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The following theorem is from my textbook:

Theorem 1.7. Let $S$ be a linearly independent subset of a vector space $V$, and let $v$ be a vector in $V$ that is not in $S$. Then $S \cup \{v\}$ is linearly dependent if and only if $v \in\operatorname{span}(S)$.

I'm having trouble understanding the proof in the reverse direction. Here's the text:

Conversely, let $v \in\operatorname{span}(S)$. Then there exist vectors $v_1, v_2,\dots, v_m$ in $S$ and scalars $b_1, b_2,\dots, b_m$ such that $v = b_1v_1 + b_2v_2+\dots+b_mv_m.$ Hence $0 = b_1v_1 + b_2v_2 + \dots+ b_mv_m + (-1)v$.

Since $v \neq v_i$ for $i = 1, 2,\dots, m$, the coefficient of $v$ in this linear combination is nonzero, and so the set $\{v_1, v_2,\dots, v_m, v\}$ is linearly dependent. Therefore $S \cup \{v\}$ is linearly dependent.

What does the proof mean when it says "Since $v \neq v_i$ for $i = 1, 2,\dots, m$"? Wasn't it enough to show already that there is a non-trivial representation of $S \cup \{v\} = 0$ to prove that it is linearly dependent?

I'm just unsure why that qualification is necessary. In the case that $v = v_i$ for $i = 1, 2,\dots, m$, then wouldn't it be easy to find non-zero scalars for $v$ and that $v_i$ to prove that it is linearly dependent?

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Saying that $v\ne v_i$ for $1,2,\dots,m$ is wrong: it could well be one of them! The important thing is that $$ 0 = b_1v_1 + b_2v_2 + \dots+ b_mv_m + (-1)v $$ and in this linear combination one of the coefficients is nonzero: precisely the coefficient of $v$.


The statement

If $v\in\operatorname{span}(S)$ then $S\cup\{v\}$ is linearly dependent

is true with no hypothesis on $S$, with the same proof.

It's the converse where linear independence of $S$ is used. Indeed, if $S\cup\{v\}$ is linearly dependent, then there are $v_1,\dots,v_m\in S$ and scalars such that $$ b_1v_1+\dots+b_mv_m+bv=0 $$ with $b,b_1,\dots,b_m$ not all zero. In this case $b=0$ would imply $b_1v_1+\dots+b_mv_m=0$ and so $b_1=\dots=b_m=0$, which is false. So $b\ne0$ and therefore, easily, $v\in\operatorname{span}(S)$.

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  • $\begingroup$ Just a thought: is the qualification necessary because if $v=v_i$, then $S \cup \{v\}$ would still just be $S$, and thus linearly independent by definition? $\endgroup$ – shinify Jan 22 '15 at 19:06
  • $\begingroup$ @shinify It's completely irrelevant: the fact is that $-1\ne0$. It's correct that $v$ cannot belong to $S$, but, as I remarked in the addition, the assertion is true also without the assumption that $S$ is linearly independent. $\endgroup$ – egreg Jan 22 '15 at 20:28

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