# More Linear Algebra Help Needed.

**URL:** <https://boards.straightdope.com/t/more-linear-algebra-help-needed/134311>\
**Category:** Factual Questions\
**Created:** [October 29, 2002, 12:02am UTC](https://boards.straightdope.com/t/more-linear-algebra-help-needed/134311 "2002-10-29T00:02:52Z")\
**Posts on this page:** 5\
**Page:** 1

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**Author:** ![Muad\_Dib](https://avatars.discourse-cdn.com/v4/letter/m/35a633/32.png) [@Muad\_Dib](https://boards.straightdope.com/u/Muad_Dib)\
**Post date:** [October 29, 2002, 12:02am UTC](https://boards.straightdope.com/t/more-linear-algebra-help-needed/134311/1 "2002-10-29T00:02:52Z")

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How would this be proven?

> [@](#):
>
> If S={ **v** [sub]1[/sub], **v** [sub]2[/sub], …, **v** [sub]r[/sub]} and S’={ **w** [sub]1[/sub], **w** [sub]2[/sub], …, **w** [sub]r[/sub]} are two sets of vectors in a vector space V, then
> 
> span{ **v** [sub]1[/sub], **v** [sub]2[/sub], …, **v** }= span{ **w** [sub]1[/sub], **w** [sub]2[/sub], …, **w** }
> 
> If and only if each vector in S is a linear combination of those in S’ and each vector in S’ is a linear combination of those in S.

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**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\
**Post date:** [October 29, 2002, 12:32am UTC](https://boards.straightdope.com/t/more-linear-algebra-help-needed/134311/2 "2002-10-29T00:32:21Z")

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Remember that a set of vectors is in its own span, and that the span of a set of vectors is the set of all linear combinations of vectors in that set. That should help you.

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**Author:** ![Ring](https://avatars.discourse-cdn.com/v4/letter/r/6a8cbe/32.png) [@Ring](https://boards.straightdope.com/u/Ring)\
**Post date:** [October 29, 2002, 2:05am UTC](https://boards.straightdope.com/t/more-linear-algebra-help-needed/134311/3 "2002-10-29T02:05:04Z")

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Suppose that span {v1,v2,?..vr} = span {w1,w2?.wk}. Since each vector vi in S belongs to span {v1,v2,?..vr}, it must by the definition of span{w1,w2,?..wk} be a linear combination of the vectors in S?. The converse must also hold.

Suppose that each vector in S is a linear combination of those in S? and conversely. Then we can express each vector v as a linear combination of the vectors w1, w2,?wk, so span{v1,v2,?vr} C\_ span{w1, w2,?wk}. But conversely we have  
span{w1,w2,?wk} C\_ span{v1,v2,?vk}, so the two sets are equal.

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**Author:** ![Ring](https://avatars.discourse-cdn.com/v4/letter/r/6a8cbe/32.png) [@Ring](https://boards.straightdope.com/u/Ring)\
**Post date:** [October 29, 2002, 2:55am UTC](https://boards.straightdope.com/t/more-linear-algebra-help-needed/134311/4 "2002-10-29T02:55:50Z")

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Jeez Us where did all those question marks come from?

Suppose that span {v1,v2,…,vr} = span {w1,w2…,wk}. Since each vector vi in S belongs to span {v1,v2,…,vr}, it must by the definition of span{w1,w2,…,wk} be a linear combination of the vectors in S’. The converse must also hold.

Suppose that each vector in S is a linear combination of those in S’ and conversely. Then we can express each vector v as a linear combination of the vectors w1, w2,…,wk, so span{v1,v2,…,vr} C span{w1, w2,…,wk}. But conversely we have span{w1,w2,…,wk} C span{v1,v2,…,vk}, so the two sets are equal.

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**Author:** ![Ring](https://avatars.discourse-cdn.com/v4/letter/r/6a8cbe/32.png) [@Ring](https://boards.straightdope.com/u/Ring)\
**Post date:** [October 29, 2002, 3:01am UTC](https://boards.straightdope.com/t/more-linear-algebra-help-needed/134311/5 "2002-10-29T03:01:23Z")

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Oh God help me, I meant to hit preview not submit. I wanted to add that the above proof is right from a textbook.