Linear algebra, is W a subspace of R^2 problem

In summary: Oh, I didn't notice that. So it's not a subspace.In summary, the subset W = {(x,y): y = ax, a is an integer} is not a subspace of the vector space V = R^2 because it violates the closure condition for vector addition when a is any integer, not a fixed integer. This can be illustrated by considering the addition of two vectors (x,mx) and (y,ny) where m and n are different integers.
  • #1
jubjub49
4
0

Homework Statement



Determine whether W is a subspace of the vector space: W={(x,y):y=ax, a is an integer} , V=R^2

Homework Equations


none

The Attempt at a Solution



Is u+v in W?
Let u = (u,au) and v = (v,av)
u+v = (u,au) + (v,av) = (u+v, au + av) = (u+v, a(u+v))
If x = u+v => u + v = (x,ax)
=> closure under addition

Is cu in W?
cu = c(u,au) = (cu,acu)
If x = cu => cu = (x,ax)
=> closure under scalar multiplication

=> W is a subspace of R^2

but my book says that W is not a subspace of R^2 and I'm not sure what I'm doing wrong. Does it have something to do with a being an integer? The problem before had something similar but with 2x instead of ax where a is an integer and it was a subspace so I thought this would have the same answer. Thanks for any help in advance, this is really confusing me now.
 
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  • #2
jubjub49 said:

Homework Statement



Determine whether W is a subspace of the vector space: W={(x,y):y=ax, a is an integer} , V=R^2


Homework Equations


none


The Attempt at a Solution



Is u+v in W?
Let u = (u,au) and v = (v,av)
u+v = (u,au) + (v,av) = (u+v, au + av) = (u+v, a(u+v))
If x = u+v => u + v = (x,ax)
=> closure under addition

Is cu in W?
cu = c(u,au) = (cu,acu)
If x = cu => cu = (x,ax)
=> closure under scalar multiplication

=> W is a subspace of R^2

but my book says that W is not a subspace of R^2 and I'm not sure what I'm doing wrong. Does it have something to do with a being an integer? The problem before had something similar but with 2x instead of ax where a is an integer and it was a subspace so I thought this would have the same answer. Thanks for any help in advance, this is really confusing me now.

I think the answer in your book is wrong and you are right. 2 is an integer.
 
  • #3
jubjub49 said:

Homework Statement



Determine whether W is a subspace of the vector space: W={(x,y):y=ax, a is an integer} , V=R^2


Homework Equations


none


The Attempt at a Solution



Is u+v in W?
Let u = (u,au) and v = (v,av)
u+v = (u,au) + (v,av) = (u+v, au + av) = (u+v, a(u+v))
If x = u+v => u + v = (x,ax)
=> closure under addition

Is cu in W?
cu = c(u,au) = (cu,acu)
If x = cu => cu = (x,ax)
=> closure under scalar multiplication

=> W is a subspace of R^2

but my book says that W is not a subspace of R^2 and I'm not sure what I'm doing wrong. Does it have something to do with a being an integer? The problem before had something similar but with 2x instead of ax where a is an integer and it was a subspace so I thought this would have the same answer. Thanks for any help in advance, this is really confusing me now.

You haven't told us what vector addition and scalar multiplication is defined as. Are they the usual operations in ##R^2##?

Also, it isn't clear whether W = {(x,y): y = ax for some fixed integer a} or whether (x,y) is in W if there is any integer m such that y = mx.
 
  • #4
LCKurtz said:
You haven't told us what vector addition and scalar multiplication is defined as. Are they the usual operations in ##R^2##?

Also, it isn't clear whether W = {(x,y): y = ax for some fixed integer a} or whether (x,y) is in W if there is any integer m such that y = mx.

Oh, good point. If a is AN integer then it is. If a is ANY integer, then it's not.
 
  • #5
I'm using the test for a subspace given as

"If W is a nonempty subset of a vector space V, then W is a subspace of V if and only if the following closure conditions hold.

1. If u and v are in W, then u+v is in W.
2. If u is in W and c is any scalar, then cu is in W."

Vector addition and scalar multiplication in R^n are defined as

"Let u = (u1,u2,u3,...,un) and v = (v1,v2,v3,...,vn) be vectors in R^n and let c be a real number. Then the sum of u and v is defined as the vector

u+v = (u1 + v1, u2 + v2, u3 + v3,..., un+vn)

and the scalar multiple of u by c is defined as the vector

cu = (cu1, cu2, cu3,...,cun)."

As for if a is a fixed integer or any integer, that is all the problem states, I'm assuming it means any integer not a fixed one though.
 
  • #6
That might be what I'm missing but what would be an example of an integer which would violate one of the closure axioms? All it really says is "a is an integer".
 
  • #7
jubjub49 said:
That might be what I'm missing but what would be an example of an integer which would violate one of the closure axioms? All it really says is "a is an integer".

What happens if you add (x,mx) and (y,ny) if m and n are different integers?
 
  • #8
jubjub49 said:
That might be what I'm missing but what would be an example of an integer which would violate one of the closure axioms? All it really says is "a is an integer".

Then they didn't really quantify the question clearly. If a is ANY integer, can you prove it's not a subspace? Do you see the difference here? If a is a single integer then the graph is a line. If a is any integer then it's a whole bunch of lines.
 
  • #9
Sure that definitely makes sense, I'm still not sure what my book actually means because there aren't any other questions that include that phrase to compare it to, which is strange because these are supposed to be review questions over the chapter. Would make sense of why the answer is like that though. Thanks to both of you for the quick replies!
 

Related to Linear algebra, is W a subspace of R^2 problem

1. What is linear algebra?

Linear algebra is a branch of mathematics that deals with the study of linear equations and their representations in vector spaces. It involves the use of matrices, vectors, and linear transformations to solve mathematical problems.

2. What is a subspace in linear algebra?

A subspace in linear algebra is a subset of a vector space that is closed under vector addition and scalar multiplication. In other words, if you add or multiply any two vectors within the subset, the result will still be within the same subset.

3. How do you determine if a subset is a subspace of a vector space?

To determine if a subset is a subspace of a vector space, you need to check if it satisfies three conditions: closure under addition, closure under scalar multiplication, and contains the zero vector. If all three conditions are met, then the subset is a subspace.

4. Is W a subspace of R^2 if it contains the zero vector?

No, containing the zero vector alone does not guarantee that W is a subspace of R^2. It also needs to satisfy the other two conditions: closure under addition and closure under scalar multiplication.

5. How do you prove that W is a subspace of R^2?

To prove that W is a subspace of R^2, you need to show that it satisfies all three conditions: closure under addition, closure under scalar multiplication, and contains the zero vector. You can also use the definition of a subspace, which states that W must be a subset of R^2 and must be closed under vector addition and scalar multiplication.

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