# Does the vector set span R3?

by jinksys
Tags: span, vector
 P: 123 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
 P: 738 True! A shortcut: can your bottom row always contain 0s for every value of c?
P: 123
 Quote by Tedjn True! A shortcut: can your bottom row always contain 0s for every value of c?
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.

 Mentor P: 21,286 Does the vector set span R3? Shorter cut: Can a set of two vectors possibly span a three-dimensional space?
 P: 297 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
P: 123
 Quote by Mark44 Shorter cut: Can a set of two vectors possibly span a three-dimensional space?
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?
P: 738
 Quote by jinksys 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 R3 and add whatever else you want to it. Then those n > 3 vectors will also span R3. However, I can also give you 3 or 4 or n vectors all in one plane so that it doesn't span R3. Observe that a plane of vectors is very, very, very, ..., very small compared to the space of all vectors in R3, 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 R3 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 R3, linear independence says that at most 3 vectors can be linearly independent in R3. Together, this leads to the idea that the dimension of R3 is equal to 3. But I'm getting ahead of myself.
P: 123
 Quote by Tedjn You only need 3. However, take any 3 vectors that span R3 and add whatever else you want to it. Then those n > 3 vectors will also span R3. However, I can also give you 3 or 4 or n vectors all in one plane so that it doesn't span R3. Observe that a plane of vectors is very, very, very, ..., very small compared to the space of all vectors in R3, 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 R3 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 R3, linear independence says that at most 3 vectors can be linearly independent in R3. Together, this leads to the idea that the dimension of R3 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?
 P: 738 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: $$\left[\begin{array}{ccc} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 2 & 2 & 2 \\ 3 & 2 & 1 \\ 5 & 2 & ? \end{array}\right]\left[\begin{array}{c} x \\ y \\ z \end{array}\right] = \left[\begin{array}{c} a \\ b \\ c \\ d \\ e \end{array}\right].$$ 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 R5 by 3 vectors in R5. From analogous reasoning as that in the above discussion, we see that this is impossible. We need at least 5 vectors in R5 to span R5. But what you are referring to is probably $$\left[\begin{array}{ccccc}0 & 1 & 2 & 3 & 5 \\ 1 & 0 & 2 & 2 & 2 \\ 0 & 1 & 2 & 1 & ?\end{array}\right]\left[\begin{array}{c}v \\ w \\ x \\ y \\ z\end{array}\right] = \left[\begin{array}{c}a \\ b \\ c\end{array}\right].$$ In this system, the variables v, w, x, y, z tell us how much of each of the 5 vectors in R3 we are supposed to use. If they span R3, 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 R3.

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