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I know minimal spanning set is independent . But how do i prove that a spanning set which is linearly independent i.e a basis is minimal spanning set ? I tried to prove by contradiction by assuming that a set which is smaller and different than the basis set is a minimal set . But this way prove seems to be very lengthy. is it possible that a smaller set containing different independent vectors (other than those is S) is minimal spanning vector ?

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    $\begingroup$ i tried 5 hours to solve this problem . $\endgroup$ – Flintoff Feb 15 '18 at 17:25
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    $\begingroup$ this is not a homework or assignment problem $\endgroup$ – Flintoff Feb 15 '18 at 17:25
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    $\begingroup$ Just from curiosity of my mind $\endgroup$ – Flintoff Feb 15 '18 at 17:26
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Let $S$ be your basis. If it was not a minimal spanning set, there would be a $s\in S$ such that $\bigl\langle S\setminus\{s\}\bigr\rangle$ is the whole space. Then, $s$ can be written as a linear combination $\alpha_1s_1+\cdots+\alpha_ns_n$ of elements of $S\setminus\{s\}$. That is, we have$$1.s-\alpha_1s_1-\cdots-\alpha_ns_n=0,$$ which is impossible, since $S$ is linearly independent.

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    $\begingroup$ but is it possible that a smaller set containing different independent vectors (other than those is S) is minimal spanning vector ? $\endgroup$ – Flintoff Feb 15 '18 at 17:39
  • $\begingroup$ The cardinality of all the bases of a vector space is the same, so it is not possible. $\endgroup$ – Laura Feb 15 '18 at 17:44
  • $\begingroup$ @neraj Are you assuming that $S$ is finite? $\endgroup$ – José Carlos Santos Feb 15 '18 at 17:44

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