1. May 8, 2005

### Josh123

Given 3 vectors:
v1=(1,1,2)
v2=(1,0,1)
v3=(2,1,3)

To determine if they span R^3, I placed these 3 vectors in an augmented matrix. I found the determinant to be 0 which means that the 3 vectors do NOT span R^3. My question is, since v1,v2,v3 are linearly dependent, are they not supposed to span R^3?

2. May 8, 2005

### HallsofIvy

Yes, if three vectors are independent, then they must span a three dimensional vector space- in this case R3. I am wondering about your use of "augmented" and then "determinant". "Augmenting" this matrix would give you a 4 by 3 matrix that does not have a determinant. Of course, you are correct that the determinant formed by the vectors (not augmented) has determinant 0. Now, are you sure these vectors are independent?

What you may be thinking of is this: to determine whether the vectors are independent or not, you could set up the matrix equation corresponding to saying that some linear combination: av1+ bv2+ cv2= 0 and set up the augmented matrix corresponding to that. If the row reduction of that gives a matrix having a row of all zeros then the vectors are NOT independent. What happens here?

3. May 8, 2005

### Josh123

I got a little mixed up with the linear dependence/independence, but I understand now. Thanx

I have another question though.. just want to make sure if my answer is correct:

Let S= Span {(1,1-3), (2,2,-6) , (-4,-4,12)}.

Is the basis of vectors = {(1,1,-3)(2,2,-6), (-4,-4,12)} ?

4. May 8, 2005

### Hurkyl

Staff Emeritus
If they (1) span that subspace, and (2) are linearly independent, then yes, that is a basis of that subspace.

5. May 8, 2005

### jackdamack10

I have a question that kind of relates tp Josh123's questions about spanning.

Let's say we have a subspace: (2,4,7) (1,3,2)(1,1,5)

Since one of them is a linear combination of the other two, does it mean that we only need two of those vectors to form the basis?

Span{(1,3,2) (1,1,5)}

6. May 8, 2005

### OlderDan

Two of those vectors will span and form the basis for a two dimensional subspace (plane) in the three dimensional space. They can be combined with appropriate multiplying coefficients to form any vector in that plane. The only way to represent a vector that is not in the plane is to have a third vector that is linearly independent of those two.

7. May 8, 2005

### HallsofIvy

If you "understand" the first problem, how could you possibly have a problem with this? Even if you don't want to do the work of setting up a matrix and row reducing, did you notice that (2,2,-6)= 2(1,1,-3) and that
(-4,-4,12)= 4(1,1,-3)?