Longest vector spanned by unit-normalized basis I'm trying to understand the length of the longest vector that a basis with certain properties could span.
Problem statement: Fix a euclidean space $\mathbb{R}^n$ and a fixed scalar $\epsilon > 0$. Consider a basis $\mathcal{B} = \{v_1,v_2,....,v_n\}$ such that $\forall i$ $||v_i||_2 = 1$ and for any $i,j$ $|v_i \cdot v_j| < \cos {\theta_o}$ where $\theta_o \le \frac{\pi}{2}$ is a fixed angle. Now, consider an arbitrary vector $p$ such that for any $i$ $|p\cdot v_i| < \epsilon$. Then, (conjecture) the maximum norm of $p$ is bounded as $||p||_2 \le C_o\cdot \sqrt{n}\cdot \epsilon$ where $C_o$ depends only on $\theta_0$.
Some approach: If $\theta_o = \pi/2$, then the bound on the norm of $p$ is $\sqrt{n}\cdot \epsilon$. Now, it is unclear if the stated bound holds if $\theta_o$ is smaller. I tried considering the projection map $\mathbb{P}_{-i}$ defined by the subspace spanned by $\mathcal{B} \setminus v_i$ and then consider the length of $\mathbb{P}_{-i} p$. Idea is to use induction but then the bound on $\mathbb{P}_{-i} p$ is $C_o\cdot \sqrt{n-1}\cdot \epsilon$ (which would be weak). I'm completely sure if the stated bound is correct. It could be that I'm missing some easy result to wrap this up. If there is some other bound which beats the $\sqrt{n}$ factor it would be interesting to see.
Additional assumption: I've also been trying to impose some additional structure on the basis vectors e.g. If for any $i$ $|\mathbb{P}_{-i} v_i \cdot v_i| < \cos {\theta_o}$. This would make the cone defined by the basis vectors more symmetric and some corner cases could be eliminated. Can we show the stated bound using this additional assumption?
 A: [Post edited]
Matters are as conjectured.
For easing the arguments below, let $n$ even.
First, notice that you can do the following construction. Let $e = (1,1,\cdots, 1)$ be the vector of all ones, and $u_i$ be the $i$th unit vector.  Then let the base vectors $v_i = a u_i + b e$. We can determine the constants $a$ and $b$ such that $v_i^2=a^2 + 2ab + nb^2= 1$ and $v_i v_j = 2ab + n b^2 = \cos(\theta_0)$ with equality for all $i \ne j$, giving
$a = \sqrt{1 - \cos(\theta_0)}$ and $b =  \frac{1}{n} (\sqrt{1 +(n-1) \cos(\theta_0)} -  \sqrt{1 - \cos(\theta_0)})$
(conjecture) With this base, one vector with the largest norm is  $p = c (\sum_{i=1}^{n/2} v_i - \sum_{j=1+n/2}^{n} v_j) $. The two halves of the base vectors can actually be assigned randomly to the two sums.
Then you have $|p v_k|=c( 1 - \cos(\theta_0))$ for all $k$, so you can actually equate this to $\epsilon$. This gives $$
c = \frac{\epsilon}{1 - \cos(\theta_0)}
$$
Further, you have
$$
p^2 = c^2 (\sum_{i=1}^{n/2} v_i - \sum_{j=1+n/2}^{n} v_j )^2 = c^2 n (1 - \cos(\theta_0))
$$
Hence you obtain
$$
||p||_2 = \frac{1}{\sqrt{1 - \cos(\theta_0)}} \sqrt{n} \epsilon = \frac{1}{\sqrt{2} \cdot \sin(\theta_0/2)} \sqrt{n} \epsilon
$$
For $\theta_0 = \pi/2$, this gives
$
||p||_2 (\theta_0 = \pi/2)= \sqrt{n}  \epsilon
$
as you already stated.
In general, the behavior of $||p||_2$ is as conjectured, with the constant $C_0 = \frac{1}{\sqrt{2} \cdot \sin(\theta_0/2)}$ depending only on $\theta_0$. $\qquad \Box$
