Are These Sets Vector Subspaces of R^3?

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Discussion Overview

The discussion revolves around determining whether two specific sets, \(V\) and \(W\), are vector subspaces of \(\mathbb{R}^3\). The focus is on analyzing the conditions that define these sets and verifying the properties required for them to qualify as subspaces, including closure under addition and scalar multiplication.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Post 1 introduces the sets \(V\) and \(W\) and states the need to prove closure properties for them to be subspaces.
  • Post 2 suggests that analyzing the equations can provide insights into the subspace properties, specifically mentioning that for set \(W\), \(x\) must equal zero.
  • Post 3 indicates that the participant has successfully proven \(W\) is a subspace but struggles with \(V\) due to its complexity.
  • Post 4 confirms that \(W\) is indeed a linear subspace and prompts further analysis of \(V\) by considering the implications of the squared terms in its defining equation.
  • Post 5 concludes that since \(x-y=0\) and \(z=0\), \(V\) must also be a subspace, although this reasoning may require further scrutiny.
  • Post 6 clarifies that the equation for \(W\) does not involve \(y\) or \(z\) and emphasizes the need to show that the subset \(\{(0, y, z)\}\) is a subspace.

Areas of Agreement / Disagreement

Participants generally agree that \(W\) is a subspace of \(\mathbb{R}^3\). However, the status of \(V\) remains less clear, with some participants asserting it is a subspace while others express uncertainty about the reasoning behind this conclusion.

Contextual Notes

The discussion highlights the complexity of proving subspace properties, particularly for set \(V\), and the reliance on specific conditions derived from the defining equations. There are unresolved aspects regarding the closure properties for both sets.

Yankel
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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.
 
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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.
 

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