From Terence Tao's Linear Algebra notes:

We sat that any collection $S$ of vectors in a vector space $V$ are linearly dependent if we can find distinct elements $v_1,...,v_n\in S$, and scalars $a_1,...,a_n$, not all equal to zero, such that $$a_1v_1+a_2v_2+...+a_nv_n=0.$$ ... In the case where $S$ is a finite set $S=\{v_1,...,v_n\}$, then $S$ is linearly dependent if and only if we can find scalars $a_1,...,a_n$ not all zero such that $$a_1v_1+...+a_nv_n=0.$$ (Why is the same as the previous definition? It's a little subtle).

I can recognize that in the first definition we have only "if ... then ..." whereas the second one is an equivalence "... if and only if ..."

But where does the difference really lie? What is the edge case that makes the first definition not an equivalence?

  • 3
    $\begingroup$ Both are supposed to be "if and only if". The difference between the infinite case and the finite case is that, in the infinite case, you pick a finite collection $v_1, \dots, v_n$ from $S$, whereas in the finite case, you use ALL of the $v_i$'s in $S$. $\endgroup$ – Kenny Wong May 3 '17 at 8:49
  • $\begingroup$ @KennyWong I don't think that's what Terence meant when he said "Why is the same as the previous definition? It's a little subtle" $\endgroup$ – user442916 May 3 '17 at 8:50
  • $\begingroup$ Well, if $S = \{ v_1, \dots, v_n \}$ is a finite set, the first definition says that $S$ is linearly dependent iff we can find $a_1, \dots, a_k$ (not all zero) such that $a_1 v_1+ \dots + a_k v_k = 0$ (possibly after renumbering the $v_i$'s). But then, if we set $a_{k+1} = \dots a_n = 0$, we have $a_1 v_1+ \dots + a_n v_n = 0$ too. Or did I miss something? $\endgroup$ – Kenny Wong May 3 '17 at 8:53
  • $\begingroup$ @KennyWong It seems to me that you violated the definition when you set $a_{k+1}=...=a_n=0$ "... and scalars $a_1,...,a_n$, not all equal to zero..." $\endgroup$ – user442916 May 3 '17 at 8:55
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    $\begingroup$ I'm pretty sure that Tao meant "if and only if". People very often write "if" when they mean "if and only if" in definitions. $\endgroup$ – Kenny Wong May 3 '17 at 9:02

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