The question is:
Let $V$ be a $\mathbb{R}$-vector space and dim$V$=n. It has an inner product $(,)$.
Let $\alpha_1,\alpha_2,\cdots,\alpha_m\in V$.
They satisfy that if there exists non-negative real numbers $\lambda_1,\lambda_2,\cdots ,\lambda_m$ such that $\lambda_1\alpha_1+\lambda_2\alpha_2+\cdots+\lambda_m\alpha_m=0$, it must be $\lambda_1=\lambda_2=\cdots=\lambda_m=0$.
I want to proof that there exists $\alpha\in V$ such that $(\alpha,\alpha_i)>0,\forall~1\le i\le m$.
My attempt is :
Use induction on m.
If m=1, it is trivial.
Suppose when m-1,it holds.So there exists $\beta_1$ such that $ (\beta_1,\alpha_j)>0,\forall~1\le j\le m-1$.
Let W=span{$\alpha_1,\alpha_2,\cdots,\alpha_{m-1}$}, W is closed .${\alpha_m}$ is a convex set and compact.
Using the Hahn-Banach theorem, there exists $\beta_2$ such that $(\beta_2,\alpha_m)>0$ and $(\beta_2,\alpha_j)=0,\forall 1\le j\le m-1$.
Then considering $\beta_2+\beta_1$. If necessary, we can multiply $\beta_2$ by a positive coefficient.
But this is wrong, because $W\cap$ {$\alpha_m$} may not be empty.
I want to solve the problem or get some hints.
Thanks!