Answer to Question #121749 in Abstract Algebra for Henry

Question

Let f be a degree two polynomial over R in n-variables and a = (a1, . . . , an) a
critical point of f. Show that there is a homogenous degree two polynomial Q
such that
f(x) − f(a) = Q(x − a) (x ∈ Rn
)

Answer 1

Consider a polynomial of two variables $x_1, x_2$

$f(x_1,x_2)=c_{11}x_1^2+c_{22}x_2^2+c_{12}x_1x_2+\\ +c_1x_1+c_2x_2+c$

$a=(a_1,a_2)$ - critical point of $f$ :

$\frac{\partial f(a)}{\partial x_1}=2c_{11}a_1+c_{12}a_2+c_1=0\\ \frac{\partial f(a)}{\partial x_2}=2c_{22}a_2+c_{12}a_1+c_2=0$

Consider a polynomial

$f(x)-f(a)=c_{11}x_1^2+c_{22}x_2^2+c_{12}x_1x_2+\\ +c_1x_1+c_2x_2+c-c_{11}a_1^2-c_{22}a_2^2-c_{12}a_1a_2-\\ -c_1a_1-c_2a_2-c=\\ =c_{11}(x_1-a_1)(x_1+a_1)+\\ +c_{22}(x_2-a_2)(x_2+a_2)+\\ +c_{12}(x_1x_2-a_1x_2+a_1x_2-a_1a_2)+\\ +c_1(x_1-a_1)+c_2(x_2-a_2)=\\ =c_{11}(x_1-a_1)(x_1+a_1)+\\ +c_{22}(x_2-a_2)(x_2+a_2)+\\ +c_{12}x_2(x_1-a_1)+c_{12}a_1(x_2-a_2)+\\ +c_1(x_1-a_1)+c_2(x_2-a_2)=\\ =(x_1-a_1)\cdot\\ \cdot(c_{11}(x_1+a_1)+c_{12}x_2+c_1)+\\ +(x_2-a_2)\cdot\\ \cdot(c_{22}x_2(x_2+a_2)+c_{12}a_1+c_2)$

Consider

$c_{11}(x_1+a_1)+c_{12}x_2+c_1=\\ =c_{11}x_1+c_{12}x_2-c_{11}a_1-c_{12}a_2=\\ =c_{11}(x_1-a_1)+c_{12}(x_2-a_2)\\ c_{22}x_2(x_2+a_2)+c_{12}a_1+c_2=\\ =c_{22}x_2-c_{22}a_2=c_{22}(x_2-a_2)$

Then

$f(x)-f(a)=(x_1-a_1)\cdot\\ \cdot(c_{11}(x_1-a_1)+c_{12}(x_2-a_2))+\\ +(x_2-a_2)\cdot\\ \cdot c_{22}(x_2-a_2)=\\ =c_{11}(x_1-a_1)^2+\\ +c_{12}(x_1-a_1)(x_2-a_2)+\\ + c_{22}(x_2-a_2)^2$

There is a homogenous degree two

Consider a polynomial of variables $x_1,x_2,...,x_n$

$f=\sum\limits_{i,j=1}^{n}c_{ij}x_ix_j+\sum\limits_{i=1}^{n}c_ix_i+c$

$a=(a_1,...,a_n)$ - - critical point of $f$ :

$\frac{\partial f(a)}{\partial x_1}=2c_{11}a_1+\\ +c_{12}a_2+...+c_{1n}a_n+c_1=0\\ ...\\ \frac{\partial f(a)}{\partial x_n}=2c_{nn}a_n+\\ +c_{n1}a_1+...+c_{n,n-1}a_{n-1}+c_n=0$

$f(x)-f(a)=\\ =\sum\limits_{i,j=1}^{n}c_{ij}x_ix_j+\sum\limits_{i=1}^{n}c_ix_i+c-\\ -\sum\limits_{i,j=1}^{n}c_{ij}a_ia_j-\sum\limits_{i=1}^{n}c_ia_i-c=\\ =\sum\limits_{i,j=1}^{n}c_{ij}(x_i-a_i)(x_j-a_j)$

There is a homogenous degree two