二次型的矩阵表达

令

$$a_{ij}=a_{ji}\left(i<j\right)$$

则 $2a_{ij}{x}_i{x}_j=a_{ij}{x}_i{x}_j+a_{ji}{x}_j{x}_i$. 于是,上述表达式变为

$$\begin{array}{rl}& f\left(x_1,x_2,\cdots ,x_n\right)\\ =& a_{11}{x}_1^2+a_{12}{x}_1{x}_2+\cdots +a_{1n}{x}_1{x}_n\\ & +a_{21}{x}_2{x}_1+a_{22}{x}_2^2+\cdots +a_{2n}{x}_2{x}_n+\cdots \\ & +a_{n1}{x}_n{x}_1+a_{n2}{x}_n{x}_2+\cdots +a_{nn}{x}_n^2\\ =& x_1\left(a_{11}{x}_1+a_{12}{x}_2+\cdots +a_{1n}{x}_n\right)+\\ & x_2\left(a_{21}{x}_1+a_{22}{x}_2+\cdots +a_{2n}{x}_n\right)+\cdots +\\ & x_n\left(a_{n1}{x}_1+a_{n2}{x}_2+\cdots +a_{nn}{x}_n\right)\\ =& \left(x_1,x_2,\cdots ,x_n\right)\left(\begin{array}{c}{a}_{11}{x}_1+a_{12}{x}_2+\cdots +a_{1n}{x}_n\\ a_{21}{x}_1+a_{22}{x}_2+\cdots +a_{2n}{x}_n\\ ⋮\\ a_{n1}{x}_1+a_{n2}{x}_2+\cdots +a_{nn}{x}_n\end{array}\right)\\ =& \left(x_1,x_2,\cdots ,x_n\right)\left(\begin{array}{cccc}{a}_{11}& a_{12}& \cdots & a_{1n}\\ a_{21}& a_{22}& \cdots & a_{2n}\\ ⋮& ⋮& & ⋮\\ a_{n1}& a_{n2}& \cdots & a_{nn}\end{array}\right)\left(\begin{array}{c}{x}_1\\ x_2\\ ⋮\\ x_n\end{array}\right)\end{array}$$

其中矩阵

$$\mathbf{A}=\left(\begin{array}{cccc}{a}_{11}& a_{12}& \cdots & a_{1n}\\ a_{21}& a_{22}& \cdots & a_{2n}\\ ⋮& ⋮& & ⋮\\ a_{n1}& a_{n2}& \cdots & a_{nn}\end{array}\right)$$

为对称矩阵. 若记

$$\mathbf{X}=\left(\begin{array}{c}{x}_1\\ x_2\\ ⋮\\ x_n\end{array}\right)$$

则二次型可简写成

$$f\left(x_1,x_2,\cdots ,x_n\right)={\mathbf{X}}^{\mathrm{T}}\mathbf{AX}.$$

由于 $a_{ij}=a_{ji}$,所以矩阵 $\mathbf{A}$ 是由二次型唯一确定的对称矩阵, 称为二次型 $f\left(x_1,x_2,\cdots ,x_n\right)$ 的矩阵. 反过来, 有了对称矩阵 $\mathbf{A}$, 二次型 $f\left(x_1,x_2,\cdots ,x_n\right)=$ ${\mathbf{X}}^{\mathrm{T}}\mathbf{AX}$ 也就唯一确定.