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Let

$$v = \begin{pmatrix} v_1 \\ \vdots \\ v_m \end{pmatrix}$$ and $$A = \begin{pmatrix} a_{11} & \cdots & a_{1n} \\ \vdots & \ddots & \vdots\\ a_{m1} & \cdots & a_{mn} \end{pmatrix}$$

If we define the operation $\Box$ as follows

$$v \Box A = \begin{pmatrix} v_1 a_{11} & \cdots & v_1 a_{1n} \\ \vdots & \ddots & \vdots\\ v_m a_{m1} & \cdots & v_m a_{mn} \end{pmatrix},$$

is there a clean way to write it down? Does the operation have any interpretation?

I was hoping it has something to do with tensor products but, after looking up about tensors, it doesn't look like it.

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Let's define the diagonal matrix $V$ as follows:

$$V := \mbox{diag} (v) = \begin{pmatrix} v_{1} & 0 & \cdots & 0\\ 0 & v_{2} & \cdots & 0\\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & \cdots & v_{m} \end{pmatrix}$$

Then:

$$v \Box A = V \cdot A , $$

where $\cdot$ is the standard product between matrices.

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  • $\begingroup$ How exactly is this generalized by the Kronecker product? $\endgroup$ – Leo Aug 29 '16 at 9:43
  • $\begingroup$ No, it isn't. Thanks a lot. $\endgroup$ – the_candyman Aug 29 '16 at 9:49

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