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What is the sigma representation of adding two or more vectors with an identical number of dimensions into one vector?

For example, something like this: $$ [x_1,y_1,z_1,\dots,n_1]+[x_2,y_2,z_2,\dots,n_2]+[x_3,y_3,z_3,\dots,n_3]+\dots+ [x_m,y_m,z_m...n_m] \\= \begin{bmatrix}\bigg(\displaystyle\sum_{i=1}^{m} x_i \bigg),\bigg( \displaystyle\sum_{i=1}^{m} y_i \bigg),\bigg(\displaystyle\sum_{i=1}^{m} z_i\bigg)..,\bigg(\displaystyle\sum_{i=1}^{m} n_i\bigg)\end{bmatrix} $$

But i am sure there's a shorter formal way of doing it

Matthew Leingang
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soundslikefiziks
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1 Answers1

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$$ \sum_i^m [x_i , y_i, \ldots , n_i] $$ will do the job, assuming your reader knows how to add vectors.

If you're doing this a lot you might want to name the vectors $v_i$.

Ethan Bolker
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  • I thought about that, but it seemed unclear to me not to include anything that would indicate an addition of the dimensions $$(x_1 +x_2+x_3...)$$ , is this a formal representation ? – soundslikefiziks Sep 27 '17 at 23:41
  • You're not adding dimensions. All the vectors have the same dimension, as you know. You aren't (and can't) name the dimension, because you're using $x, y, \ldots , n$ to name the coordinates, rather than going for double subscripts. (Euler would approve - makes for readability.) If you go for double subscripts then it's clear that you are just adding the rows (or columns) of a matrix. – Ethan Bolker Sep 27 '17 at 23:48
  • Yes, i meant the addition of the identical dimensions, so what you are suggesting would look like this ? : $$\sum_{i=1}^{m} [[v_i]_1,[v_i]_2,[v_i]_3..,[v_i]_n]$$ – soundslikefiziks Sep 28 '17 at 00:06
  • Yes, probably without the inner brackets. Perhaps $v_{i,j}$. But your $x,y$ etc is OK. – Ethan Bolker Sep 28 '17 at 00:15
  • But wouldn't this still force me to elaborate with the addition of : $$=[x_1,y_1,z_1,\dots,n_1]+[x_2,y_2,z_2,\dots,n_2]+[x_3,y_3,z_3,\dots,n_3]+\dots+ [x_m,y_m,z_m...n_m]$$ – soundslikefiziks Sep 28 '17 at 00:36
  • I don't think so. Apparently you don't either since you've accepted the answer. I wouldn't use $n$ for the last coordinate. Perhaps $x, y, \ldots, z$ or $a,b, \ldots, z$ (as might Euler). – Ethan Bolker Sep 28 '17 at 13:43