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$f命题：$设实二次型[fleft( {{x_1}, cdots ,{x_n}} ight) = sumlimits_{i = 1}^n {{{left( {{a_{i1}}{x_1} + cdots + {a_{in}}{x_n}} ight)}^2}} ]

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

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