The axioms that define an inner product in a vector space $V$ are
Positivity $\langle v,v\rangle\ge 0$ for all $v\in V$
Definiteness $\langle v,v\rangle=0\iff v=0$
Additivity in first slot $\langle u+v,w\rangle=\langle u,w\rangle+\langle v,w\rangle$ for all $u,v,w\in V$
Homogeneity in first slot $\langle\lambda u,v\rangle=\lambda\langle u,v\rangle$ for all $\lambda\in\Bbb F$
Conjugate symmetry $\langle u,v\rangle=\overline{\langle v,u\rangle}$ for all $u,v\in V$
Now the exercise ask
Prove that for a non-zero real vector space if we change the first axiom (positivity) by
- Exists some $v\in V$ such that $\langle v,v\rangle>0$
then the set of functions that define an inner product on $V$ is the same.
Im completely stuck in this exercise. I tried to prove it by contradiction: WLOG suppose that exists two linearly independent $x,y\in V$ with $\langle x,x\rangle >0$ and $\langle y,y\rangle<0$ such that
$$\langle x,x\rangle+\langle y,y\rangle=0$$
But from here Im unable to show a contradiction. Some help will be appreciated, thank you.
Best Answer
If you have linearly independent vectors $x$ and $y$ with $\langle x,x\rangle>0>\langle y,y\rangle$, consider $\langle z,z\rangle$ for $z$ on the line segment joining $x$ to $y$.