1. The problem statement, all variables and given/known data Does the vector set span R3? (1,-1,2) and (0,1,1) 2. Relevant equations I'm assuming I set up a matrix... 1 0 a -1 1 b 2 1 c then solve for rref? If my bottom row doesn't contain all zeros, does this mean the vectors do not span r3? 3. The attempt at a solution
I'm goes to guess not, right? So unless it does, the vectors don't span r3? Let's say I was able to solve this matrix and got answers for c1, c2, and c3, does that mean the vectors span r3? Sorry if these are elementary questions, my book is awful.
You need three vectors to span R3, you have two so the answer is no. To your second question, if you have three vectors and rref, the set spans R3 if you have three pivots. If your last row is only zeros then the set does not span R3. Its a years since I took Linear Algebra so I give no guaranties
Ok, that makes sense. So do I need at least three or exactly three? What if I am asked if four different vectors span R3?
You only need 3. However, take any 3 vectors that span R^{3} and add whatever else you want to it. Then those n > 3 vectors will also span R^{3}. However, I can also give you 3 or 4 or n vectors all in one plane so that it doesn't span R^{3}. Observe that a plane of vectors is very, very, very, ..., very small compared to the space of all vectors in R^{3}, so if I choose 3 vectors at random, I am almost certainly going to get three that span the space. The reason that you need at least 3 to span R^{3} is as you stated in your first post. The reason you only need 3 is because of an example, namely [1 0 0], [0 1 0], and [0 0 1]. A related concept to span you will encounter soon if not already is that of linear independence. This is complementary to the idea of span, because while span says at least 3 vectors are needed to span R^{3}, linear independence says that at most 3 vectors can be linearly independent in R^{3}. Together, this leads to the idea that the dimension of R^{3} is equal to 3. But I'm getting ahead of myself.
So say I pick 5 vectors at random, and they all happen to span R3, when I construct my augmented matrix which may look like this... 0 1 0 a 1 0 1 b 2 2 2 c 3 2 1 d 5 2 e e If the vectors span R3, does that mean the bottom two columns will consist of zeros when I am done solving for rref?
If you meant the bottom two rows consisting of all zeros, not quite. Since you can choose anything for d and e, there's no way that row reduction will always give you two rows of zeros at the bottom. Recall what this augmented matrix is short for: [tex]\left[\begin{array}{ccc} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 2 & 2 & 2 \\ 3 & 2 & 1 \\ 5 & 2 & ? \end{array}\right]\left[\begin{array}{c} x \\ y \\ z \end{array}\right] = \left[\begin{array}{c} a \\ b \\ c \\ d \\ e \end{array}\right].[/tex] This can be treated as a linear combination of the columns, how many of each column we use is given by x, y, and z. So, you are trying to reach anything in R^{5} by 3 vectors in R^{5}. From analogous reasoning as that in the above discussion, we see that this is impossible. We need at least 5 vectors in R^{5} to span R^{5}. But what you are referring to is probably [tex]\left[\begin{array}{ccccc}0 & 1 & 2 & 3 & 5 \\ 1 & 0 & 2 & 2 & 2 \\ 0 & 1 & 2 & 1 & ?\end{array}\right]\left[\begin{array}{c}v \\ w \\ x \\ y \\ z\end{array}\right] = \left[\begin{array}{c}a \\ b \\ c\end{array}\right].[/tex] In this system, the variables v, w, x, y, z tell us how much of each of the 5 vectors in R^{3} we are supposed to use. If they span R^{3}, then no matter what a, b, and c are, we never get an inconsistent system. In row reducing the augmented matrix (which has 6 columns), we never get all zeros in the first 5 columns of any row. Or else we've seen from above that a bad choice of c means our system is inconsistent. But this is not true for any 5 vectors, only 5 vectors that span R^{3}.