Let $\{v_1, \dotsc, v_m\} \in \mathbb{C}^{2^n}$ be a set of orthonormal vectors. Define a map $R_m$ from $2^n \times 2^n$ to $m \times m$ matrices as follows:
$$R_m(M) := \sum_{i,j=1}^m (v_i^*M v_j) E_{ij}$$
where $E_{ij} := e_i e_j^*$ is the all-zeroes matrix with entry 1 at position $(i,j)$. In other words, $R_m$ performs a unitary change of basis and then takes the top-left $3 \times 3$ block of the resulting matrix.

For any $n \geq 2$, show that no matter how the vectors $\{v_i\}$ are chosen,
$$\mathrm{U}(3) \not\subseteq \bigl\{ R_3(U_1 \otimes \dotsb \otimes U_n) : U_i \in \mathrm{U}(2) \bigr\}.$$
That is, as matrices $U_i$ range over all possible $2 \times 2$ unitaries, their tensor product will never contain the set of all $3 \times 3$ unitaries in some basis.

More generally, this should hold even if $2 \times 2$ unitaries are replaced by $d \times d$ unitaries and $m = 3$ is replaced by any $m > d$. The motivation for this question comes from "encoded universality" in quantum computing.

**Note:** I originally asked this question on [math.stackexchange][1] but did not receive any answer.


  [1]: http://math.stackexchange.com/questions/1112208/why-a-tensor-product-of-2-times-2-unitaries-cannot-implement-a-3-times-3-uni