Recovering a matrix after multiplication by its transpose

Question

Let $A$ be a $p \times k$ matrix (with $p > k$) that has full column rank, i.e., $\mbox{rank}(A) = k$. Suppose we have

$$B = A A^t$$

$A$ and $B$ have the same degree of freedom. How can the original values of $A$ be calculated from $B$?

Answer 1

We can't in general.

After all, $B=AA^T=(-A)(-A)^T$.

To compute a particular $A$ that satisfies the factorization, consider the Cholesky factorization.

Answer 2

We assume that the considered matrices are real.

If $B=AA^T$, then, necessarily $B$ is symmetric $\ge 0$ and $rank(B)=k$..

In fact, $A$ and some admissible $B$ have not the same degree of freedom. Indeed, if $A$ is a solution, then, for every $O\in O(k)$, $AO$ is also a solution . Moreover, assume that $AO_1=AO_2$; since $A$ has full column rank, its Moore-Penrose inverse is $A^+=(A^TA)^{-1}A^T$ and satisfies $A^+A=I_k$. That implies $O_1=O_2$ and $A$ depends on $k(k-1)/2$ more parameters than $B$.

Conclusion. To recover $A$, you need $k(k-1)/2$ more informations about $A$.