1. ## simetric function problem

5)there is $\displaystyle f:R^{n}xR^{n}->R$ which is defined by $\displaystyle f(x,y)=n\sum_{i=1}^{n}(x_{i}-m_{x})(y_{i}-m_{y})$ where
$\displaystyle m_{y}=\frac{y_{1}+..+y_{n}}{n}$
$\displaystyle m_{x}=\frac{x_{1}+..+x_{n}}{n}$
$\displaystyle x=(x_{1}+..+x_{n}) y=(y_{1}+..+y_{n})$
A)check if this function is a simetric bilenear.
B)find the representative matrices by the standart basis of $\displaystyle R^{n}$
C)find the jordan form of the matrices you found in part B
D)check if $\displaystyle n||x||^{2}>=f(x,x)>=0$ for every $\displaystyle x\in R^{n}$
??
regarding A:
f is simetric if f(x,y)=f(y,x) so i need to prove that
$\displaystyle f(x,y)=n\sum_{i=1}^{n}(x_{i}-m_{x})(y_{i}-m_{y})=n\sum_{i=1}^{n}(y_{i}-m_{y})(x_{i}-m_{x})=f(y,x)$
but when i put the given $\displaystyle m_{y}$and $\displaystyle m_{x}$ i get two different expressions
?

2. ## Re: simetric function problem

a) since each $\displaystyle x_i-m_x,y_i-m_y$ is a real number, isn't it true that $\displaystyle (x_i-m_x)(y_i-m_y) = (y_i-m_y)(x_i-m_x)$ ? you also need to check linearity in x and y.

3. ## Re: simetric function problem

i made a typing mistake in the questions
$\displaystyle x=(x_{1},..,x_{n})$ $\displaystyle y=(y_{1},..,y_{n})$

ok i understand part A
regarding B)

for standart basis (1,0..,0) .. (0,0..,1)

so for each vector in this basis

$\displaystyle m_{y}=\frac{0+..+1+..0+0}{n}$ $\displaystyle m_{x}=\frac{0+..+1+..0+0}{n}$ and

$\displaystyle y_{i}=x_{j}=1$

$\displaystyle f(e_i,e_j)=n\sum_{i=1}^{n}(1-\frac{1}{n})(1-\frac{1}{n})=n\sum_{i=1}^{n}(\frac{n-1}{n})^{2}=n^{2}(\frac{n-1}{n})^{2}=(n-2)^{2}$
in transformations each column of the representation matrices would be T(e1)
what to do here

4. ## Re: simetric function problem

first of all, your sum isn't correct. if you want to calculate f(ei,ej) you should sum over some other index, like k (because k can be anything from 1 to n, but i and j are "constants").

for k ≠ i,j, we have the term 1/n^2, for k = i or j, we have the term (1 - 1/n)(-1/n) = (1-n)/n^2 (when i ≠ j), so that f(ei,ej) = n[(n-2)/n^2 + (2-2n)/n^2] = -n/n = -1.

when i = j, we have n[(n-1)/n^2 + (n-1)^2/n^2] = n-1. so for n = 2, for example, you should get the matrix:

[1 -1]
[-1 1].