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Let $c_1,c_2,c_3,c_4,c_5$ be vectors in $R^4$

I'm trying to show that the set (call it set A) {$c_1,c_2,c_3,c_4,c_5$} spans $R^4$ if and only if the set (say, set B) of vectors

{$c_1+c_2,c_2+c_3,c_3+c_4,c_4+c_5,c_5+c_1$} spans $R^4$

I tried a proof by contradiction to show that there can't exist a vector b that is formed by a linear combination of vectors from B but not from A, but that doesn't seem to be the case. Am I messing something up somewhere/is there something I could be doing differently, possibly with matrices?

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2 Answers

up vote 3 down vote accepted

Hint:

$ (c_1 + c_2) = (c_1) + (c_2)$

$ (c_1) = \frac {1}{2} \left[ (c_1 + c_2) - (c_2+c_3)+(c_3+c_4) - (c_4+c_5) + (c_5 + c_1)\right]$

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Sorry, I'm not quite sure how that's supposed to help. Could you elaborate a bit further, please? –  user1903336 Jan 29 '13 at 8:05
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@user1903336 Can you show that $Span B \subset Span A$ using the first equation (and it's cyclic versions)? Similarly show that $Span A \subset Span B$ using the second equation. Hence conclude that $Span A = Span B$. –  Calvin Lin Jan 29 '13 at 8:09
    
My fundamentals are very fuzzy, sorry. Do you mind explaining yourself a bit more? –  user1903336 Jan 29 '13 at 8:15
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If $b \in Span B$, then $b = \sum b_i ( c_i + c_{i+1})$ (where $c_6 = c_1)$ for some coefficients $b_i$. Then, $b = \sum (b_i + b_{i-1} )c_i$ shows that $b \in Span A$. Thus, $Span B \subset Span A$. Do the second statement in a similar way. –  Calvin Lin Jan 29 '13 at 8:19
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The hard part would be finding the equations stated in the hint. The rewrite should be obvious, by just expanding and combining terms. –  Calvin Lin Jan 29 '13 at 8:46
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You can actually prove something stronger, that A and B always generate the same vector space, by showing that every vector in A can be written as a linear combination of vectors in B, and every vector in B is a linear combination of vectors in A.

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Can you explain why that would work/how I would go about doing that? –  user1903336 Jan 29 '13 at 8:16
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