Suppose we have a matrix $A\in\mathbb{R}^{n\times n}$ (we assume $A$ is always full rank) whose singular value decomposition is $A=U\Sigma V^\top$. And we are given another diagonal matrix $D$ which is of the same size as $\Sigma$. My question is that: is there a fast way to compute $U D V^\top$ without computing $U$ and $V$ out directly?
Some context of the question: if the diagonal elements of $D$ are all ones, there is a fast Newton-Schulz iteration (equation (5.22) in [1]): $$ X_{k+1}=\frac{1}{2} X_k\left(3 I-X_k^2\right), \quad X_0=A . $$ that converges to the sign of matrix $A$, namely $UDV^\top=UV^\top$, and I'm wondering if this result could be extended.
References:
[1] Higham, Nicholas J. Functions of matrices: theory and computation. Society for Industrial and Applied Mathematics, 2008.
