证明严格对角占优矩阵经一步 Gauss 消元后仍为严格对角矩阵

问题

设矩阵 \(\mathbf{A} = (a_{ij}) \in \mathbf{R}^{n \times n}\) 满足 \(a_{11} \neq 0\),且其经一步 Gauss 消元后成为

\[ \begin{pmatrix} a_{11} & \pmb{a}_{1}^{T} \\ \pmb{0} & \pmb{A}_2 \\ \end{pmatrix} \]

证明:若 \(\pmb{A}\) 为严格对角占优矩阵,则经一次 Gauss 消元之后的 \(\pmb{A}^{(1)}\) 也是严格对角矩阵.

解答

注意到矩阵 \(\pmb{A}^{(1)}\)\(i\)\(j\) 列元素为

\[a_{ij}^{(1)} = a_{ij} - a_{i1} \frac{a_{1j}}{a_{11}}, \quad i, j = 2, 3, ..., n.\]

要证严格对角占优矩阵 \(\mathbf{A}\) 经过一步 Gauss 消元后的矩阵 \(\mathbf{A}^{(1)}\) 仍是严格对角占优矩阵,即证

\[\left| a_{ii} - a_{i1} \frac{a_{1i}}{a_{11}} \right| > \sum_{j = 2, \\ j \neq i}^{n} \left| a_{ij} - a_{i1} \frac{a_{1j}}{a_{11}} \right|\]

然后,对不等式进行放缩,把左边缩小为

\[\left| a_{ii} \right| - \left| a_{ii} \right| \frac{\left| a_{1i} \right|}{\left| a_{11} \right|}\]

右边放大为

\[\sum_{j = 2, \\ j \neq i}^{n} \left( \left| a_{ij} \right| + \frac{\left| a_{1j} \right|}{\left| a_{11} \right|} \left| a_{i1} \right| \right)\]

放缩后的左边减去右边得

\[ \begin{equation}\begin{split} & \left| a_{ii} \right| - \left| a_{ii} \right| \frac{\left| a_{1i} \right|}{\left| a_{11} \right|} - \sum_{j = 2, \\ j \neq i}^{n} \left| a_{ij} \right| - \sum_{j = 2, \\ j \neq i}^{n} \frac{\left| a_{1j} \right|}{\left| a_{11} \right|} \left| a_{i1} \right| \\ & = \left| a_{ii} \right| - \sum_{j = 2, \\ j \neq i}^{n} \left| a_{ij} \right| - \sum_{j = 2}^{n} \frac{\left| a_{1j} \right|}{\left| a_{11} \right|} \left| a_{i1} \right| \nonumber \\ & > \left| a_{i1} \right| - \left| a_{i1} \right| \sum_{j = 2}^{n} \frac{\left| a_{1j} \right|}{\left| a_{11} \right|} \\ & > 0 \end{split}\end{equation}\]

所以,\(\mathbf{A}^{(1)}\) 仍是严格对角占优矩阵。