Given two matrices $A$ and $B$, the question is whether $AA^\ast \preceq B^\ast B$ implies any relation between the singular values of the matrices.
For example, $$\bar{\sigma}(A)\le \underline{\sigma}(B),$$ where $\bar{\sigma}(\cdot)$ and $\underline{\sigma}(\cdot)$ are the maximal and minimal singular values of a matrix, respectively.
$AA^*\leq B^*B$ implies
$\big(AA^*\big)^\frac{1}{2}\leq \big(B^*B\big)^\frac{1}{2}$
where the LHS has the same singular values as $A$ and the RHS has the same singular values as $B$.
Proof: If $0 \le A \le B$ then $\sqrt{A} \le \sqrt{B}$
with singular values in the usual ordering $\sigma_1\geq \sigma_2\geq ...\geq \sigma_n$
using the left singular vectors of $A$
$U^{(k)} := \mathbf u_1\mathbf u_1^*+\mathbf u_2\mathbf u_2^* + ... + \mathbf u_k\mathbf u_k^*$
thus
$\sum_{j=1}^k \sigma_j^{(A)}=\text{trace}\Big(U\big(AA^*\big)^\frac{1}{2}\Big)\leq \text{trace}\Big(U\big(BB^*\big)^\frac{1}{2}\Big)\leq \sum_{j=1}^k \sigma_j^{(B)}$
where the second inequality follows from von Neuman Trace Inequality and the first inequality comes from $\big(AA^*\big)^\frac{1}{2}\leq \big(BB^*\big)^\frac{1}{2}$ i.e.
$\text{trace}\Big(U\big(AA^*\big)^\frac{1}{2}\Big) = \mathbf u_1^*\big(AA^*\big)^\frac{1}{2}\mathbf u_1+\mathbf u_2^*\big(AA^*\big)^\frac{1}{2}\mathbf u_2 + ... + \mathbf u_k^*\big(AA^*\big)^\frac{1}{2}\mathbf u_k^*$
The conclusion is that
$\Sigma^{(A)}\preceq_w \Sigma^{(B)}$
where $\preceq_w$ denotes weak majorization.
