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 $\|a_i\|_2 = 1, i=1,\dots,m$.

- Is there a name for such a decomposition? This is not Cholesky, although it looks similar.
- 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.
- 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).

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