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?
$\endgroup$4 Answers
$\begingroup$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 $R^2$.
$\endgroup$ 4 $\begingroup$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?
$\endgroup$ 7 $\begingroup$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.
$\endgroup$ $\begingroup$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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