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