1
$\begingroup$

Assume that I am given the following matrices: $A = \begin{bmatrix} a_{1,1} & \dots & a_{1,n_a} \\ \vdots & \ddots & \vdots \\ a_{m_a,1} & \dots & a_{m_a,n_a} \end{bmatrix}$ and $B = \begin{bmatrix} b_{1,1} & \dots & b_{1,n_b} \\ \vdots & \ddots & \vdots \\ b_{m_b,1} & \dots & b_{m_b,n_b} \end{bmatrix}$. I need to consider the following matrix and I am looking for a way to relate it to $A$ and $B$. \begin{bmatrix} a_{1,1} & a_{1,1} & \dots & a_{1,1} & a_{1,2} & \dots & a_{1,2} & \dots & a_{1,n_a} & a_{1,n_a} & \dots & a_{1,n_a} \\ a_{2,1} & a_{2,1} & \dots & a_{2,1} & a_{2,2} & \dots & a_{2,2} & \dots & a_{2,n_a} & a_{2,n_a} & \dots & a_{2,n_a} \\ \vdots & \vdots & \ddots & \vdots & \vdots & \ddots & \vdots & \ddots & \vdots & \vdots & \ddots & \vdots\\ a_{m_a,1} & a_{m_a,1} & \dots & a_{m_a,1} & a_{m_a,2} & \dots & a_{m_a,2} & \dots & a_{m_a,n_a} & a_{m_a,n_a} & \dots & a_{m_a,n_a}\\ b_{1,1} & b_{1,2} & \dots & b_{1,n_b} & b_{1,1} & \dots & b_{1,n_b} & \dots & b_{1,1} & b_{1,2} & \dots & b_{1,n_b} \\ b_{2,1} & b_{2,2} & \dots & b_{2,n_b} & b_{2,1} & \dots & b_{2,n_b} & \dots & b_{2,1} & b_{2,2} & \dots & b_{2,n_b} \\ \vdots & \vdots & \ddots & \vdots & \vdots & \ddots & \vdots & \ddots & \vdots & \vdots & \ddots & \vdots\\ b_{m_b,1} & b_{m_b,2} & \dots & b_{m_b,n_b} & b_{m_b,1} & \dots & b_{m_b,n_b} & \dots & b_{m_b,1} & b_{m_b,2} & \dots & b_{m_b,n_b} \end{bmatrix}

What would be good name and symbol for such an operation? I am thinking about cartesian product.

$\endgroup$

1 Answer 1

3
$\begingroup$

In terms of operations that already exist, we can write this as the block matrix $$ \pmatrix{A \otimes X\\ X \otimes B} $$ Where $X$ is the row-vector $(1,\dots,1)$ and $\otimes$ denotes the Kronecker product.

A nice shorthand for this operation (if the above is not sufficiently compact) could be $\frac{A}{B}$, which has no common second meaning in this context.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .