Let $\rho, \sigma$ be finite dimensional positive semidefinite matrices with trace less than or equal to 1. Let $\lambda(1)$ be the smallest nonnegative number (possibly infinity) such that
$$\rho \preceq \lambda(1)\sigma,$$
where $A\preceq B$ means that $B-A$ is positive semidefinite. We are now required to find $\lambda(n)$ which is the smallest nonnegative number such that
$$\rho^{\otimes n} \preceq \lambda(n)\sigma^{\otimes n},$$
where $A^{\otimes n} = A\otimes A... \otimes A$ and there are $n$ terms in the tensor product. For $n\rightarrow\infty$, can we say anything about $\lambda(n)$ in terms of $\lambda(1)$? It seems like $\lambda(n)$ cannot be smaller than $\lambda(1)^n$ (one can construct trivial examples to see this) but are there any matrices $\rho$ and $\sigma$ for which $\lambda(n) < \lambda(1)^n$?
No such matrices exist and indeed $\lambda(n) = \lambda(1)^n$. To see this note that we should have $\mathrm{supp}(\rho) \subseteq \mathrm{supp}(\sigma)$ when $\lambda(1) < \infty$, where $\mathrm{supp}$ denotes the orthogonal complement of the kernel. For such cases we see that $\rho \leq \lambda(1) \sigma$ is equivalent to the condition (the inverse is taken on the support of $\sigma$) $$ \sigma^{-1/2} \rho \sigma^{-1/2} \leq \lambda(1) I $$ which is in turn equivalent to $$ \|\sigma^{-1/2} \rho \sigma^{-1/2}\| \leq \lambda(1) $$ where $\| \cdot \|$ is the operator norm. Then for PSD matrices it is straightforward to show via the spectral theorem that $\|A^{\otimes n}\| = \|A\|^n$. If we take the smallest such $\lambda(1)$, i.e., $\lambda(1) = \|\sigma^{-1/2} \rho \sigma^{-1/2}\|$ then, $$ \lambda(n) = \|(\sigma^{\otimes n})^{-1/2} \rho^{\otimes n} (\sigma^{\otimes n})^{-1/2}\| = \|(\sigma^{-1/2} \rho \sigma^{-1/2})^{\otimes n}\| = \lambda(1)^n. $$