j ≠ i.*

Hence,
$$\frac{\left|{a}_{ji}\right|}{\left|{a}_{jj}\right|{p}_{ji}^{\left(1\right)}-{\displaystyle \sum _{k\ne j,i}\left|{a}_{jk}\right|{p}_{ki}^{\left(1\right)}}}=\frac{\left|{a}_{jj}\right|{p}_{ji}^{\left(1\right)}-{\displaystyle \sum _{k\ne j,i}\left|{a}_{jk}\right|{\upsilon}_{ki}^{\left(0\right)}}}{\left|{a}_{jj}\right|{p}_{ji}^{\left(1\right)}-{\displaystyle \sum _{k\ne j,i}\left|{a}_{jk}\right|{p}_{ki}^{\left(1\right)}}}\le 1,$$
by the definition of *h*_{i}^{(1)}, we have 0 ≤ *h*_{i}^{(1)} ≤ 1, *i* Î *N.*

Since ${h}_{i}^{\left(1\right)}=\underset{j\ne i}{\text{max}}\ufe5b\frac{\left|{a}_{ji}\right|}{\left|{a}_{jj}\right|{p}_{ji}^{\left(1\right)}-{\displaystyle \sum _{k\ne j,i}\left|{a}_{jk}\right|{p}_{ki}^{\left(1\right)}}}\ufe5c,i\in N,$ we have
$${h}_{i}^{\left(1\right)}\ge \frac{\left|{a}_{ji}\right|}{\left|{a}_{jj}\right|{p}_{ji}^{\left(1\right)}-{\displaystyle \sum _{k\ne j,i}\left|{a}_{jk}\right|{p}_{ki}^{\left(1\right)}}},\text{\hspace{0.5em}}i.e.,\text{\hspace{0.5em}}{p}_{ji}^{\left(1\right)}{h}_{i}^{\left(1\right)}\ge \frac{\left|{a}_{ji}\right|+{\displaystyle \sum _{k\ne j,i}\left|{a}_{jk}\right|{p}_{ki}^{\left(1\right)}{h}_{i}^{\left(1\right)}}}{\left|{a}_{jj}\right|}={\upsilon}_{ji}^{\left(1\right)},j,i\in N,j\ne i.$$
By 0 ≤ *h*_{i}^{(1)} ≤ 1, *i* Î *N,* we have *p*_{ji}^{(1)} ≥ *υ*_{ji}^{(1)} ≥ 0, *j, i* Î *N, j ≠ i.* From the definitions of *υ*_{ji}^{(1)}, *p*_{ji}^{(2)}, we obtain *υ*_{ji}^{(1)} ≥ *p*_{ji}^{(2)} ≥ 0, *j, i* Î *N, j ≠ i.*

In the same way as above, we can also prove that
$${p}_{ji}^{\left(2\right)}\ge {\upsilon}_{ji}^{\left(2\right)}\ge \cdots {p}_{ji}^{\left(t\right)}\ge {\upsilon}_{ji}^{\left(t\right)}\ge \cdots \ge 0,1\ge {h}_{i}^{\left(t\right)}\ge 0,j,i\in N,j\ne i,t=2,3,\mathrm{\dots .}$$

The proof is completed. □

Using the same technique as the proof of Lemma 2.2 in [11], we can obtain the following lemma.

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