Intersection of positive half-spaces Let $V$ be a Euclidean space with inner product $(\cdot,\cdot)$. Let $\{\eta_i\}_{i=1}^{n}$ be a basis for $V$ , $n$ is the dimension of $V$. I want to prove that there exists a vector $t \in V$ such that $(t,\eta_i)>0$ for all $i.$ I got an idea that if I define
$$\displaystyle\delta_i=\eta_i-\sum\limits_{\substack{1\leq j\leq n\\j\neq i}}\frac{(\eta_i,\eta_j)}{(\eta_j,\eta_j)}\eta_j$$
then vector $t = \sum_{i=1}^nr_i\delta_i$ for each $r_i>0$ would solve the problem. However, the only case I can prove the problem in this approach is for $n=2$. Does this still work for $n \ge 3$? Any help would be appreciated.
 A: If the construction worked for all $n$-tuples of positive real numbers, like it does when $n = 2$, then by continuity we would necessarily have $(\delta_i,\eta_k) \geqslant 0$ for all $i,k$. But for $i \neq k$, we find
$$(\delta_i,\eta_k) = (\eta_i,\eta_k) - \sum_{j\neq i} \frac{(\eta_i,\eta_j)}{(\eta_j,\eta_j)}(\eta_j,\eta_k) = - \sum_{j\notin \{i,k\}}\frac{(\eta_i,\eta_j)}{(\eta_j,\eta_j)}(\eta_j,\eta_k),$$
and that can be negative. So, for $n \geqslant 3$, you generally cannot pick arbitrary $r_i > 0$ to have $t = \sum_{i=1}^n r_i\delta_i$ satisfy $(t,\eta_i) > 0$ for all $i$. It may be possible that you can always pick some $n$-tuple $(r_1,\dotsc,r_n)$ of positive numbers such that the condition holds, but showing that - if it holds - seems to be complicated.
There are however simple ways to prove the existence of such $t\in V$. For example, you can use the dual basis to $\{ \eta_i\}_{i=1}^n$ to construct such a $t$. Or you can look at the linear map $A\colon V \to \mathbb{R}^n$ (or to $\mathbb{C}^n$ if you consider a complex vector space $V$) given by
$$A(x) = \begin{pmatrix} (x,\eta_1) \\ \vdots \\ (x,\eta_n)\end{pmatrix}.$$
Since $\{\eta_i\}_{i=1}^n$ is a basis of $V$, we have $\ker A = \{0\}$, and by the rank formula it follows that $A$ is surjective, hence bijective. Then we can take for example
$$t = A^{-1}\begin{pmatrix} 1\\1\\\vdots\\1\end{pmatrix}.$$
