Linear Algebra: Determine Span of {(1, 0, 3), (2, 0, -1), (4, 0, 5), (2, 0, 6)}

[Jake 7174]

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.

 

[Mark44]

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

This is more work than you need to do. The fourth vector is a multiple of the first, so it suffices to check whether the first three vectors in your set are linearly independent.

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 |

How did Z turn into 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

The x-z plane is a subset of $\mathbb{R}^3$, so the set of four vectors spans a plane in $\mathbb{R}^3$.

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.

Right. The vectors span a plane in space.

 

[Jake 7174]

This is more work than you need to do. The fourth vector is a multiple of the first, so it suffices to check whether the first three vectors in your set are linearly independent.
How did Z turn into 1?
The x-z plane is a subset of $\mathbb{R}^3$, so the set of four vectors spans a plane in $\mathbb{R}^3$.

Right. The vectors span a plane in space.

Z turns to 1 because of the way a TI89 handles rref of an augmented matrix.

Thank you for your help.