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Suppose

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

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!

Improve this question
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    $\begingroup$ 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
    Commented Mar 11, 2018 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
    Commented Mar 11, 2018 at 20:10
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    $\begingroup$ @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
    Commented Mar 12, 2018 at 12:38
  • $\begingroup$ @NathanielJohnston: good point, thanks! I confess that I hadn't actually written it out. $\endgroup$
    – Tobias Fritz
    Commented Mar 12, 2018 at 15:58

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