# Subspace (question)

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 $\mathbb{R}^{3}$?

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.

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quasar987
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jackdamack10 said:
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).

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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 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
Homework Helper
Gold Member
jackdamack10 said:
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.

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