二次型:含有n个变量x1,x2,...xnx_1,x_2,...x_nx1,x2,...xn的二次齐次函数:
f(x1,x2,...xn)=a11x12+a12x1x2+a13x1x3+a14x1x4...+a1nx1xnf(x_1,x_2,...x_n)=a_{11}x_1^2+a_{12}x_1x_2+a_{13}x_1x_3+a_{14}x_1x_4...+a_{1n}x_1x_nf(x1,x2,...xn)=a11x12+a12x1x2+a13x1x3+a14x1x4...+a1nx1xn
+a21x2x1+a22x22+a23x1x3+a24x2x4...+a2nx2xn+a_{21}x_2x_1+a_{22}x_2^2+a_{23}x_1x_3+a_{24}x_2x_4...+a_{2n}x_2x_n+a21x2x1+a22x22+a23x1x3+a24x2x4...+a2nx2xn
…
+an1xnx1+an2xnx2+an3xnx3+an4xnx4...+annxn2+a_{n1}x_nx_1+a_{n2}x_nx_2+a_{n3}x_nx_3+a_{n4}x_nx_4...+a_{nn}x_n^2+an1xnx1+an2xnx2+an3xnx3+an4xnx4...+annxn2
称为二次型。 二次型可以用矩阵表示记作:
f=XTAXf=X^TAXf=XTAX
其中X=[x1x2...xn]X=\begin{bmatrix}
x_1\\
x_2\\
...\\x_n
\end{bmatrix}X=⎣⎢⎢⎡x1x2...xn⎦⎥⎥⎤, A=[a11a12...a1na21a22...a2n...an1an2...ann]A=\begin{bmatrix}
a_{11}&a_{12}&...&a_{1n}\\
a_{21}&a_{22}&...&a_{2n}\\
...\\a_{n1}&a_{n2}&...&a_{nn}
\end{bmatrix}A=⎣⎢⎢⎡a11a21...an1a12a22an2.........a1na2nann⎦⎥⎥⎤,可以通过先计算AXAXAX,再计算XTAXX^TAXXTAX来证明:
AX=[a11x1+a12x2+...+a1nxna21x1+a22x2+...+a2nxn...an1x1+an2x2+...+annxn]AX=\begin{bmatrix}
a_{11}x_1+a_{12}x_2+...+a_{1n}x_n\\
a_{21}x_1+a_{22}x_2+...+a_{2n}x_n\\
...\\a_{n1}x_1+a_{n2}x_2+...+a_{nn}x_n
\end{bmatrix}AX=⎣⎢⎢⎡a11x1+a12x2+...+a1nxna21x1+a22x2+...+a2nxn...an1x1+an2x2+...+annxn⎦⎥⎥⎤在左乘XTX^TXT得到原式:
[x1x2...xn][a11x1+a12x2+...+a1nxna21x1+a22x2+...+a2nxn...an1x1+an2x2+...+annxn]\begin{bmatrix}
x_1\\
x_2\\
...\\x_n
\end{bmatrix}\begin{bmatrix}
a_{11}x_1+a_{12}x_2+...+a_{1n}x_n\\
a_{21}x_1+a_{22}x_2+...+a_{2n}x_n\\
...\\a_{n1}x_1+a_{n2}x_2+...+a_{nn}x_n
\end{bmatrix}⎣⎢⎢⎡x1x2...xn⎦⎥⎥⎤⎣⎢⎢⎡a11x1+a12x2+...+a1nxna21x1+a22x2+...+a2nxn...an1x1+an2x2+...+annxn⎦⎥⎥⎤
我们来求一下二次型的导数:
df(x1,x2,...xn)dx1=2a11x1+a12x2+a13x3+a14x4...+a1nxn\frac{df(x_1,x_2,...x_n)}{dx_1}=2a_{11}x_1+a_{12}x_2+a_{13}x_3+a_{14}x_4...+a_{1n}x_ndx1df(x1,x2,...xn)=2a11x1+a12x2+a13x3+a14x4...+a1nxn
+a21x2+0+0+0...+0+a_{21}x_2+0+0+0...+0+a21x2+0+0+0...+0
…
+an1xn+0+0+0...+0+a_{n1}x_n+0+0+0...+0+an1xn+0+0+0...+0
=2a11x1+(a12+a21)x2+(a13+a31)x3+(a14+a41)x4...+(a1n+an1)xn=2a_{11}x_1+(a_{12}+a_{21})x_2+(a_{13}+a_{31})x_3+(a_{14}+a_{41})x_4...+(a_{1n}+a_{n1})x_n=2a11x1+(a12+a21)x2+(a13+a31)x3+(a14+a41)x4...+(a1n+an1)xn,其余的还有对x1,x2...xnx_1,x_2...x_nx1,x2...xn的求导。
正定二次型:设二次型f=XTAXf=X^TAXf=XTAX,如果对任何X≠0X\neq0X̸=0,都有f(X)>0f(X)>0f(X)>0,则称fff为正定二次型,并称对称矩阵AAA是正定的。
由于AAA是对称矩阵,所以其导数为:
df(x1,x2,...xn)dx1=2a11x1+2a12x2+2a13x3+2a14x4...+2a1nxn\frac{df(x_1,x_2,...x_n)}{dx_1}=2a_{11}x_1+2a_{12}x_2+2a_{13}x_3+2a_{14}x_4...+2a_{1n}x_ndx1df(x1,x2,...xn)=2a11x1+2a12x2+2a13x3+2a14x4...+2a1nxn
df(x1,x2,...xn)dx2=2a21x1+2a22x2+2a23x3+2a24x4...+2a2nxn\frac{df(x_1,x_2,...x_n)}{dx_2}=2a_{21}x_1+2a_{22}x_2+2a_{23}x_3+2a_{24}x_4...+2a_{2n}x_ndx2df(x1,x2,...xn)=2a21x1+2a22x2+2a23x3+2a24x4...+2a2nxn
............
所以▽f(X)=2AX\bigtriangledown f(X)=2AX▽f(X)=2AX
举个例子:
其对其求导可得矩阵:
▽f(X)=[2x−4y−4x+z...−6z+y]\bigtriangledown f(X)=\begin{bmatrix}
2x-4y\\
-4x+z\\
...\\-6z+y
\end{bmatrix}▽f(X)=⎣⎢⎢⎡2x−4y−4x+z...−6z+y⎦⎥⎥⎤