inner product not by standart basis

we choose a basis B={v1=(1,0)v2=(1,1)} from $\displaystyle R^2$

and define $\displaystyle (x,y)_B=x_1y_1+x_2y_2$

why does the prof say that

$\displaystyle x=x_1v_1+x_2v_2$

$\displaystyle y=y_1v_1+y_2v_2$

how did he get those formulas?

and i cant se the structure of our chosen basis here?

how to see it in thos formulas

?

Re: inner product not by standart basis

Quote:

Originally Posted by

**transgalactic** we choose a basis B={v1=(1,0)v2=(1,1)} from $\displaystyle R^2$ and define $\displaystyle (x,y)_B=x_1y_1+x_2y_2$ why does the prof say that $\displaystyle x=x_1v_1+x_2v_2$ $\displaystyle y=y_1v_1+y_2v_2$

Although I don't know what is the final question, taking into account that $\displaystyle B$ is a basis of $\displaystyle \mathbb{R}^2$ , every $\displaystyle x\in \mathbb{R}^2$ can be uniquely expressed in the way $\displaystyle x=x_1v_1+x_2v_2$ . Same considerations for $\displaystyle y$ .

Re: inner product not by standart basis

if x=(x1,x2) then x could be represented as $\displaystyle x=x_1v_1+x_2v_2$

only if v1=(1,0) v2=(0,1)

but here we have a different basis not a standart one.

so our coefficient is not x1 x2

??

Re: inner product not by standart basis

Perhaps this simple example will help you. Consider $\displaystyle x=(5,2)$ and $\displaystyle y=(2,1)$ . Prove that $\displaystyle x=3v_1+2v_2$ and $\displaystyle y=1v_1+1v_2$ . According to the given definition of the inner product we have $\displaystyle (x|y)=3\cdot 1+2\cdot 1=5$ .

Re: inner product not by standart basis

ok so if i will go by your example

(x,y)_B=x1y1+x2y2=5*2+2*1=12

x=(5,2)=5(1,0)+2(1,1)=(7,2)

i try to follow the logic but i cant understand this term

Re: inner product not by standart basis

Quote:

Originally Posted by

**transgalactic** x=(5,2)=5(1,0)+2(1,1)=(7,2)

$\displaystyle (5,2)\neq (7,2)$

Re: inner product not by standart basis

i know

i got here some thing wrong

just cant understand this basis of inner product

i am looking for a tutorial with basic examples