Inner product space - Linear algebra 2 - Positive attribute

Question

Lets say I have:
$<(a_1,a_2),(b_1,b_2)>$ = $2a_1a_2 + 4b_1b_2$
I am going to check positive in order to refute Inner product space
I get $2a_1a_2+4a_1a_2$ = $6a_1a_2$
Now, can I say that if $a_1<0$ and $a_2>0$ so I get that refute? I am stuck an hour trying to think about it.
Thanks.

Answer 1

Let $V$ be the usual real vector space of ordered pairs of real numbers. Your reasoning shows that the mapping $(\alpha,\beta)=((a_1,a_2),(b_1,b_2)) \mapsto 2a_1a_2 + 4b_1b_2$ cannot be an inner product for this space. It cannot be, since the positivity axiom (axiom number four) fails.

All the axioms are listed here: https://math.stackexchange.com/tags/inner-products/info

But the relevant one here is:

4. $\quad (\forall v\in V):\langle v,v\rangle\geqslant0$ and $\langle v,v\rangle=0\iff v=0$.