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?
Best Answer
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 $