I came across this notation in a textbook of Algebra.

With respect to the definition of linear independence in a Vector Space $V$. We define a subset $S = \{\alpha_i \ | \ i \in I\} \subset V$ as lin. ind over some field if we have: $$\sum_{j=1}^n {a_j \alpha_{i_j}} = 0 \implies a_j = 0 \ \ \ for \ \ j = 1,...n$$

I got confused with this notation: $\alpha_{i_j}$ i.e. alpha sub "i" sub "j"

is the summation equal to:

$$\sum_{j=1}^n {a_j \alpha_{i_j}} = \sum_{j=1}^n\sum_{i \in I}{a_j \alpha_i}$$

  • $\begingroup$ What's $I{}{}{}{}$? $\endgroup$
    – MJD
    Jan 31, 2014 at 2:56
  • $\begingroup$ $I$ is an indexing set $\endgroup$
    – Quester
    Jan 31, 2014 at 3:05
  • $\begingroup$ Something isn't sitting right with me, as they're using $j$ as a summation index and $j$ as a free variable in the $\forall$... $\endgroup$
    – apnorton
    Jan 31, 2014 at 3:18
  • $\begingroup$ I corrected the question to show the way it is written. I guess it was bad notation using the index j as a variable with $\forall$ sign $\endgroup$
    – Quester
    Jan 31, 2014 at 3:42

1 Answer 1


$I$ is not necessarily the set $\{1, 2, 3, \ldots n\}$. The $\alpha_i$ are vectors, but they might not be indexed with natural numbers, because in general $S$ might not be countable. So the elements of $S$ are indexed with some set, possibly some uncountable set, and each $\alpha_i$ is a vector from $S$, where $i\in I$.

Then we select some finite subset of the vectors from $S$, first by choosing a finite subfamily of the index set $I$, namely $\{i_1, i_2, \ldots, i_n\}\subset I$, so that each of the $i_j$ is an element of $I$. Then we pick the elements of $S$ that correspond to this finite subset of $I$, that's $\{\alpha_{i_1}, \alpha_{i_2}, \ldots, \alpha_{i_n}\}\subset S$.

Having selected a finite subset $\{\alpha_{i_1}, \alpha_{i_2}, \ldots, \alpha_{i_n}\}$ of the vectors from $S$, we form a linear combination of them, which is $$ \sum_{j=1}^n a_j\alpha_{i_j}.$$

In particular, this does not mean the double sum that you suggested.

I share user @anorton's puzzlement about the $\forall j$, and wonder if the text contains an error, of if you made a transcription error.


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