# Thread: linear functional inner product

1. ## linear functional inner product

Consider the vector space R^2 with the inner product given by:
<(x1,x2),(y1,y2)>=2x1y1-x2y1-x1y1+8x2y2.(note: don't verify this is an inner product)
Given the functional fi:R^2 --> R, fi(x1,x2)=6x1-x2, we know that there is a unique vector v=(y1,y2) such that fi(u)=<u,v> for every u=(x1,x2) in R^2.
Find the vector v. (recall you can verify the correctness if you want)

Now this is what I have done but i'm not sure if its right:
Since (x1,x2)=6x1-x2 in R, then : <(x1,x2),(6,-1)>=6x1-x2 therefore our
v=(6,-1) in R, now I want to find v in R^2 so: <(x1,x2),(y1,y2)>=x1y1+x2y2 = 2x1y1-x2y1-x1y2+8x2y2
then 0=x1y1-x2y1-x1y2+7x2y2= x1(y1-y2)+x2(-y1+7y2)
<(x1,x2),(y1-y2,-y1+7y2)>=(plug in 6,-1: (6-(-1),-6+7(-1))=(7,-13) in R^2. done
Is this correct? any help would be great. thank you!

2. Originally Posted by jax
Consider the vector space R^2 with the inner product given by:
<(x1,x2),(y1,y2)>=2x1y1-x2y1-x1y1+8x2y2.
That is not an inner product. I suppose:

$<(x_1,x_2),(y_1,y_2)>=2x_1y_1-x_2y_1-x_1y_2+8x_2y_2=$

$(x_1,x_2) \begin{pmatrix}{2}&{-1}\\{-1}&{8}\end{pmatrix}\begin{pmatrix}{y_1}\\{y_2}\end {pmatrix}=x^tGy\;,\quad x=\begin{pmatrix}{x_1}\\{x_2}\end{pmatrix},\;y=\be gin{pmatrix}{y_1}\\{y_2}\end{pmatrix}$

... then : <(x1,x2),(6,-1)>=6x1-x2
I don't know why do you use the usual inner product. You have to find a vector $y$ such that $f(x)=$ for all $x\in \mathbb{R}^2$. But

$f(x)=6x_1-x_2=(x_1,x_2)\begin{pmatrix}{6}\\{-1}\end{pmatrix}=x^tb\;,\quad b=\begin{pmatrix}{6}\\{-1}\end{pmatrix}$

Then,

$f(x)=\Leftrightarrow x^tb=x^tGy\Leftrightarrow x^t(b-Gy)=0$.

and this for all $x\in \mathbb{R}^2$ so necessarily $b-Gy=0$

Equivalently,

$\begin{pmatrix}{2}&{-1}\\{-1}&{8}\end{pmatrix}\begin{pmatrix}{y_1}\\{y_2}\end {pmatrix}=\begin{pmatrix}{6}\\{-1}\end{pmatrix}$

Now, solve the system.

Regards.