Are These Sets Vector Subspaces of R^3?

[Yankel]

Dear all,

I am trying to find if these two sets are vector subspaces of R^3.

\[V=\left \{ (x,y,z)\in R^{3}|(x-y)^{2}+z^{2}=0 \right \}\]

\[W=\left \{ (x,y,z)\in R^{3}|(x+1)^{2}=x^{2}+1 \right \}\]

In both cases the zero vector is in the set, therefore I just need to prove closure to addition and to scalar multiplication. Can you kindly assist ?

Thank you.

 

[GJA]

Hi Yankel,

Your analysis is off to a good start when you noted that $0 = (0,0,0)$ is a member of both sets. It is helpful to look at the equations to see if we can infer from them any information about these subspaces of $\mathbb{R}^{3}.$ In the case of $W$, for example, try filling in the details of the argument that allows us to conclude $x=0.$ From there it may be easier to establish/refute the remaining subspace properties we must check. Feel free to ask any follow up questions.

 

[Yankel]

Thank you. I have managed to prove W is a subspace using your tip that x=0 (got it why it is). I can't prove the other set. Can't bring the condition to a simpler form.

 

[GJA]

Nicely done. Yes, $W$ is a linear subspace of $\mathbb{R}^{3}.$ To analyze $V$, note that each of the terms (thinking of $(x-y)$ as one term) is squared and that the sum of these two squared terms is zero. Can you conclude anything from thinking along these lines?

 

[Yankel]

Oh...silly me. x-y = 0 and z=0.

So V is also a subspace, has to be then.

 

[HOI]

For W, the equation that must be satisfied does not involve y or z. $(x+ 1)^2= x^2+ 2x+ 1= x^2+ 1$ so 2x= 0 and x= 0. The problem is to show that the subset or $R^3$, \{(0, y, z)\}$, with y and z any real numbers, is a subspace.