$$M := \begin{bmatrix} M_{11} & \cdots &M_{1d} \\ \vdots & \ddots & \vdots \\ M_{d1} & \cdots & M_{dd} \end{bmatrix}$$

is a $d \times d$ block matrix such that

$$M_{jk} = \sum_{i=1}^{r}(A_i)_{jk} B_i$$

for some $A_i \in M_d(\mathbb{C})$, $B_i \in M_{n}(\mathbb{C})$ and $d,n,r >2$. Now, let

$$M^{\square} := \begin{bmatrix} M^T_{11} & \cdots &M^T_{1d} \\ \vdots & \ddots & \vdots \\ M^T_{d1} & \cdots & M^T_{dd} \end{bmatrix}$$

where $M^T_{jk} = \sum_{i=1}^{r}(A_i)_{jk} B^T_i$. Is the following inequality true?

$$\frac{\mbox{rank}(M^{\square})}{\mbox{rank}(M)} \leq r$$

For $r=1,2$ this statement is true!

  • 3
    $\begingroup$ For the benefits of other readers, the same matrices written with Kronecker product notation: $M = \sum_{i=1}^r A_i \otimes B_i, \, M' = \sum_{i=1}^r A_i \otimes B_i^T$. $\endgroup$ – Federico Poloni Mar 11 '18 at 7:59
  • $\begingroup$ The operation that you're performing there is also called 'partial transposition'. It's not hard to show that partial transposition preserves rank, by simply computing that the partial transpose of a rank one matrix again has rank one. Just write the rank one matrix explicitly as $(\sum_i a_i\otimes b_i)(\sum_j x_j \otimes y_j)^T$ and compute the partial transpose. $\endgroup$ – Tobias Fritz Mar 11 '18 at 20:10
  • 1
    $\begingroup$ @Tobias - That's not true -- the partial transpose of the standard rank-1 maximally-entangled state in $M_2 \otimes M_2$ has rank 4. The partial transpose swaps the $b_i$ and $y_j$ terms in your sum, so you can no longer necessarily split up the sum over $i$ and the sum over $j$. $\endgroup$ – Nathaniel Johnston Mar 12 '18 at 12:38
  • $\begingroup$ @NathanielJohnston: good point, thanks! I confess that I hadn't actually written it out. $\endgroup$ – Tobias Fritz Mar 12 '18 at 15:58

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.