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I've been researching for a while and trying to wrap my head around spanning of vector spaces completely (by visualizing them in R3) before moving on to Linear Independence, Basis' and anything else taught after subspaces. (had no calc 3 unfortunately =/)Based on what I'm reading, the span of a set of vectors is every possible linear combination of those vectors. After reading/looking at this figure:

enter image description here

I think I understand it for two vectors in R3. It looks like any two arbitrary vectors (that arent scalar multiples of eachother) in R3 will span a never ending plane through the origin.

Does this mean that the span of any 3 arbitrary vectors in the vector space R3 will form a never ending 3d shape spanning all of R3? (as long as two of them arent scalar multiples of eachother (EDIT: or as long as they don't span R2?)

also, what if you have any random set of more than 3 vectors (that qualify as being in R3), is the span of that set also all of R3? (as long as the vectors dont end up being scalar multiples of eachother so that the set spans a line)(EDIT: or as long as they don't span R2?)

while I'm at it, does this mean that any 4 (or more?) vectors in R4 that arent multiples of any of the other 4 span all of R4? (EDIT:if not R2 or R3?)

if this is all true, I think I had a big epiphany and everything now makes sense to me..... such as being able to visualize linear dependent and linear independent.. I'm guessing that a set of vectors is L.D. if at least one vector can be written as a scalar multiple of another vector in the set. and LI means they are all unique, as in they have no scalar multiples of eachother, which also means that if the only solution to a set of 3 vectors in R3 is (0,0,0) or the homogeneous solution, or the trivial solution.... then that set spans R3? and the same applies to sets of vectors in R4, such as if there is 4 or 5 vectors satsifying R4 rules, and the only solution to their system of equations is homogeneous, they are LI, they span R4, and the 4x4 systems determinant is not equal to 0....

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  • $\begingroup$ can I get a confirmation/good job or not so fast mr.? $\endgroup$ – J L Nov 9 '14 at 3:36
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    $\begingroup$ What is the span of the three vectors $\{(1,1,0), (1,0,0), (0,1,0)\}$, none of which are scalar multiples of each other? $\endgroup$ – Joshua Mundinger Nov 9 '14 at 3:40
  • $\begingroup$ oh.. it looks like the z axis in each of those is 0 so they will only span a plane through the origin.. so R2... $\endgroup$ – J L Nov 9 '14 at 3:55
  • $\begingroup$ so does this mean that the set of 3 or more vectors where each vector is in R3 will span R3 unless their is a row of zeros in the augmented coeficient matrix they form, in which it spans R2, or if their is 2 rows of zeros it spans R1 (a line through the origin)? $\endgroup$ – J L Nov 9 '14 at 4:03
  • $\begingroup$ or basically as long as a set of 3 or more R3 vectors don't span R1(a line) or R2 (a plane through the origin) they will span R3. Is this correct to assume? $\endgroup$ – J L Nov 9 '14 at 4:15

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