stuck_in_mud Posted March 22, 2004 Share Posted March 22, 2004 hello, i'm studying vectors, and i have come across spanning, linear independency and bases. i have two bases in v3® which are 1) (1,0,0) , (0,1,0) , (0,0,1) 2) (1,0,3) , (2,1,4) , (1,0,0) i have shown these to be linearly independent as this is one of the points that must be met for a base to occur. The second point is to show that they are spans of their own set, for this i know they must have some linear combination. This is the bit that is confusing me, how can i show the linear combinations, basically how do i prove the second point of having a base. Any ideas? Link to comment Share on other sites More sharing options...

Dave Posted March 23, 2004 Share Posted March 23, 2004 You've actually done most of the work by yourself. The linear independence is the hardest bit really. Basically if a set of vectors span a vector space, this means that if you take any element in the vector space, you can use your set to form that particular element in the vector space. I'll do the standard basis for you (the first one) and leave you to do the other one. Consider the general case, for some element (a,b,c) in R^{3}. If (1,0,0),(0,1,0),(0,0,1) span the vector space, then we have: p(1,0,0) + q(0,1,0) + r(0,0,1) = (a,b,c) We want to show that the constants p, q and r actually exist, otherwise the vectors wouldn't span the space. I think it's pretty obvious that the co-efficients actually exist (p = a, q = b, r = c) so the vectors span R^{3}. The second set of vectors can be proven to span by a similar method. Hope this helps. Link to comment Share on other sites More sharing options...

matt grime Posted March 29, 2004 Share Posted March 29, 2004 If you have a set of n linearly independent vectors in R^n they span. Link to comment Share on other sites More sharing options...

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