Does a positive linear combination of $n$ linearly independent vectors always have an acute angle with at least one vector? (edit: I seem to have caused some confusion by calling my vectors basis vectors, which (I suppose) implied that they are orthonormal.)
Does a linear combination of $n$ linealy independent vectors (not necesserily orthogonal) with only non-negative coefficients always have an acute angle with at least one of the vectors?
I am writing my bachelor's thesis about simple Lie algebras and root systems and I needed this quick linear algebra statement. It seems really obvious (in for example 2 or 3 dimensions), but I cannot seem to find its proof. I was thinking of induction to the dimension $n$, but did not manage to complete a proof. I feel a bit embarrassed to no be able to find the solution to such a simple problem, but perhaps it is less obvious than I think. Any help is appreciated!
 A: I don't know if anyone cares, but I found the proof with the help of my supervisor.
Let $(p_i)_{1\leq i\leq n}$ be the linearly indpendent vectors. Let $r$ is a positive linear combination of them, say $r=\sum_{i=1}^n\lambda_ip_i$ with $\lambda_i\geq 0$. 
Instead, suppose that $(r,p_i)\leq 0$ for all $p_i$. From linear algebra we know that a basis $(p_i^*)_{1\leq i \leq n}$, called the dual basis, exists such that $(p_i^*,p_j)=\delta_{ij} \ \forall_{1\leq i,j\leq n}$. In this basis we have $r=\sum_{i=1}^n\mu_ip_i^*$. Now for all $j$ $0\geq (r,p_j)=(\sum_{i=1}^n\mu_ip_i^*,p_j)=\sum_{i=1}^n\mu_i\delta_{ij}=\mu_j.$ At the same time $0<(r,r)=(\sum_{i=1}^n\mu_ip_i^*,\sum_{j=1}^n\lambda_jp_j)=\sum_{i=1}^n\sum_{j=1}^n\mu_i\lambda_j(p_i^*,p_j)=\sum_{i=1}^n\mu_i\lambda_i.$ This obviously impossible as $\mu_i\lambda_i\leq 0$ for all $i$. So by contradiction there must have been a $p_i$ such that $(r,p_i)>0$.
A: One strategy to show this would be as follows -- I'm assuming acute means strictly between $-\pi/2$ and $\pi/2$. Firstly let's assume everything is normalized ($a \neq 0$ otherwise `angle' is not defined) -- norm wont affect the angle. Let $\{x_1,x_2, \dots, x_n\}$ be an orthonormal basis and let $a = \sum_i c_i x_i$ where $c_i \geq 0$ and are such that $a$ has unit norm. 
Now consider the angle between $a$ and $x_1$, i.e., $\theta_1 = \mathrm{arccos}(a \cdot x_1) = \mathrm{arccos}(c_1)$. Notice that as $0 \leq c_1 \leq 1$ we must have $\theta_1 \in [-\pi/2, \pi/2]$ and $\theta_1 = \pi/2$ or $-\pi/2$ only when $c_1 = 0$. Therefore if we want $a$ to not have an acute angle with the basis vector $x_1$ then we must have $c_1 = 0$. 
Repeating this argument for all of the other basis vectors we find that we have to set each $c_i = 0$ in order to avoid forming an acute angle with the basis vector $x_i$. But then we must have $a=0$ and so we can't find a vector which doesn't have an acute angle with at least one of the basis vectors.
A: Yes. Suppose $e_1,\ldots,e_n$ are basis vectors and $v=\sum\alpha_k e_k$ with all $\alpha$'s positive. Then $0\lt\langle v,v\rangle=\sum\langle e_k,v\rangle\alpha_k$ so we must have $\langle e_k,v\rangle\gt 0$ for at least one $k$, which implies the angle between those vectors is between $-\pi/2$ and $\pi/2$.
