49886

$\bf命题：$设实二次型\[f\left( {{x_1}, \cdots ,{x_n}} \right) = \sum\limits_{i = 1}^n {{{\left( {{a_{i1}}{x_1} + \cdots + {a_{in}}{x_n}} \right)}^2}} \]

证明二次型的秩等于$A = {\left( {{a_{ij}}} \right)_{n \times n}}$的秩

证明：我们容易知道
\[f\left( {{x_1}, \cdots ,{x_n}} \right) = \sum\limits_{i = 1}^n {x'{\alpha _i}{\alpha _i}^\prime x} = x'\left( {\sum\limits_{i = 1}^n {{\alpha _i}{\alpha _i}^\prime } } \right)x\]
其中${{\alpha _i} = {{\left( {{a_{i1}}, \cdots ,{a_{in}}} \right)}^\prime }}$，$x = {\left( {{x_1}, \cdots ,{x_n}} \right)^\prime }$，从而$f$的矩阵为
\[\sum\limits_{i = 1}^n {{\alpha _i}{\alpha _i}^\prime } = \left( {{\alpha _1}, \cdots ,{\alpha _n}} \right)\left( {\begin{array}{*{20}{c}}
{{\alpha _1}^\prime }\\
\vdots \\
{{\alpha _n}^\prime }
\end{array}} \right) = A'A\]
而$r\left( {A'A} \right) = r\left( A \right)$，故即证