# Showing that Something is a Subspace of R^3

## 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.

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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Orodruin
Staff Emeritus
Homework Helper
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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.

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?

Orodruin
Staff Emeritus
Homework Helper
Gold Member
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.

• Prof. 27
Changed parts in bold
Later changes are in italics

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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

Thanks. That makes sense I think.

Mark44
Mentor
Changed parts in bold
Later changes are in italics

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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