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

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

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

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

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

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!

Suppose \begin{equation} M= \begin{bmatrix} M_{11} & \cdots &M_{1d} \\ \vdots & \ddots & \vdots \\ M_{d1} & \cdots & M_{dd} \end{bmatrix} \end{equation} 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 \begin{equation} M^{\prime}= \begin{bmatrix} M^T_{11} & \cdots &M^T_{1d} \\ \vdots & \ddots & \vdots \\ M^T_{d1} & \cdots & M^T_{dd} \end{bmatrix}, \end{equation} where $M^T_{jk} = \sum_{i=1}^{r}(A_i)_{jk} B^T_i$. Is it true that $\frac{R(M^{\prime})}{R(M)} \leq r$, where $R$ denotes the matrix rank? (for $r=1,2$ this statement is true!)

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

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

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

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

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

Suppose \begin{equation} M= \begin{bmatrix} M_{11} & \cdots &M_{1d} \\ \vdots & \ddots & \vdots \\ M_{d1} & \cdots & M_{dd} \end{bmatrix} \end{equation} is a $d \times d$ block matrix such that $M_{jk} = \sum_{i=1}^{r}(A_i)_{jk} B_i$, for some $n \times n$ complex matrices $A_i, B_i$, and $d,n,r >2$. Now, let \begin{equation} M^{\prime}= \begin{bmatrix} M^T_{11} & \cdots &M^T_{1d} \\ \vdots & \ddots & \vdots \\ M^T_{d1} & \cdots & M^T_{dd} \end{bmatrix}, \end{equation} where $M^T_{jk} = \sum_{i=1}^{r}(A_i)_{jk} B^T_i$. Is it true that $\frac{R(M^{\prime})}{R(M)} \leq r$, where $R$ denotes the matrix rank? (for $r=1,2$ this statement is true!)

Suppose \begin{equation} M= \begin{bmatrix} M_{11} & \cdots &M_{1d} \\ \vdots & \ddots & \vdots \\ M_{d1} & \cdots & M_{dd} \end{bmatrix} \end{equation} 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 \begin{equation} M^{\prime}= \begin{bmatrix} M^T_{11} & \cdots &M^T_{1d} \\ \vdots & \ddots & \vdots \\ M^T_{d1} & \cdots & M^T_{dd} \end{bmatrix}, \end{equation} where $M^T_{jk} = \sum_{i=1}^{r}(A_i)_{jk} B^T_i$. Is it true that $\frac{R(M^{\prime})}{R(M)} \leq r$, where $R$ denotes the matrix rank? (for $r=1,2$ this statement is true!)