- #1
nayfie
- 50
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Hello :)
I've been doing a lot of work on subspaces but have come across this question and need a bit of help!
[itex]W = {(x, y) \in R^{2} | x^{2} + y^{2} = 0}[/itex]
1. 0 ∈ W
2. ∀ u,v ∈ W; u+v ∈ W
3. ∀ c ∈ R and u ∈ W; cu ∈ W
Check for 0 vector
[itex]x^{2} + y^{2} = 0[/itex]
[itex]0^{2} + 0^{2} = 0[/itex]
[itex]0 = 0[/itex]
Check closure under scalar addition
Let [itex]u = x^{2} + y^{2} = 0[/itex]; let [itex]v = a^{2} + b^{2} = 0[/itex]
[itex]u + v = (x^{2} + a^{2}) + (y^{2} + b^{2}) = 0 + 0 = 0[/itex]
Check for closure under scalar multiplication
[itex]ku = (kx)^{2} + (ky)^{2} = 0[/itex]
[itex]= k^{2}(x^{2} + y^{2}) = 0[/itex]
[itex]x^{2} + y^{2} = \frac{0}{k^{2}}[/itex]
[itex]x^{2} + y^{2} = 0[/itex]
-----------------------------------
I have shown that the zero vector is in the set, and that it is closed under scalar multiplication, however; I'm not sure whether or not it is closed under scalar multiplication.
I have shown that u + v = 0, but, u + v does not have the same form as u and v individually, so I don't think u + v is part of the set?
I've been doing a lot of work on subspaces but have come across this question and need a bit of help!
Homework Statement
[itex]W = {(x, y) \in R^{2} | x^{2} + y^{2} = 0}[/itex]
Homework Equations
1. 0 ∈ W
2. ∀ u,v ∈ W; u+v ∈ W
3. ∀ c ∈ R and u ∈ W; cu ∈ W
The Attempt at a Solution
Check for 0 vector
[itex]x^{2} + y^{2} = 0[/itex]
[itex]0^{2} + 0^{2} = 0[/itex]
[itex]0 = 0[/itex]
Check closure under scalar addition
Let [itex]u = x^{2} + y^{2} = 0[/itex]; let [itex]v = a^{2} + b^{2} = 0[/itex]
[itex]u + v = (x^{2} + a^{2}) + (y^{2} + b^{2}) = 0 + 0 = 0[/itex]
Check for closure under scalar multiplication
[itex]ku = (kx)^{2} + (ky)^{2} = 0[/itex]
[itex]= k^{2}(x^{2} + y^{2}) = 0[/itex]
[itex]x^{2} + y^{2} = \frac{0}{k^{2}}[/itex]
[itex]x^{2} + y^{2} = 0[/itex]
-----------------------------------
I have shown that the zero vector is in the set, and that it is closed under scalar multiplication, however; I'm not sure whether or not it is closed under scalar multiplication.
I have shown that u + v = 0, but, u + v does not have the same form as u and v individually, so I don't think u + v is part of the set?
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