- #1
Jake 7174
- 80
- 3
Homework Statement
Determine whether the set spans ℜ3. If the set does not span ℜ3 give a geometric description of the subspace it does span.
s = {(1, 0, 3), (2, 0, -1), (4, 0, 5), (2, 0, 6)}
Homework Equations
The Attempt at a Solution
I am having trouble with the second part of this problem (geometric shape). First off here is what I did to show that S does not span ℜ3.
First I define A, B, C, D as scalars and say;
A + 2B + 4C + 2D = X
0 + 0 + 0 + 0 = Y
3A - B + 5C + 6D = Z
I set up the augmented matrix
| 1 2 4 2 : X |
| 0 0 0 0 : Y |
| 3 -1 5 6 : Z |
rref (using ti89 titanium) ~
| 1 0 2 2 : 0 |
| 0 1 1 0 : 0 |
| 0 0 0 0 : 1 |
I can conclude from this result that S does not span ℜ3.
Now for the geometric shape.
I immediately see that the first and last vectors in the set are scalar multiples which means that they aer parallel.
Next I note that for all vectors y=0. My first thought is that the geometric shape is the plane formed by the x z axis to check I create a new set in ℜ2 on the xz plane
W = {(1, 3), (2, -1), (4, 5), (2, 6)}
Then repeat the above process to see is W spans xz plane
My augmented matrix becomes
| 1 2 4 2 : X |
| 3 -1 5 6 : Z |
rref (using ti89 titanium) ~
| 1 0 2 2 : (X + 2Z) / 7 |
| 0 1 1 0 : (3X - Z) / 7 |
From this I can say that W spans xz in ℜ2.
Is this correct? If not, where am I lacking in my understanding? My big hold up is the only example we have seen is a line. I am not convinced that this is not a line in space.
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