Parametric equation in subspace

[negation]

Homework Statement

The following describes a subset S of R3, you are asked to decide if the subset is a subspace of R3.

x = 1-4t
y = -2-t
z = -2-t

The Attempt at a Solution

R3 = {(1-4t, -2-t, -2-t) | t element of all Real number}If S is a subset, at least one must be true.

1) must contain zero vector
2) Sum of any 2 members in S can be found in S
3) scalar multiple of any member in S can be found in S

y + z = -2-t + (-2-t) =/=0

S is not a subset of R3.

 

[pasmith]

Homework Statement

The following describes a subset S of R3, you are asked to decide if the subset is a subspace of R3.

x = 1-4t
y = -2-t
z = -2-t

The Attempt at a Solution

R3 = {(1-4t, -2-t, -2-t) | t element of all Real number}


If S is a subset, at least one must be true.

S plainly is a subset of \mathbb{R}^3. For S to be a subspace, all three of the following must be true:

1) must contain zero vector
2) Sum of any 2 members in S can be found in S
3) scalar multiple of any member in S can be found in S

On the other hand, if S is not a subspace then at least one of those is false.

y + z = -2-t + (-2-t) =/=0

That doesn't help you.

Concentrate on the first requirement: is there a t \in \mathbb{R} for which 1 - 4t = -2 - t = 0?

 

[negation]

S plainly is a subset of \mathbb{R}^3. For S to be a subspace, all three of the following must be true:



On the other hand, if S is not a subspace then at least one of those is false.



That doesn't help you.

Concentrate on the first requirement: is there a t \in \mathbb{R} for which 1 - 4t = -2 - t = 0?

Yes there is a t = 1

Why doesn't that help me? Zero vector right?

 

[HallsofIvy]

I don't know what you mean by "Zero vector, right?". If you mean you are trying to show that the zerol vector is not in this set then you need to look at (1- 4t, -2- t, -2- t)= (0, 0, 0). That was why pasmith asked "is there a t for which 1- 4t= -2- t= 0?" Showing that y+ z is not identically 0 doesn't tell you anything.

And no, t= 1 does NOT make 1- 4t= -2- t= 0.

 

[negation]

I don't know what you mean by "Zero vector, right?". If you mean you are trying to show that the zerol vector is not in this set then you need to look at (1- 4t, -2- t, -2- t)= (0, 0, 0). That was why pasmith asked "is there a t for which 1- 4t= -2- t= 0?" Showing that y+ z is not identically 0 doesn't tell you anything.

And no, t= 1 does NOT make 1- 4t= -2- t= 0.

There's no t value for which 1-4t = -2-t = 0

That was nonsensical-just feeling really stressed out with all the work.

 

[HallsofIvy]

Yes, there is no value of t for which 1- 4t= -2- t= 0 which means that (1- 4t, -2- t, -2- t) is never equal to (0, 0, 0). The zero vector is not in this set.

 

[negation]

Yes, there is no value of t for which 1- 4t= -2- t= 0 which means that (1- 4t, -2- t, -2- t) is never equal to (0, 0, 0). The zero vector is not in this set.

Understood.