Is This Set a Valid Subspace of R^4?

In summary: For vectors, v= (x,y,z) and for numbers v= (2,3,5). 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?
  • #1
NewtonianAlch
453
0

Homework Statement


Show that the set:

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

is a subspace of R[itex]^{4}[/itex]


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 [itex]\lambda[/itex] = 0, the vector becomes (0,0,0,0)[itex]^{T}[/itex] - 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 [itex]\alpha[/itex] and [itex]\beta[/itex] respectively.

u + v = ?

3) For closure under multiplication, isn't this obviously already closed? Heck it's being multiplied by a scalar quantity already.
 
Physics news on Phys.org
  • #2
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.
 
  • #3
The scalar would be (a + b)

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

Shouldn't it be (a + b)*(4,0,2,-2)[itex]^{T}[/itex] though ? Because if we do u + v, the vectors would add as well as the scalars.
 
  • #4
NewtonianAlch said:
The scalar would be (a + b)

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

Shouldn't it be (a + b)*(4,0,2,-2)[itex]^{T}[/itex] 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.
 
  • #5
Dang, just when I thought I was getting the hang of this stuff. No, it doesn't become (4,0,2,-2)[itex]^{T}[/itex]

It becomes (2a + 2b, 0, a+b, -a - b)[itex]^{T}[/itex]

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

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

Thanks a lot for your help Dick.
 
  • #6
NewtonianAlch said:
The scalar would be (a + b)

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

Shouldn't it be (a + b)*(4,0,2,-2)[itex]^{T}[/itex] 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.
 

FAQ: Is This Set a Valid Subspace of R^4?

What is a subspace?

A subspace is a subset of a vector space that satisfies the three properties of a vector space: closure under addition, closure under scalar multiplication, and containing the zero vector.

How do you prove a set is a subspace?

To prove a set is a subspace, you must show that it satisfies the three properties of a vector space: closure under addition, closure under scalar multiplication, and containing the zero vector. This can be done by showing that the set contains all possible combinations of its vectors using addition and scalar multiplication.

What is closure under addition?

Closure under addition means that when two vectors from a set are added together, the resulting vector is also in the set. In other words, the set is closed under the operation of vector addition.

What is closure under scalar multiplication?

Closure under scalar multiplication means that when a vector from a set is multiplied by a scalar, the resulting vector is also in the set. In other words, the set is closed under the operation of scalar multiplication.

What is the zero vector and why is it important in proving a set is a subspace?

The zero vector is a vector with all of its components equal to zero. It is important in proving a set is a subspace because it is required to be in the set for it to be closed under addition and scalar multiplication. Additionally, the zero vector is necessary for the existence of the additive identity element, which is a property of vector spaces.

Similar threads

Vector Subspaces: Determining U as a Subspace of M4x4 Matrices
Replies
8
Views
1K
Can a Finite Set of Vectors Form a Subspace?
Replies
4
Views
2K
I Showing that a set of differentiable functions is a subspace of R
Replies
19
Views
5K
I Vector subspace and basis vectors in the context of data science
Replies
6
Views
2K
[Linear Alg] Determining which sets are subspaces of R[x]
Replies
7
Views
2K
MHB Are These Sets Vector Subspaces of R^3?
Replies
5
Views
1K
Finding Linearly Independent Vectors in Subspaces
Replies
6
Views
2K
Vector Subspaces: Understanding Closure Properties
Replies
2
Views
2K
I When is a subset a subspace in a vector space?
Replies
7
Views
2K
[Linear Algebra] Another question on subspaces
Replies
15
Views
1K

Hot Threads

Recent Insights

Back
Top