Theorem: Let V be a vector space and $S = (v_1, . . . , v_n)$ a spanning sequence of V . Prove that a minimal spanning subsequence of S is linearly independent and a basis of V.

This problem confuses me a bit. What is a minimal spanning subsequence of S?. I tried to google but minimum spanning tree comes out.


Assume to the contrary that a minimal spanning subsequence X of S is linearly dependent, so $\exists v_j \in X$ such that $v_j \in span(X-v_j)$. This implies that there exists a smaller spanning sub-sequence embedded in X which contradicts our assumption of X being a minimal spanning sub sequence. Therefore, X is linearly independent.

If we append a $v_j \in S-X $ in X, then X must become linearly dependent, otherwise if linearly independent then X is not a spanning sub-sequence of S. Since X spans S and S spans V, then X spans V, so X is a basis of V.


1 Answer 1


Suppose that minimal spanning set is not linearly independent, then there's an element $\;v_i\in\{v_1,...,v_n\}\;$ which is a linear combination of the other ones, But then


contradicting the minimality of $\;\{v_1,...,v_n\}\;$

  • $\begingroup$ thanks. Check my new approach. What do you think? $\endgroup$ May 22, 2016 at 8:53
  • $\begingroup$ @TheMathNoob I think it is correct, though the notation used is a little non-standard and can be confusing. I think you meant $$\exists\,v_j\in X\;\;s.t.\;\;v_j\in\text{Span}\left(X\setminus\{v_j\}\right)$$ $\endgroup$
    – DonAntonio
    May 22, 2016 at 8:58
  • $\begingroup$ yes, I was looking for that symbol. Thanks for the clarification $\endgroup$ May 22, 2016 at 8:59
  • $\begingroup$ @TheMathNoob Any time. Nice to see you could basically do it by yourself. :) $\endgroup$
    – DonAntonio
    May 22, 2016 at 9:00

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