Decomposition of matrix occuring in problem of finding $n+1$ vectors in $\mathbb{R}^n$ with pairwise equal inner product 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.

Adding to the existing answer, you can also extract your vectors by orthogonally diagonalizing $M$. The matrix $M$ has rank $n$ with two eigenvalues: The eigenvalue $\lambda := \frac{n+1}{n}$ of multiplicity $n$ and the eigenvalue $0$ of multiplicity one. Let $U$ be an orthogonal matrix such that
$$ U \begin{pmatrix} \lambda I_{n \times n} & 0_{n \times 1} \\ 0_{1 \times n} & 0 \end{pmatrix} U^T = M. $$
The first $n$ columns of $U$ form an orthonormal basis for the eigenspace of $M$ associated to $\lambda$ which is
$$\{ (x_1, \dots, x_{n+1})^T \, | \, x_1 + \dots + x_{n+1} = 0 \}. $$
Write
$$ U = \begin{pmatrix} \hat{U}_{n \times n} & x_{n \times 1} \\ y_{1 \times n} & z \end{pmatrix}, V = \begin{pmatrix} \hat{V}_{n \times n} & v_{n+1} \end{pmatrix}.  $$
The equation $V^T V = M$ then becomes
$$ \begin{pmatrix} \hat{V}^T \hat{V} & \hat{V}^T v_{n+1} \\ v_{n+1} V^T & v_{n+1}^T v_{n+1} \end{pmatrix} = \lambda \begin{pmatrix}  \hat{U} \hat{U}^T & \hat{U} y^T \\ y \hat{U}^T & y \hat{y}^T \end{pmatrix}. $$
Hence if you take $\hat{V} = \sqrt{\lambda} \hat{U}^T, v_{n+1} = \sqrt{\lambda} y^T$ you get a solution. In other words, if you take the $n + 1$ rows of the eigenvector matrix $U$ (whose columns are the eigenvectors), forget the last coordinates and transpose, you get a solution. Since the eigenspace assosicated to the eigenvalue $\lambda$ is $n$-dimensional, the matrix
$$ \begin{pmatrix} \hat{U} \\ y \end{pmatrix} $$ is determined up to an orthogonal transformation of an $n$-dimensional subspace which corresponds to the freedom in rotating the columns of $V$ in an $n$-dimensional space.

Example: For $n = 2$, one possible decomposition is
$$ M = \begin{pmatrix} 1 & -1/2 & -1/2 \\ -1/2 & 1 & -1/2 \\ -1/2 & -1/2 & 1 \end{pmatrix} = \underbrace{\begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{3}} \\ 0 & -\frac{2}{\sqrt{6}} & \frac{1}{\sqrt{3}} \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{3}} \end{pmatrix}}_{U} \begin{pmatrix} \frac{3}{2} & 0 & 0 \\ 0 & \frac{3}{2} & 0 \\ 0 & 0 & 0 \end{pmatrix} 
\underbrace{\begin{pmatrix} \frac{1}{\sqrt{2}} & 0 & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{6}} & -\frac{2}{\sqrt{6}} & \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} \end{pmatrix}}_{U^T} $$
and the corresponding vectors are:
$$ v_1 = \sqrt{\frac{3}{2}} \begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \end{pmatrix}^T = \begin{pmatrix} \frac{\sqrt{3}}{2} \\ \frac{1}{2} \end{pmatrix}, \\
v_2 = \sqrt{\frac{3}{2}} \begin{pmatrix} 0 & -\frac{2}{\sqrt{6}} \end{pmatrix}^T = \begin{pmatrix} 0 \\ -1 \end{pmatrix}, \\
v_3 = \sqrt{\frac{3}{2}} \begin{pmatrix} -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \end{pmatrix}^T = \begin{pmatrix} -\frac{\sqrt{3}}{2} \\ \frac{1}{2} \end{pmatrix} $$

