$M = AA^t$ where $A$ has unit norm columns

Question

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 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).

Answer 1

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 $m> \frac{k^2+k-2}{2(k-1)}=\frac{k+2}{2}$.