Showing that Something is a Subspace of R^3

In summary, the question asks to determine if the given set is a subspace of R^3. The set is defined as W = {(x,y,z) : x,y,z ∈ ℝ; x = y + z}. After showing that the components of the sum of two vectors in W are dependent on each other, it is concluded that W is a subspace of R^3.
  • #1
Prof. 27
50
1

Homework Statement


The question asks to show whether the following are sub-spaces of R^3. Here is the first problem. I want to make sure I'm on the right track.

Problem: Show that W = {(x,y,z) : x,y,z ∈ ℝ; x = y + z} is a subspace of R^3.

Homework Equations


None

The Attempt at a Solution


Okay, so here is my attempt:
Since y,z ∈ ℝ
y+z ∈ ℝ because ℝ is closed under addition.
So x ∈ ℝ.
:Break:
Let u = (x,y,z) be some vector in W.
Let v = (a,b,c) be some vector in W.
(x,y,z) + (a,b,c) = (x+a, y+b, z+c)
x+a ∈ ℝ because ℝ is closed under addition and x,a ∈ ℝ by the previous logic.
y+b ∈ ℝ because ℝ is closed under addition and y,b ∈ ℝ.
z+c ∈ ℝ because ℝ is closed under addition and z,c ∈ ℝ.
x and a are still dependent on y,z and b,c respectively.
So u + v ∈ W
:Break:
Let k be some scalar on ℝ
Let (x,y,z) be some vector u on ℝ.
k(x,y,z) = (kx,ky,kz)
Since ℝ is closed under multiplication and x,y,z ∈ ℝ (as already demonstrated), kx ∈ ℝ, ky ∈ ℝ, kz ∈ ℝ
kx = ky + kz, so kx remains dependent on ky, kz.
So ku ∈ W
:Break:
So u is a subspace of ℝ.

Am I missing anything? All the examples I've found are extremely poor and dissimilar to the specified problem.
Also, since W has 3 real valued components, isn't W technically defined on ℝ^3, not ℝ^2? How can it be a subspace if this is the case?
 
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  • #2
Prof. 27 said:
Am I missing anything?
Yes. You are missing the crucial part. That the result of the addition is a vector in R^3 follows directly from the fact that R^3 is a vector space. You need to show that the sum is still in the subspace. It is not enough to say that the components are dependent, you must verify the dependence to be the one defining the subspace.
 
  • #3
Okay, so x + a = y + z + b + c, which satisfies the dependence of both x,a on the other respective components. So (x+a,y+b,z+c) is in W?
 
  • #4
Right, but the other way around. You want to deduce that (x+a) = (y+b) + (z+c) is satisfied. It is satisfied because u and v are in W. Hence, u+v is in W.
 
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Likes Prof. 27
  • #5
Changed parts in bold
Later changes are in italics

-----------------------------------------------------------------------------------------------
Since y,z ∈ ℝ
y+z ∈ ℝ because ℝ is closed under addition.
So x ∈ ℝ.
:Break:
Let u = (x,y,z) be some vector in W.
Let v = (a,b,c) be some vector in W.
(x,y,z) + (a,b,c) = (x+a, y+b, z+c)
x+a ∈ ℝ because ℝ is closed under addition and x,a ∈ ℝ by the previous logic.
y+b ∈ ℝ because ℝ is closed under addition and y,b ∈ ℝ.
z+c ∈ ℝ because ℝ is closed under addition and z,c ∈ ℝ.
x and a are still dependent on y,z and b,c respectively, because x + a = y + z + b + c = (y+z) + (b+c) by associativity; furthermore, x + a = (y+b) + (z+c) by associativity.
So u + v ∈ W
:Break:
Let k be some scalar on ℝ
Let (x,y,z) be some vector u in W.
k(x,y,z) = (kx,ky,kz)
Since ℝ is closed under multiplication and x,y,z ∈ ℝ (as already demonstrated), kx ∈ ℝ, ky ∈ ℝ, kz ∈ ℝ
kx = ky + kz, so kx remains dependent on ky, kz.
So ku ∈ W
:Break:
Since every vector in W is also on ℝ^3, since x,y,z ∈ ℝ (as demonstrated previously) and every vector in W has closure with respect to addition and scalar multiplication,
W is a subspace of ℝ^3
 
  • #6
Thanks. That makes sense I think.
 
  • #7
Prof. 27 said:
Changed parts in bold
Later changes are in italics

-----------------------------------------------------------------------------------------------
Since y,z ∈ ℝ
y+z ∈ ℝ because ℝ is closed under addition.
So x ∈ ℝ.
You don't need to show this -- it is given information. The set consists of triples (x, y, z), where all three components are real.
Prof. 27 said:
:Break:
Let u = (x,y,z) be some vector in W.
Let v = (a,b,c) be some vector in W.
(x,y,z) + (a,b,c) = (x+a, y+b, z+c)
x+a ∈ ℝ because ℝ is closed under addition and x,a ∈ ℝ by the previous logic.
y+b ∈ ℝ because ℝ is closed under addition and y,b ∈ ℝ.
z+c ∈ ℝ because ℝ is closed under addition and z,c ∈ ℝ.
x and a are still dependent on y,z and b,c respectively, because x + a = y + z + b + c = (y+z) + (b+c) by associativity; furthermore, x + a = (y+b) + (z+c) by associativity.
So u + v ∈ W
You have a lot of extra, and unneeded cruft in here (such as x+a ∈ ℝ and so on). All you need to show is that if u and v are in W, and defined as above, then the sum u + v satisfies the requirement that the first coordinate is equal to the sum of the second and third coordinate.
Prof. 27 said:
:Break:
Let k be some scalar on ℝ
Let (x,y,z) be some vector u in W.
k(x,y,z) = (kx,ky,kz)
Since ℝ is closed under multiplication and x,y,z ∈ ℝ (as already demonstrated), kx ∈ ℝ, ky ∈ ℝ, kz ∈ ℝ
Again, there is extra stuff in here that isn't needed, and shouldn't be included, such as saying kx ∈ ℝ, etc. Just show that if k is a scalar, and u is in W, then ku is also in W; i.e., that the first coordinate of ku is equal to the sum of the second and third coordinates. Anything else is just useless baggage.
Prof. 27 said:
kx = ky + kz, so kx remains dependent on ky, kz.
So ku ∈ W
:Break:
Since every vector in W is also on ℝ^3, since x,y,z ∈ ℝ (as demonstrated previously) and every vector in W has closure with respect to addition and scalar multiplication,
W is a subspace of ℝ^3
 

1. What is a subspace of R^3?

A subspace of R^3 is a subset of the three-dimensional Euclidean space that satisfies three conditions: closure under vector addition, closure under scalar multiplication, and contains the zero vector.

2. How do you show that something is a subspace of R^3?

To show that something is a subspace of R^3, you must first prove that it satisfies the three conditions mentioned above. This can be done by showing that the set is closed under vector addition and scalar multiplication, and that it contains the zero vector. Additionally, you must also show that it is a subset of R^3.

3. What are some examples of subspaces of R^3?

Some examples of subspaces of R^3 include the x-y plane, the y-z plane, the x-z plane, and the line passing through the origin and a point in R^3.

4. Can a subspace of R^3 be a line or a plane?

Yes, a subspace of R^3 can be a line or a plane. As long as it satisfies the three conditions mentioned above, it can be considered a subspace.

5. What is the importance of proving that something is a subspace of R^3?

Proving that something is a subspace of R^3 is important because it allows us to determine if a set of vectors has certain properties that are essential in many areas of mathematics and science. It also helps in solving systems of equations and understanding linear transformations.

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