Vector Subspaces: Understanding Closure Properties

[nayfie]

Hello :)

I've been doing a lot of work on subspaces but have come across this question and need a bit of help!

Homework Statement

W = {(x, y) \in R^{2} | x^{2} + y^{2} = 0}

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

x^{2} + y^{2} = 0

0^{2} + 0^{2} = 0

0 = 0

Check closure under scalar addition

Let u = x^{2} + y^{2} = 0; let v = a^{2} + b^{2} = 0

u + v = (x^{2} + a^{2}) + (y^{2} + b^{2}) = 0 + 0 = 0

Check for closure under scalar multiplication

ku = (kx)^{2} + (ky)^{2} = 0

= k^{2}(x^{2} + y^{2}) = 0

x^{2} + y^{2} = \frac{0}{k^{2}}

x^{2} + y^{2} = 0

-----------------------------------

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?

 

[nayfie]

Just took another look at this question and might have solved it.

u = (a, b), v = (c, d), u + v = (a + c, b + d)

(a + c)^{2} + (b + d)^{2} = 0

a^{2} + 2ac + c^{2} + b^{2} + 2bd + d^{2} = 0

But, we know from the constraints of the subspace that;

a^{2} + b^{2} = 0; c^{2} + b^{2} = 0

So if we cancel those out, we get;

2ac + 2bd = 0

So in general, scalar addition breaks the constraint.

Does this even make sense or do I need more sleep?

 

[Mark44]

Yes, you need more sleep.

Think about what your set looks like - W = {(x, y) \in R²| x² + y² = 0}.

This set consists of a single point at (0, 0). If u \in W and v \in W, what must u and v be? It should be easy to show that u + v \in W.