Given two matrices $A$ and $B$ of size $a \times n$ and $b \times m$ consider the following operation $A \dagger B$ whose result is an $a b^n \times n m$ matrix. $A \dagger B$ is a block matrix with $a$ blocks $(A \dagger B)_{i,j}$ of size $b^n \times n m$ each constructed as follows. For each vector $j \in \{1, \dots, b\}^n$ create a row $(A_{i,1} \cdot B_{j[1]}, A_{i,2} \cdot B_{j[2]}, \dots, A_{i,n} \cdot B_{j[n]})$ where $B_t$ is the $t$-th row of $B$.

Is there a name for this matrix operation?

  • 5
    $\begingroup$ How is this operation arising? Or are you hoping to call it the "Yaroslavtsev super-slam"? $\endgroup$ Sep 29, 2015 at 16:33
  • 3
    $\begingroup$ I just need the fact that $rank(A \dagger B) = rank(A)rank(B)$ which is easy to show directly. However, it would be helpful to know if this operation and its properties are already known so that I can just cite an appropriate source. I like your "super-slam" idea though :) $\endgroup$ Sep 29, 2015 at 16:50
  • 3
    $\begingroup$ I'm in favor of the yaroslavtsev super-slam. You can cite this MO question:) $\endgroup$
    – Thomas Rot
    Sep 29, 2015 at 18:42
  • 6
    $\begingroup$ If you just need rank multiplicativity, there is a more economic one - tensor product (of size $ab\times nm$, with $(A\otimes B)_{\langle i,k\rangle,\langle j,l\rangle}=A_{i,j}B_{k,l}$) $\endgroup$ Sep 29, 2015 at 20:09
  • 1
    $\begingroup$ I've posted a question about the latter $\endgroup$ Sep 30, 2015 at 6:32


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