Let $M \in \mathbb{R}^{k\times k}$ positive definite with $\operatorname{tr} M = m$, where $m$ is an integer such that $m \geq k$. I have found a way (using [this](https://mathoverflow.net/a/280203/89544) answer) to decompose $M = AA^t$ with $A \in \mathbb{R}^{k \times m}$ such that $A = (a\_1, \dots, a\_m), a\_i \in \mathbb{R}^k$ and $\|a\_i\|\_2 = 1, i=1,\dots,m$.


1. Is there a name for such a decomposition? This is not Cholesky, although it looks similar.
2. Is this decomposition unique? We can always take $\hat{A} := AD$ where $D$ is a permutation matrix with $\pm 1$ entries. Then $\hat{A}\hat{A}^t = M$ and $\hat{A}$'s columns have unit norm. I am not sure if there is any other obstruction to uniqueness.
3. In my numerical experiments, I find that the some columns of $A$ are identical (up to a sign). Any reason for that?


Example
=======


Let $M =diag(1.5,1.5)$. One can verify that $M=AA^t$ for


$$
A = 
\begin{pmatrix}
\sqrt{3/4}& \sqrt{3/4}& 0 \\
-1/2& 1/2 & 1\\
\end{pmatrix}
$$
P.S. The assumption on the trace above is necessary because $\text{tr} M = \text{tr} AA^t = \text{tr}A^tA$ and $A^tA\in \mathbb{R}^{m \times m}$ has unit diagonal.


Reference
=========


Using Raphael's answer below I was able to find the reference:


Peter A. Fillmore, On sums of projections, Journal of functional analysis 4, 146-152 (1969).



$M = AA^t$ where $A$ has unit norm columns
This decomposition is equivalent to write $M$ as a sum of rank one orthogonal projection $$ M = \sum\_{i=1}^m a\_i a\_i^\* $$
with $\|a\_i\|=1$. Indeed for any $x$ we have $$(Mx)\_{s} = \sum\_{i\leq m,t\leq k} A\_{si}A^T\_{it}x\_t = \sum\_{i\leq m} (a\_i)\_s \langle a\_i,x\rangle
$$
Remark that in form it is easy to see the invariance by permutation with $\pm 1$ entries and that $\text{Tr}(M)=m$.


We can consider the application $\phi:(\mathbb{S}^{k-1})^m\rightarrow \mathbb{R}^{k\times k}$, $\phi(a\_1,\cdots,a\_m)=AA^T=M$. Because $(\mathbb{S}^{k-1})^m$ is a manifold of dimension $m(k-1)$ and the subset of symetric matrices of trace $m$ is a manifold of dimension $\frac{k(k+1)}{2}-1$. It is clear that we don't have unicity in the general case if $m> \frac{k^2+k-2}{2(k-1)}=\frac{k+2}{2}$.


