0
$\begingroup$

How can I show that these two formulas for $\langle v,w \rangle$ might define inner products on $\Bbb R^2$?

1.) $v_1 w_2 + v_2 w_1$

2.) $2v_1 w_1 + (v_1-v_2)(w_1-w_2)$

I know that for number 1 it does not define an inner product and for number 2 it does define an inner product but why? I also know in order to define an inner product it must satisfy the conditions of bilinearity, symmetry, and positivity but I am confused on how I can show that?

  • 0
    I mean $v_1w_2+v_2w_1 $ mean $v_1.w_2+v_2.w_1$2012-09-23

2 Answers 2

2

1) let $(v,w) \equiv v_1 w_2 + v_2 w_1$, then let check bilinearity $(\alpha v, w) = \alpha v_1 w_2 + \alpha v_2 w_1 = \alpha (v_1 w_2 + v_2 w_1) = \alpha (v,w)$ $(v, \beta w) = v_1 \beta w_2 + v_2 \beta w_1 = \beta (v_1 w_2 + v_2 w_1) = \beta (v,w)$ $(v+q,w) = v_1 w_2 + q_1 w_2 + w_1 v_2 + w_1 q_2 = (v,w) + (q,w)$ and so on

2) symmetry $(v,w) = v_1 w_2 + v_2 w_1 = w_1 v_2 + w_2 v_1 = (w,v)$

3) positivity $(v,v) = v_1 v_2 + v_2 v_1 = 2 v_1 v_2$ which is not always $>0$. your definition failed here.

in contrast:

1) let $(v,w) \equiv 2 v_1 w_1 + (v_1 -v_2)(w_1-w_2) $, positivity $(v,v) = 2 v_1 v_1 + (v_1 - v_2)(v_1-v_2) = 2 v_1^2 + (v_1-v_2)^2 > 0$ for any $v \ne 0 $

  • 0
    I do not know what a metric tensor is but nonetheless thank you very much!!!2012-09-23
1
  1. What is $< x,x>$ where $x=(1,-1)\in \mathbb{R}^2$?
  2. i). $=2x_1^2+(x_1-x_2)^2\ge 0\quad\forall x=(x_1,x_2)\in\mathbb{R}^2$. Also note that $=0 \Leftrightarrow x=0$.

ii)$=$, as multiplication of real numbers is commutative.

iii)$=2u_1(v_1+aw_1)+(u_1-u_2)(v_1+aw_1-v_2-aw_2)=2u_1v_1+(u_1-u_2)(v_1-v_2)+a[2u_1w_1+(u_1-u_2)(w_1-w_2)=+a$

  • 0
    Yup, I understand now thank you very much!2012-09-23