49886
2024-07-02 20:24:36 223人阅读
$\bf命题:$设实二次型
f(x1,?,xn)=∑i=1n(ai1x1+?+ainxn)2 <script id="MathJax-Element-1" type="math/tex; mode=display">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}} </script>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(x1,?,xn)=∑i=1nx′αiαi′x=x′(∑i=1nαiαi′)x <script id="MathJax-Element-2" type="math/tex; mode=display">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</script>
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
∑i=1nαiαi′=(α1,?,αn)???α1′?αn′???=A′A <script id="MathJax-Element-3" type="math/tex; mode=display">\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</script>
而$r\left( {A‘A} \right) = r\left( A \right)$,故即证
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