Proving Subspace Membership in R^2 using Vector Dot Product

In summary: Thus, I don't believe you understand that point, and thus I can't give you full credit for this problem.In summary, the conversation involves a question about whether the set S = { x is in R^2 | x dot a = 0} is a subspace, with the vectors x and a being discussed. The conversation also includes a discussion of how to show that S is closed under addition and scalar multiplication, with various proofs being suggested and critiqued. The final conclusion is that the proof of scalar multiplication is lacking and the use of proper notation and logical steps is emphasized.
  • #1
frasifrasi
276
0
The question I am looking at asks,

Let a be the vector (-1,2) in R^2. Is the set S = { x is in R^2 | x dot a = 0}
a subspace?

--> x and a are vectors...


Can anyone explain how to show this?

I was thinking that since the zero vector is in R^2, this must also be a subspace.

Thank you.
 
Physics news on Phys.org
  • #2
Since 0 is in S, you can't conclude that S that S is not a subspace.

Is it a subspace? To aswer this, you must show that S is closed under addition and scalar multiplication. It is pretty direct, try it.

If you run into problem, I can help you.
 
  • #3
Oh yes.
Ok, but how would I proceed.

Do i just look at 0 -- k(0) = (k0).

I am not sure how to apply the axiums, can you/someone demonstrate?

Thank you.
 
  • #4
I'll do closure under addition for you.

Let x and y be in S. We want to show that x+y is in S also.

well (x+y) dot a = (x dot a) + (y dot a) = 0 + 0 = 0,

so x+y is in S!

(I used the distributivity property of the scalar product)
 
  • #5
For the constant,

k(x dot a) = (kx dot a)
k(0) = 0
0 = 0

Is this the proper way?
Thanks for the help.
 
  • #6
The exposition is not very logical.

Try again using my template.

"Let x be in S and k be a constant. We want to show..."
 
  • #7
Why do you keep looking at 0? Not only does just proving that 0 is in the set not prove it is a subspace, you don't need to prove that 0 is in the set!

You need to prove two things: the set is closed under addition and the set is closed under scalar multiplication.

Suppose u and v are two vectors such that [itex]u\cdot a= 0[/itex] and [itex]v\cdot a= 0[/itex] where a is a fixed vector. Can you prove that u+ v is also in that set- that [itex](u+v)\cdot a= 0[/itex]?

Suppose u is a vector such that [itex]u\cdot a= 0[/tex] and [itex]\alpha[/itex] is any real number. Can you prove that [itex]\alpha[/itex] is also in that set- that [itex]\alpha u\cdot a= 0[/itex]?

After you have proved the second of those, you don't need to prove 0 is in the set- it follows from the fact that 0v= 0 for any v.
 
  • #8
so, au dot a = 0

u dot a = 0
so, u(0) = 0.

Is that enough?

thank you for the patience everyone.
 
  • #9
frasifrasi said:
so, au dot a = 0

u dot a = 0
so, u(0) = 0.

Is that enough?

thank you for the patience everyone.
I have no idea what you mean by this. You started with a a vector so I don't know what you mean by "au dot a= 0". Did you accidently also use a to mean a number?

What do you mean by u(0)? u is a vector, not a function.

If I try to translate, I get "If ku dot a= 0, then u dot a= 0".
That's exactly the opposite of what I suggested: Show that if u dot a= 0, then ku dot a= 0 for any real number k.

And, of course, you must prove that if u dot a= 0 and u dot v= 0, then (u+ v) dot a= 0.
 
  • #10
like the poster above mentioned,

(u+v) dot a = 0
then

(u dot a) + (v dot a) = 0
0 + 0 = 0 so, that is closed under addition.

For scalar multiplication, yes I meant k:

k(u dot a) = k.0 = 0

ku dot a = k.0 = 0

Is this the right way to prove it?
 
  • #11
Maybe you understand, but I wouldn't know because your exposition lacks words or at least "<==>" symbols for it to make any sense.

In your proof of scalar multiplication, the critical part of the proof is the fact that ku dot a = k(u dot a). I see you nowhere make that connection explicitely.
 

Related to Proving Subspace Membership in R^2 using Vector Dot Product

1. What is linear algebra?

Linear algebra is a branch of mathematics that deals with the study of linear equations, linear transformations, and vector spaces. It involves the use of matrices and vectors to represent and solve systems of linear equations, as well as the study of properties and operations of these mathematical objects.

2. What is a subspace in linear algebra?

A subspace is a subset of a vector space that is closed under addition and scalar multiplication. In other words, any linear combination of vectors in a subspace will also be in that subspace. Subspaces can be thought of as smaller vector spaces within a larger vector space.

3. How do you determine if a set of vectors forms a subspace?

In order for a set of vectors to form a subspace, it must satisfy three conditions: closure under addition, closure under scalar multiplication, and contain the zero vector. This means that when any two vectors in the set are added together, the result must also be in the set, and when any vector in the set is multiplied by a scalar, the result must also be in the set. Additionally, the set must contain the zero vector, which is a vector with all components equal to zero.

4. What is the difference between a subspace and a span?

A subspace is a subset of a vector space that satisfies certain conditions, while a span is the set of all possible linear combinations of a given set of vectors. In other words, a subspace is a type of vector space, while a span is a set of vectors that can be used to create a subspace.

5. How is linear algebra used in real-world applications?

Linear algebra has a wide range of applications in various fields, including physics, engineering, computer science, and economics. It is used to solve systems of equations, analyze data and patterns, and model real-world situations. Some specific applications include image and signal processing, machine learning, and optimization problems.

Similar threads

  • Calculus and Beyond Homework Help
Replies
4
Views
1K
  • Calculus and Beyond Homework Help
Replies
15
Views
2K
  • Calculus and Beyond Homework Help
Replies
8
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
530
  • Calculus and Beyond Homework Help
Replies
0
Views
515
  • Calculus and Beyond Homework Help
Replies
10
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
  • Calculus and Beyond Homework Help
Replies
10
Views
2K
  • Calculus and Beyond Homework Help
Replies
8
Views
1K
  • Calculus and Beyond Homework Help
Replies
5
Views
6K
Back
Top