Definition talk:Linearly Independent/Set

In case of infinite dimensional vector spaces it is actually essential that $\{a_1, \ldots, a_k\} \subseteq S$, not equality. This is also the definition (you can get away with equality iff $S$ is finite; a lin. indep. set needn't be finite). --Lord_Farin 04:32, 23 May 2012 (EDT)

The case $\R^n$ formally also needs to incorporate infinite sets (though they will all be linearly dependent, we still want to define that they are...). --Lord_Farin 04:33, 23 May 2012 (EDT)


 * Yes I had half a braincell on that when I was looking at it this morning. Again, it's a case of "blame the source work" which only considers finite cases. It's wrong anyway, I'll have to go back and revisit it. I was short of time this morning. --prime mover 09:01, 23 May 2012 (EDT)