Does the following type of "SVD" exist? SVD of $A$ gives $U$, $\Sigma$ and $V$ such that $A = U \Sigma V$.
I am interested in a different problem. Given an $A$ and $\Sigma_1,\ldots,\Sigma_{n-1}$ diagonal matrices, such that we know that
$$A = U_1 \Sigma_1 U_2 \Sigma_2 U_3 \Sigma_3 \ldots U_n$$ for some unitary matrices $U_1,\ldots,U_n$, we need to recover $U_1,\ldots,U_n$.
EDIT: I can choose $\Sigma_i$ freely a priori, if that helps make the recovery of $U$ possible. I also don't mind if the $U$'s are identified up to a multiplication by $-1$ or $1$.
 A: In general this can't be done.  For example, in the $2 \times 2$ case with $V = \pmatrix{0 & 1\cr 1 & 0\cr}$ and $\Sigma_3  = \Sigma_1$,
$$ I \Sigma_1 I \Sigma_2 V \Sigma_1 V = V \Sigma_1 V \Sigma_2 I  \Sigma_1 I $$
In addition, you can always multiply the $U$'s by scalars of absolute value $1$ whose product is $1$.
A: Let's look at $3 \times 3$ matrices. 
Each column of each $U$ is a unit vector, so $U$ is in $O(3)$. The dimension of $O(3)$ is 3. The entries of $A$ give you nine degrees of freedom. So the map from 
$$
(U_1, \ldots, U_n) \mapsto A,
$$
for a fixed set of $\Sigma_i$s, is a smooth map from a $3n$ manifold to $\mathbb R^9$. If $n > 3$, it cannot possibly be injective (by the implicit function theorem), hence it's impossible to recover the $U_s$ from $A$. The same argument generalizes to larger matrices, of course. 
I now notice that you said "unitary", but the dimension of the unitary group is also quadratic in the number $k$ of rows. That means that regardless of $k$ there's always an $n$ (and it's not very large!) such that when you have $n$ factors, you've got more unknowns than your $k^2$ knowns, and the problem's unsolvable. 
