It is impossible to find $n+1$ mutually orthogonal unit vectors in $\mathbb{R}^n$.
However, a simple geometric argument shows that the central angle between any two legs of a simplex goes as $\theta = \mathrm{arccos}(-1/n)$. This approaches $90$ degrees as $n \rightarrow \infty$, so since there are $n+1$ vertices of a simplex in $n$-dimensional space, we can conclude
Given $\epsilon > 0$, there exists a $n$ such that we can find $n+1~$ approximately mutually orthogonal vectors in $\mathbb{R}^n$, up to tolerance $\epsilon$. (Unit vectors $u$ and $v$ are said to be approximately orthogonal to tolerance $\epsilon$ if their inner product satisfies $\langle u,v \rangle < \epsilon$)
My question is a natural generalization of this - if we can squeeze $n+1$ approximately mutually orthogonal vectors into $\mathbb{R}^n$ for $n$ sufficiently large, how many more vectors can we squeeze in? $n+2$? $n+m$ for any $m$? $2n$? $e^n$?
Edit: I think one can squeeze at least $n+m$ for any $m$, via the following construction. Given $\epsilon$, one finds the $k$ such that you can have $k+1$ $\epsilon$-approximate mutually orthogonal unit vectors in $\mathbb{R}^k$. Call these vectors $v_1, v_2, ..., v_k$. Then you could squeeze $mk+m$ vectors in $\mathbb{R}^{mk}$, by using the vectors $$\begin{bmatrix} v_1 \\ 0 \\ \vdots \\ 0 \end{bmatrix}, \begin{bmatrix} v_2 \\ 0 \\ \vdots \\ 0 \end{bmatrix}, \begin{bmatrix} v_{k+1} \\ 0 \\ \vdots \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ v_1 \\ \vdots \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ v_2 \\ \vdots \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ v_{k+1} \\ \vdots \\ 0 \end{bmatrix}, \dots \begin{bmatrix} 0 \\ 0 \\ \vdots \\ v_1 \\ \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ \vdots \\ v_2 \\ \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ \vdots \\ v_{k+1} \\ \end{bmatrix}. $$
So, setting $n = mk$, we have found an $n$ such that we can fit $n + m$ $\epsilon$-orthogonal unit vectors in $\mathbb{R}^n$.
