主对角占优矩阵

矩阵\(A=\left( \begin{matrix}{}\text{a}_{11}&\text{a}_{12}&\cdots&\text{a}_{1n}\\\text{a}_{21}&\text{a}_{22}&\cdots&\text{a}_{2n}\\\vdots&\vdots&\ddots&\vdots\\\text{a}_{n1}&\text{a}_{n2}&\cdots&\text{a}_{nn}\\\end{matrix} \right)\)满足

1)\(|a_{ii}|>\sum_{j\ne i}^n{|a_{ij}|}\)，则\(\det A\ne0\).

2)\(a_{ii}>\sum_{j\ne i}^n{|a_{ij}|}\)，则\(\det A>0\).

证明：

\(\det A\ne 0\ \ \Longleftrightarrow \ \ \text{线性方程组}AX=0\text{只有零解}\)

故假设存在非零解

\[X=\left( \begin{matrix}{} x_1& x_2& \cdots& x_n\\ \end{matrix} \right) ^T \]

记\(|x_i|=\max_j\left\{ |x_j| \right\}\)，

则|\(a_{ii}x_i|\le \sum_{j\ne i}^{}{|a_{ij}||x_j|\le}|a_{ij}||x_i|\ \ \Rightarrow \ \ |a_{ii}|\le \sum_{j\ne i}^{}{|a_{ij}|}\)，矛盾.

记

\[f\left( x \right) =\det A=\left| \begin{matrix} \text{a}_{11}& x\text{a}_{12}& \cdots& x\text{a}_{1n}\\ x\text{a}_{21}& \text{a}_{22}& \cdots& x\text{a}_{2n}\\ \vdots& \vdots& \ddots& \vdots\\ x\text{a}_{n1}& x\text{a}_{n2}& \cdots& \text{a}_{nn}\\ \end{matrix} \right|\]

由1)知\(|x|\le1\)时\(f(x)\ne0\)，且\(f(0)>0\)，有连续性可知\(f(1)>0\).