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I was wondering how to write vector notation with an index which is not included in the vector. In sets we can write,

$$ A=\{0,1,2,3,4\},$$

then if we don't want to include the element $\{0\}$ we can write,

$$A \setminus \{0\}=\{1,2,3,4\}.$$

Is there a way to write that for a vector?

i.e. for vector $a=(a_0,a_1,a_2,a_3,a_4)$ can we write $a\setminus a_0=(a_1,a_2,a_3,a_4)$?

Thank you very much.

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    $\begingroup$ I'm not aware of any standard notation. $\endgroup$
    – copper.hat
    Nov 21, 2014 at 3:10

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There is a way to do this, but it's not convenient or useful. However, I thought I'd share it.

Consider the vector $a=(a_0,a_1,a_2)$. What is this vector? It's a set of ordered pairs. Specificially, this vector in $\mathbb{R}^3$ is defined as $\left\{(0,a_0),(1,a_1),(2,a_2)\right\}$. Let's suppose we wanted the vector $(a_0,a_1)\in\mathbb{R}^2$. We could write this vector as $a \setminus (2,a_2)$.

No one really thinks of vectors as sets as I've described, so it'd be confusing for almost everybody, but it can be done. :)

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  • $\begingroup$ I understand. Thank you very much! $\endgroup$
    – Nikki Mino
    Nov 21, 2014 at 4:10
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Maybe you're looking for projections rather than a difference operation?

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  • $\begingroup$ I am not entirely sure what you mean but for example the vector is composed of probabilities, i.e., $$a=(a_0,a_1,a_2,a_3,a_4) \quad s.t. \quad \sum_{i=0}^n a_i = 1.$$ The first element will be $a_1 = 1$ while the rest are zero. I want to state it in a more general form by not including the element with probability 1. Thank you. $\endgroup$
    – Nikki Mino
    Nov 21, 2014 at 3:52
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    $\begingroup$ I believe they mean defining the projection $\pi$ mapping $(a_0,a_1,\ldots)\mapsto (a_1,a_2,\ldots)$ - that is, dropping the first element, which happens to be a projection. It's a more abstract (and standard) way to do this. $\endgroup$
    – Milo Brandt
    Nov 21, 2014 at 4:22
  • $\begingroup$ I understand. I just wanted to make a statement similar to my comment above and not really something close to a projection. Thank you very much for you help. $\endgroup$
    – Nikki Mino
    Nov 21, 2014 at 9:15

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