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I feel so lost in this section regarding vector spaces, sub spaces, spanning sets and basis.

I understand the basic concepts regarding vectors where basically a vector gives magnitude and direction. Multiplying a scalar by a vector basically scales the vector by the amount multiplied, and adding vectors is a matter of adding the components of the vectors to give a new vector.

I don't understand how all of this ties to the above examples. I'm working problems and find myself kind of understanding what I'm doing, but then I hit a problem like the one below:

Determine whether the set $S$ spans $R^2$. If the set does not span $R^2$, then give a geometric description of the subspace that it does span.

$S=\{(1,3),(-2,-6),(4,12)\}$

I created a linear combination of these vectors and found that they produce an infinite amount of solutions, but I don't know what to do with this information nor what it means.

I think this is because my underlying conceptual understanding is not strong. Can someone help me to understand what is happening in all of this?

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A set of vectors in $R^2$ either spans a line or all of $R^2$ (ignoring the trivial case of the set consisting of just the $0$ vector). In your particular example, you are spanning a line. Notice how all three vectors are scalar multiples of each other, all of them lie on the line $y = 3x$. To check if a set of vectors in $R^2$ span the space it suffices to check that any two vectors in the set are not scalar multiples of each other, if that is true then the set spans $R^2$.

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  • $\begingroup$ All of $R^2$ means everything on the $x$ and $y$ axis, correct? You're saying in my example I only span a line, not the entire space? $\endgroup$ – hax0r_n_code Apr 5 '14 at 2:50
  • $\begingroup$ I think you have the right idea, you just stated it incorrectly. Everything on the x and y-axis is everything on the two lines, which wouldn't be the entire space. Imagine you drew the standard four quadrant picture with the x and y-axises intersecting. Everything means that whole thing shaded in. In your case the span is just the line y = 3x, and not the whole 2D space. $\endgroup$ – Grid Apr 5 '14 at 2:52
  • $\begingroup$ In other words, if a set vectors spans $R^2$, given any arbitrary vector $(a,b)$ in $R^2$ you would be able to write it as a linear combination of the set of vectors in question. In your particular example, you can only 'form' vectors of the form $(a,3a)$ because the span is the line $y = 3x$. $\endgroup$ – Grid Apr 5 '14 at 2:54
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Put it this way - these vectors span R² if I can get any other vector of R² from a linear combination of these.

Instead of what you tried (which is a good idea for the other part of the problem particularly) think about what would be required for me to get all vectors from linearly combining these vectors. Are you familiar with linear independence? What does linear independence mean in our case of R²?

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  • $\begingroup$ If a set is linearly independent that means there is only the trivial solution (I tried to recite from memory). $\endgroup$ – hax0r_n_code Apr 5 '14 at 2:47
  • $\begingroup$ It sounds like you are remembering the gist. The set is linearly independent if the only way to express the zero vector as a linear combination of the vectors in the set is if they are all multiplied by zero (the trivial solution.) With your vectors, I can say that (0,0) = 2(1,3) + (-2,-6) + 0(4,12) so this set is linearly dependent. Now, by definition, a set of m vectors will span a vector space of dimension n for m > n if and only if that set is linearly independent. $\endgroup$ – Jonathan Hebert Apr 5 '14 at 3:09
  • $\begingroup$ The above statement is false. You don't need linear independence to span a space. For instance, the set {(1,0), (0,1), (1,1)} spans $R^2$, but is not a linearly independent set. If you have exactly as many vectors in the set as the dimension of the space the vectors are in, then yes you need the set to be linearly independent to span the space. $\endgroup$ – Grid Apr 5 '14 at 3:17
  • $\begingroup$ *The set must contain a subset of size n that I'd linearly independent $\endgroup$ – Jonathan Hebert Apr 5 '14 at 3:18
  • $\begingroup$ Sorry, yes Grid I just caught that. I don't mean to suggest that a set can stop spanning a space by adding vectors. :-) $\endgroup$ – Jonathan Hebert Apr 5 '14 at 3:19
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Mark two points P, Q in the plane not collinear with the origin, O. Join them with O, to get OP, OQ and extend to infinity both ways. Now from an arbitrary vector (point) R draw lines parallel to OP (resp. OQ) until it intersects OQ (resp. OP). Call the intersection points P' and Q'. So the vector OR is the sum of OP' and OQ' which are scalar multiples of OP and OQ respectively. (Draw the picture).

This process shows that with any two linearly independent vectors (i.e, not scalar multiples of each other) in the plane we can obtain every vector of the plane as their linear combination.

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It will not span R2 because span(S) consist only of line y=3x ie.span(S) is one dimensional not two dimensional.geometrically it represents the line y=3x where all combination of vectors of S lies.

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