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I am wondering if there is a special name for an $m\times n$ matrix $A=(a_{i,j})$, with $a_{i,j}\in\{0,1\}$ that will pick $m$ unique components from a vector $v\in\mathbb{R}^n$ ($m\le n$), it is similar to a projection matrix, except that the image of such matrix is the subspace $\mathbb{R}^m$ instead of $\mathbb{R}^n$. Is decimation matrix correct?

For example, given a vector $v=(v_i)_{1\le i\le n}\in\mathbb{R}^n$, such matrix may return $w=(v_i)_{i=1,3,\ldots,2\lfloor \frac{n-1}{2}\rfloor+1}$ or $(v_i)_{1\le i\le 4}$ or any subset of the vector components.

Closely related question (without the coordinate position advancing requirement): Matrix with exactly one 1 in each row

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It is unclear what you say. Please give an example of such a matrix. Is maybe $m\leq n$? In any case quite possibly there is no name for the class of matrices you have in mind. – Marc van Leeuwen Oct 6 '12 at 19:04
Such a matrix is a projection only if $a_{ij}=0$ for all $i\neq j$. Is this the type of matrix you have? – N. S. Oct 6 '12 at 19:16
Thanks, I've given an example. N.S: no it's not a projection matrix, I've clarified that. – user43812 Oct 7 '12 at 9:33
up vote 0 down vote accepted

I suppose you are talking about matrices with a single $1$ in each row, and their positions advancing to the right for successive rows. I would call this a "coordinate-selection matrix", but that is in no means standard terminology. If you drop the advancing requirement, you can also permute or repeat any selected coordinates. If your matrix randomly skips one out of every ten columns, then "decimation matrix" would indeed be an appropriate term.

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I like this term, thanks! – user43812 Oct 8 '12 at 7:29

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