Denote $[S]$ as the span of $S$.
Let $S$ be a subset of a vector space $V$ and $\alpha \in V$. Then $\alpha \in [S]$ if and only if $[S]=[S\backslash \{\alpha\}]$.
Characterization of the span of a set
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linear-algebra
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3Incorrect as written. If $V=R^2$ and $S=\{(1,0),(0,1)\}$ and $\alpha = \{(1,0)\}$, $\alpha \in V$ but $[S]=R^2=V$ and $[S-\{\alpha\}]=[\{(0,1)\}] \neq R^2=V$. – 2012-12-11
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2You certainly don't want $\alpha\in V$ as hypothesis and the showing "$\alpha\in V$ if and only if ...". You probably mean $\alpha\in[S]$ before the "if and only if", but even then it does not make real sense; please say what you want to ask. – 2012-12-11
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1What does $S-\{\alpha\}$ mean? (a) $S\setminus\{\alpha\}$ or (b) $\{v-\alpha: v\in S\}$? – 2012-12-11
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0My bad.I edited the question above sir. – 2012-12-11
1 Answers
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Maybe you meant to conclude "$\alpha\in[S]$ if and only if $[S]=[S\cup\{\alpha\}]$", which is correct. But you cannot expect people to read your mind, so please state more carefully.
For the edited question "$\alpha \in V$. Then $\alpha \in [S]$ if and only if $[S]=[S\backslash \{\alpha\}]$" it is certainly incorrect; for one thing if $\alpha\notin S$ then $[S]=[S\backslash \{\alpha\}]$ holds trivially, regardless of whether $\alpha\in [S]$, while if $\alpha\in S$ then $\alpha \in [S]$ holds trivially, regardless of whether $[S]=[S\backslash \{\alpha\}]$.
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0My bad.I edited the question above sir. – 2012-12-11
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0Thank you sir. I appreciate this post. – 2012-12-11