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Proving that a minimal spanning set is linearly independent.

Show that the two statements are equivalent:

  1. $\textbf{S}$ spans $\textbf{V}$ but any proper subset of $\textbf{S}$ does not span $\textbf{V}$.
  2. $\textbf{S}$ is linearly independent and any subset of $\textbf{V}$ properly containing $\textbf{S}$ is linearly dependent.

What I have tried for the first part is a proof by contradiction, can someone please tell me if this is right or not:-

Let $\textbf{S} = \{v_1,v_2,\ldots,v_n\}$ Assume that $\textbf{S}$ is linearly dependent. This implies that there must be at least one vector say $v_k$ with $1 \leq k \leq n$ such that $v_k$ can be expressed as a linear combination of $S'=\{v_1,\ldots,v_{k-1},v_{k+1},\ldots,v_n\}$ but then Span{$\textbf{S}$}=Span$\{v_1,\ldots,v_{k-1},v_{k+1},\ldots,v_n\}$=V. This is a contradiction since $S' \subset \textbf{S}$ Hence assumption is false and $\textbf{S}$ is linearly independent. Is this correct so far?

How would I go about showing the second part? In my opinion it's not always true, since you can have a vector say $v_{n+1}$ which is a linear combination of the vectors in S, and then a subset of V say $\textbf{S}\cup v_{n+1}$ will not be linearly independent.

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  • $\begingroup$ why do you assume $S$ is finite? $\endgroup$ – Ittay Weiss Sep 11 '14 at 20:19
  • $\begingroup$ for the purpose of the question we can assume that $S$ is finite. $\endgroup$ – Pablo Sep 11 '14 at 20:21
  • $\begingroup$ You have proved half of (1)$\implies$(2) correctly, but the last word in condition 2 should be “dependent”. $\endgroup$ – egreg Sep 11 '14 at 20:22
  • $\begingroup$ So that would be 2. $\textbf{S}$ is linearly independent and any subset of $\textbf{V}$ containing $\textbf{S}$ is linearly dependent? Would would the case be when that subset is $\textbf{S}$? $\endgroup$ – Pablo Sep 11 '14 at 20:25
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    $\begingroup$ Any subset of $V$ properly containing $S$ is linearly dependent. Note that $V$ contains $S$ and is surely not linearly independent. $\endgroup$ – egreg Sep 11 '14 at 20:27
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I assume that property 2 reads

$S$ is linearly independent and any subset of $V$ properly containing $S$ is linearly dependent.

Property 2 as written in the question is false for every set $S$, because $S\cup\{0\}$ is a subset of $V$ containing $S$ and is linearly dependent.

I'll also assume as known that, if $S$ is a subset of $V$, $v\in S$ and $v$ is a linear combination of $S\setminus\{v\}$, then $\operatorname{Span}(S)=\operatorname{Span}(S\setminus\{v\})$.


(1)$\implies$(2) Assume (1) and suppose $S$ is linearly dependent. This means that there is $v\in S$ such that $v$ is a linear combination of $S\setminus\{v\}$. But then $S\setminus\{v\}$ is a spanning set of $V$, contradicting (1). [This is already in the question and is correct.]

Let $T$ be a subset of $V$ properly containing $S$. If $v\in T\setminus S$, then $v$ is a linear combination of $S$, which implies $v$ is a linear combination of $T\setminus\{v\}$, so $T$ is linearly dependent.

(2)$\implies$(1) Assume (2). Let $v\in V$; if $v\notin S$, then $S\cup\{v\}$ is linearly dependent by hypothesis. Therefore there are $v_1,v_2,\dots,v_n\in S$ and scalars $\alpha,\alpha_1,\dots,\alpha_n$, not all zero, such that $$ \alpha v+\alpha_1v_1+\dots+\alpha_nv_n=0 $$ But $\alpha=0$ would also imply $\alpha_1=\alpha_2=\dots=\alpha_n=0$, contrary to the hypotheses. Hence $\alpha\ne0$ and $v\in\operatorname{Span}(S)$.

It remains to show that no proper subset of $S$ is a spanning set. Suppose $T\subsetneq S$ is a spanning set. Then $S$, which properly contains a spanning set, is linearly dependent.

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