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 Quote by JameB Sorry, I don't mean to be rude but I'm just kinda lost. This is my first course in linear algebra and that's why I'm pretty confused. What I was trying to do was to find a basis.
I don't mean to be rude but this is your 6th post in this thread and that is the first mention of finding a basis!

 The theorem states that vectors are said to form basis for V if a) if they span V b) they are linearly independent
That's not a theorem, that is the definition of "basis". You can add a "c" to this- the number of vectors in the set is the dimension of the space. And then there is a theorem that says any two implies the third.

 I know that to prove b) I need to put it in a matrix, reduce and if I get a matrix with trivial solution meaning everything equals to 0 then it's a trivial solution and it's linearly independant.
Well, no, you don't need to do that- that's one method. What you need to do is think about the definition: a set of vectors spans a space if and only if, any vector, y, in the space can be written as a linear combination of the vectors in the set. If the set is $\{v_1, v_2, \cdot\cdot\cdot, v_n\}$ then there exist scalars, $a_1, a_2, \cdot\cdot\cdot, a_n$ such that $y= a_1v_1+ a_2v_2+ \cdot\cdot\cdot+ a_nv_n$.

 But I'm having trouble with checking if they span or not... that's the whole point of this post :) The examples in my book don't show step by step solution so that's why I'm lost. e.g. Show that S = {v1, v2, v3, v4} where v1 = [1 0 0 1], v2 = [0 1 -1 2], v3 = [0 2 2 1], v4 = [1 0 0 1] is a basis for R4. (R4 is written as R subscript 4 meaning it's referring to the row spaces?...) to show that S is linearly independant, I formed the eqn a1v1 + a2v2 + a3v3 + a4v4 = 0 turned it into a augmented matrix and got its RREF. This means that it's linearly independent.
Then you have a problem- either you did the problem wrong or you gave the vectors incorrectly here. Those vector are NOT independent because $v_1$ and $v_2$ are the same: $(1)v_1+ (-1)v_2= 0$.

 NOW, to show that S spans R4, I let x = [a, b, c, d] be a vector in R4 what should I do next? thanks!
If those vectors were actually the ones you were given, you do nothing now. Since there are 4 vectors and the are NOT indpendent, they cannot span R4. If you gave the fouth vector incorrectly and the ones you were given were independent, you still would not have to do anything more- four independent vectors must span R4.

In either case, showing linear independence or spanning, you are, in effect, solving a system of equations. You are either trying to solve $a_1v_1+ a_2v_2+ a_3v_3= 0$ (for linear independence) or $a_1v_1+ a_2v_2+ a_3v_3+ a_4v_4= y$ for any y in the space (for spanning). You could do either by setting up the "augmented" matrix with the right side as the "fifth column". Of course, with all "0"s in the fifth column, no row operations will change those: $a_1= a_2= a_3= a_4= 0$ is an obvious solution. Showing that you can reduce the first four columns to the identity matrix is enough to show that is the only solution. Similarly, to show that the set spans the space you add the components of the vector y as the fifth column. You would row-reduce the matrix in exactly the same way, just doing the row operations on the fifth column on the y- components rather than "0"s. But the actual value of the coefficients is not important, only that you can find them. So the values you wind up with in the fifth column is not important, only that you can reduce the first four columns to the identity matrix- exactly what you did to show independence. They really are exactly the same thing.

When someone said you could look at the determinant of the matrix, earlier, you asked what to do if the matrix is not square. In that case, there is not work to do- such a set of vectors cannot be a basis (though it may still span the space or be independent). The number of columns of the matrix, the number of components in each vector, is the dimension of the vector space. The number of rows is the number of vectors in the set. And those must be equal in order to have a basis. If there are fewer vectors than the dimension- if there are more columns than row- they might be independent but cannot span the space. If there are more vectors than the dimension- if there are more rows than columns, they might span the space but cannot be independent.