Suppose that $D\in\mathbb{R}^{m\times m}$ is orthogonal, $\Sigma \in \mathbb{R}^{m\times n}$ only has elements on the main diagonal, and $V\in\mathbb{R}^{n\times n}$ is orthogonal, with $n < m$. Now consider $$ A=U\Sigma V^T. $$ This reminds me of an SVD for A, but I am not sure if it is actually its SVD.
I know that $U\Sigma$ and $U\Sigma V^T$ have the same singular values, since
$$U\Sigma V^T V \Sigma^T U^T = U\Sigma \Sigma^T U^T$$
but can I say more on the singular values of $U \Sigma$?
There exist diagonal matrices $D,\Sigma_+$ (with $D$ square) such that $\Sigma = D\Sigma_+$, where the diagonal entries of $D$ are all $\pm 1$ and the diagonal entries of $\Sigma_+$ are non-negative. It follows that $$ A = (UD) \Sigma_+D^T $$ is a singular value decomposition. So, we see that the singular values of $A$ are the entries of $\Sigma_+$, which are the the absolute values of the diagonal entries of $\Sigma$.