Is Answer to Subspace Question Acceptable?

[jackdamack10]

I just wanted to know if my answer is acceptable.

Q: S={(x,y,z) E \mathbb{R}^{3} l x^2 + y^2 +z ^2 =0}
Is it a subspace of \mathbb{R}^{3}?

My answer:

It is a subspace if x=0, y =0, z= 0

Let u=(0,0,0) u2=(0,0,0) and k be a scalar

u + u2 = (0,0,0) Closed under addition

ku = k(0,0,0) = (0,0,0) Closed under scalar multiplication



Is my answer complete? I'm not sure if I'm allowed to assume that the values of x,y,z are 0.

Thank you in advance

 

[quasar987]

Is my answer complete? I'm not sure if I'm allowed to assume that the values of x,y,z are 0.

Actually, S={(x,y,z) E \mathbb{R}^{3} l x^2 + y^2 +z ^2 =0} = {(0,0,0)}

I.e. (0,0,0) is the only element in that set. It is made clear if you recall that x²+y²+z²=R² describe a sphere of radius R. Here R=0. The only point "on" a sphere of radius 0 is (0,0,0).

So, with this in mind, instead of the tentative "It is a subspace if x=y=z=0", start the proof with: "The only element of S is (0,0,0)", and then go on to show that the 3 conditions for S to be a subspace of R^3 are satisfied (like you did).

 

[jackdamack10]

Ok thanks.

What if I have a multiplication:

S={(x,y,z) E R^3 l xy =0}.
To prove that this is a subspace (or not a subspace), the only way I know to do so is by assuming a value for x, y and z before checking if it's closed under scalar mutiplication adn addition. Is there another way?

 

[Sunshine]

You can use a=(a_1,a_2,a_3), b=(b_1,b_2,b_3) and since either x or y must be 0 to satisfy your condition of xy=0, you can have the condition that either a_1 or a_2 must be zero, and similarly for b_1 and b_2. Then show that a and b are closed under scalar addition and multiplication, and you have proofed it for all of the values in the subspace

 

[quasar987]

Ok thanks.

What if I have a multiplication:

S={(x,y,z) E R^3 l xy =0}.
To prove that this is a subspace (or not a subspace), the only way I know to do so is by assuming a value for x, y and z before checking if it's closed under scalar mutiplication adn addition. Is there another way?

You must never assume a particular value of (x,y,z). You must always work your ways through the 3 conditions while assuming the most general form possible for (x,y,z).

First condition: Is (0,0,0) in S? Consider (x=0,y=0,z=0). Then xy=0, so (0,0,0) is in S. *check*

Second condition: Consider two vectors \vec{u}_1 = (x_1,y_1,z_1) and \vec{u}_2 = (x_2,y_2,z_2) in S. Because they are in S, they have the property that x_1 y_1 = 0 and x_2 y_2 = 0. According to the definition of addition in \mathbb{R}^3, we have \vec{u}_1+\vec{u}_2 = (x_1+x_2,y_1+y_2,z_1+z_2). Now the condition: Does (x_1+x_2)(y_1+y_2)=0?. Let's see: In view of the "axiom of distributivity" in \mathbb{R}, (x_1+x_2)(y_1+y_2)=x_1y_1+x_1y_2+x_2y_1+x_2y_2. We know by hypothesis that x_1y_1 = x_2y_2=0. But what about the other two terms? It could be that x_1\neq 0, \ y_1 = 0, \ x_2 = 0, \ y_2 \neq 0. In this case, x_1y_1 = 0 and x_2y_2 = 0 are indeed satisfied but x_1y_2 \neq 0. Conclusion: \vec{u}_1+\vec{u}_2 \notin S \ \forall \vec{u}, \ \vec{u}_2 \ \Rightarrow S is not a subspace of \mathbb{R}^3 because condition 2 (closed under addition) is not met.