Obtaining $A$ from $A$$A^t$$A$ Let the matrix $B$=$AA^TA$ be given to us where A is a mxn real matrix.Than how can we obtain $A$ from $B$ ? Can we do the same thing if A is a complex matrix ?
I have no idea how to do this problem.Please help and thanks in advance.
 A: Suppose $A$ has a singular value decomposition $U\Sigma V^T$ where $U$ and $V$ are orthogonal, and $\Sigma$ positive semidefinite diagonal.
In that case,
$$AA^TA=U\Sigma^3 V^T$$
Now think reverse. A SVD decomposition of $B$ has to share $U$ and $V^T$ with $A$, because the decomposition is unique up to permutation of singular values (that we usually sort). You must simply split
$$B=U\Gamma V^T$$
and compute
$$\Sigma=\sqrt[3]{\Gamma}$$
where this is meant as taking the cube root of every singular value. Because they are positive, this is not ambiguous.
Note that this gives you one solution, but it isn't unique, because the equation $\sigma_i^3=\gamma_i$ has $3$ solutions (for each singular value, a different root can be chosen).
A: Orion's method works when $m=n$. Yet a similar method works for any $m,n$. In such a case, $\Sigma$ is a $m\times n$ matrix and $AA^TA=U\Sigma\Sigma^T\Sigma V^T$. Fortunately, the matrix  $\Sigma\Sigma^T\Sigma $ has exactly the same form than $\Sigma$ ; indeed it suffices to change each $\Sigma_{i,i}$ with ${\Sigma_{i,i}}^3$. Thus $U\Sigma\Sigma^T\Sigma V^T$ is "the" SVD of $B=AA^TA$. We finish the construction as Orion did.
