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

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).

• @FedericoPoloni apart from positivity the OP is imposing a limit on $\mid M \mid$ by $|a_i|_2=1$. so it is unlike that every arbitrary positive matrix $M$ can be decomposed in the way OP required. he actually impose a boundedness condition on M. Oct 30 '20 at 18:56
• @AliTaghavi Positive semidefinite + trace=m defines a bounded set. Once one assumes positivity I don't see any obvious obstructions to that kind of factorization. Oct 30 '20 at 19:11
• In item 2, when you say "unique", do you mean "up to replacing $A$ by $AC$ where $C$ is orthonormal"? (If not, the decomposition is heavily non-unique even for $m=k$ and $M = I_m$.) Oct 30 '20 at 19:23
• @AliTaghavi: I don't know what you mean. What I'm saying is that if $m = k$ and $M = I_m$, then $A$ can be any orthonormal $m\times m$-matrix. Oct 30 '20 at 22:35
• What about $D$ a permurtation matrix? Nov 8 '20 at 7:57

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