91影视

$q\left(x\right)=x^{T}Ax$for all $x鈭�R$is a quadratic form where $A={\left[a_{ij}\right]}_{n脳n}$is a symmetric matrix.

$e_i=(0,0,鈥�,0,1,0,鈥�,0{)}^T鈭�R^n$is a vector with 1 at i-th coordinate and 0 elsewhere. For $e_i鈭�R^n$, we have

$$\begin{gathered} q\left(e_i\right)=e_i^{T}A{e}_i \\ =(0,0,鈥�,0,1,0,鈥�,0{)}^{T}A(0,0,鈥�,0,1,0,鈥�,0) \end{gathered}$$

$$\begin{gathered} =\left[0....1_{\left(i\right)}....0\right]\left[\begin{array}{c}{a}_1\\ a_2\\ .\\ .\\ .\\ a_n\end{array}\right]\left[\begin{array}{c}0\\ ..\\ ..\\ 1_{\left(i\right)}\\ ..\\ 0\end{array}\right]1(where{a}_{i}isthei-throwofA) \\ =a_i\left[\begin{array}{c}0\\ ..\\ ..\\ 1_{\left(i\right)}\\ ..\\ 0\\ \end{array}\right] \\ =\left[a_{i1}{a}_{i2}....a_{ii}....a_{in}\right]\left[\begin{array}{c}0\\ ..\\ ..\\ 1_{\left(i\right)}\\ ..\\ 0\\ \end{array}\right] \\ =a_{i1}脳0+a_{i2}脳0+L+a_{ii}脳1+L+a_{in}脳0 \\ =a_{ii} \end{gathered}$$

Now, for$1猢�i,j猢�n,i鈮�j$

$x^T={\left(e_i+e_j\right)}^T=\left[0...1_{\left(i\right)}....0\right]+\left[0...1_{\left(j\right)}....0\right]=\left[0....1_{\left(i\right)}....01_{\left(j\right)}....0\right]$

Thus, we have

$q\left(e_i+e_j\right)={\left(e_i+e_j\right)}^{T}A\left(e_i+e_j\right)$

$=\left[0.....1_{\left(i\right)}....01_{\left(j\right)}.....0\right]A\left[\begin{array}{c}0\\ ..\\ 1_{\left(i\right)}\\ 0\\ 1_{\left(j\right)}\\ 0\end{array}\right]$$$\begin{gathered} =\left[0.....1_{\left(i\right)}....01_{\left(j\right)}.....0\right]A\left[\begin{array}{c}{a}_1\\ a_2\\ .\\ .\\ a_n\end{array}\right]\left[\begin{array}{c}0\\ ..\\ 1_{\left(i\right)}\\ 0\\ 1_{\left(j\right)}\\ 0\end{array}\right]\left(where{a}_{i}isthei-throwofA\right) \\ =(a_i+a_j)\left[\begin{array}{c}0\\ .\\ 1_{\left(i\right)}\\ 0\\ 1_{\left(j\right)}\\ ..\\ 0\end{array}\right] \end{gathered}$$

$=\left[a_{i1}+a_{j1}{a}_{i2}+a_{j2}.....a_{ii}+a_{jj}....a_{in}+a_{jn}\right]\left[\begin{array}{c}0\\ .\\ 1_{\left(i\right)}\\ 0\\ 1_{\left(j\right)}\\ ..\\ 0\end{array}\right]$

$$\begin{gathered} =\left(a_{i1}+a_{j1}\right)脳0+...+\left(a_{ii}+a_{ji}\right)脳1+...+\left(a_{ij}+a_{jj}\right)脳1+...+\left(a_{in}+a_{jn}\right)脳0 \\ =a_{ii}+a_{ji}+a_{ij}+a_{jj}......\left(2\right) \end{gathered}$$

Since A is symmetric matrix, thus, form $1猢�i,j猢�n,i鈮�j,$we have$a_{ij}=a_{ji}$ and from (1) and (2), we get

$a_{ij}=a_{ji}=\frac{1}{2}\left[q\left(e_i+e_j\right)-a_{ii}-a_{jj}\right]=\frac{1}{2}\left[q\left(e_i+e_j\right)-q\left(e_i\right)-q\left(e_j\right)\right]$

simple matrix multiplication is used to prove the point.