MHB Are These Sets Vector Subspaces of R^3?

  • Thread starter Thread starter Yankel
  • Start date Start date
  • Tags Tags
    Subspaces Vector
Yankel
Messages
390
Reaction score
0
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.
 
Physics news on Phys.org
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.
 
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.
 
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?
 
Oh...silly me. x-y = 0 and z=0.

So V is also a subspace, has to be then.
 
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.
 
Thread 'Determine whether ##125## is a unit in ##\mathbb{Z_471}##'
This is the question, I understand the concept, in ##\mathbb{Z_n}## an element is a is a unit if and only if gcd( a,n) =1. My understanding of backwards substitution, ... i have using Euclidean algorithm, ##471 = 3⋅121 + 108## ##121 = 1⋅108 + 13## ##108 =8⋅13+4## ##13=3⋅4+1## ##4=4⋅1+0## using back-substitution, ##1=13-3⋅4## ##=(121-1⋅108)-3(108-8⋅13)## ... ##= 121-(471-3⋅121)-3⋅471+9⋅121+24⋅121-24(471-3⋅121## ##=121-471+3⋅121-3⋅471+9⋅121+24⋅121-24⋅471+72⋅121##...
Back
Top