Is This Set a Valid Subspace of R^4?

[NewtonianAlch]

Homework Statement

Show that the set:

S = {x \in R^{4}| x = \lambda(2,0,1,-1)^{T} for some \lambda \in R

is a subspace of R^{4}

The Attempt at a Solution

For the subspace theorem to hold, 3 conditions must be met:

1) The zero vector must exist
2) Closed under addition
3) Closed under scalar multiplication

1) If \lambda = 0, the vector becomes (0,0,0,0)^{T} - therefore that's the zero vector.

2) Closure under addition is what I'm a bit confused about.

If we define two new vectors, u and v and two scalars \alpha and \beta respectively.

u + v = ?

3) For closure under multiplication, isn't this obviously already closed? Heck it's being multiplied by a scalar quantity already.

 

[Dick]

Well, if u=a*(2,0,1,-1)^T and v=b*(2,0,1,-1)^T can you show u+v has the form something*(2,0,1,-1)^T? What's the something? Sure 3) is kind of obvious, but you still have to say why.

 

[NewtonianAlch]

The scalar would be (a + b)

So (a + b)*(2,0,1,-1)^{T}

Shouldn't it be (a + b)*(4,0,2,-2)^{T} though ? Because if we do u + v, the vectors would add as well as the scalars.

 

[Dick]

The scalar would be (a + b)

So (a + b)*(2,0,1,-1)^{T}

Shouldn't it be (a + b)*(4,0,2,-2)^{T} though ? Because if we do u + v, the vectors would add as well as the scalars.

The vectors DO NOT add as well as the scalars. a*(2,0,1,-1)^T=(2a,0,a,-a)^T. b*(2,0,1,-1)^T=(2b,0,b,-b)^T. Add them. You don't get (a+b)*(4,0,2,-2)^T, do you? This is the distributive rule with the vector part constant.

 

[NewtonianAlch]

Dang, just when I thought I was getting the hang of this stuff. No, it doesn't become (4,0,2,-2)^{T}

It becomes (2a + 2b, 0, a+b, -a - b)^{T}

So that would actually be (a + b)*(2,0,1,-1)^{T} when you factor it out.

Interesting...I have to go over the notes again.

Thanks a lot for your help Dick.

 

[HallsofIvy]

The scalar would be (a + b)

So (a + b)*(2,0,1,-1)^{T}

Shouldn't it be (a + b)*(4,0,2,-2)^{T} though ? Because if we do u + v, the vectors would add as well as the scalars.

No, that's not even true for numbers: av+ bv= (a+b)v whether v is a vector or a number.