I was toying with my intuition that there are always $n+1$ unit vectors in $\mathbb{R}^n$ such that every pair $v_i$ and $v_j$ ($i\neq j$) has the same angle between them. As those vectors are normalized this is equivalent to $$v_i\cdot v_j=\begin{cases} 1,& i=j \\m, & i\neq j \end{cases}$$ with $|m|<1$. So if $V$ is the $n\times (n+1)$ matrix whose columns are formed by the $v_i$ we have $$M:=V^TV=\begin{pmatrix}1 & m & \cdots &m \\m&1&\cdots&m\\ \vdots &&\ddots&\vdots \\ m&m&\cdots&1 \end{pmatrix}$$ $M$ is a $(n+1)\times (n+1)$ matrix and must be singular (as it is the product of two rank $n$ matrices). The sum of its rows is $(nm+1,nm+1,\ldots,nm+1)$. For $m=-\frac{1}{n}$ this is the null vector giving the required dot product $m$ as a function of $n$. E.g., for $n=2$, $m=-\frac{1}{2}$ corresponding to the expected angle of $120°=\arccos{-\frac{1}{2}}$. Now I'm asking is there a matrix decomposition so that I can get $V$ back from the product $M$? (I know that $V$ is not unique since I can freely rotate the vectors $v_i$ about an arbitrarily chosen fixed axis without changing their pairwise dot product.) I tried spectral decomposition of $M$ but that gives $n+1$ dimensional vectors and the $n$ nonzero eigenvalues of $M$ are all equal to $\frac{n+1}{n}$ which means that any linear combination of eigenvectors corresponding to that eigenvalue is also an eigenvector making the dot product between those eigenvectors rather arbitrary.
Decomposition of matrix occuring in problem of finding $n+1$ vectors in $\mathbb{R}^n$ with pairwise equal inner product
eigenvalues-eigenvectorslinear algebramatrix decompositionvector-spaces
Best Answer
In ${\mathbb R}^n$, having $n+1$ unit vectors such that the angle between any two of them be equal
implies that the vectors individuate the vertices of a regular n-simplex inscribed in the unit sphere.
That means that the cosine of the angle (your $m$) shall be $-1/n$.
$n$ of the vectors will be independent, while the (n+1)-th one will necessarily be dependent.
Therefore the $(n+1) \times (n+1)$ matrix $$ M = V^{\,T} V $$ will have null determinant.
But a $n \times n$ diagonal submatrix of it ($M'$)will be full-rank, symmetric (Hermitean) and positive definite, being a matrix of inner products.
So it admits a Cholesky decomposition $$ M' =L L^{\,T} $$ which is unique and therefore we shall have $$ M' = L\,L^{\,T} = V'^{\,T} V'\quad \Rightarrow \quad V' = L^{\,T} $$ where $V'$ is the $n \times n$ matrix of $n$ of the (column) vectors.
The additional $n+1$ vector $v$ will be derived by solving $$ V'^{\,T} v = Lv = - \frac{1}{n}\left( {\begin{array}{*{20}c} 1 \\ 1 \\ \vdots \\ 1 \\\end{array}} \right) $$
For example, for ${\mathbb R}^3$ we get $$ \begin{array}{l} M' = \left( {\begin{array}{*{20}c} 1 & { - 1/3} & { - 1/3} \\ { - 1/3} & 1 & { - 1/3} \\ { - 1/3} & { - 1/3} & 1 \\ \end{array}} \right) = \\ = L\,L^{\,T} = \left( {\begin{array}{*{20}c} 1 & 0 & 0 \\ { - \frac{1}{3}} & {\frac{{2\sqrt 2 }}{3}} & 0 \\ { - \frac{1}{3}} & { - \frac{{\sqrt 2 }}{3}} & {\frac{{\sqrt 3 \sqrt 2 }}{3}} \\ \end{array}} \right)\left( {\begin{array}{*{20}c} 1 & { - \frac{1}{3}} & { - \frac{1}{3}} \\ 0 & {\frac{{2\sqrt 2 }}{3}} & { - \frac{{\sqrt 2 }}{3}} \\ 0 & 0 & {\frac{{\sqrt 3 \sqrt 2 }}{3}} \\ \end{array}} \right) \\ \end{array} $$ and $$ V = \left( {\begin{array}{*{20}c} 1 & { - \frac{1}{3}} & { - \frac{1}{3}} & { - \frac{1}{3}} \\ 0 & {\frac{{2\sqrt 2 }}{3}} & { - \frac{{\sqrt 2 }}{3}} & { - \frac{{\sqrt 2 }}{3}} \\ 0 & 0 & {\frac{{\sqrt 3 \sqrt 2 }}{3}} & { - \frac{{\sqrt 3 \sqrt 2 }}{3}} \\ \end{array}} \right) $$