Proving Vector Subspaces of R^3

In summary, the conversation discusses the question of whether two given sets are vector subspaces of $\mathbb{R}^{3}$. It is noted that both sets contain the zero vector and therefore the only requirement is to prove closure to addition and scalar multiplication. After some discussion and suggestions, it is concluded that the sets do satisfy the requirements and are indeed subspaces of $\mathbb{R}^{3}$.
  • #1
Yankel
395
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.
 
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  • #2
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.
 
  • #3
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.
 
  • #4
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?
 
  • #5
Oh...silly me. x-y = 0 and z=0.

So V is also a subspace, has to be then.
 
  • #6
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.
 

1. What is a vector subspace of R^3?

A vector subspace of R^3 is a subset of R^3 that satisfies the properties of a vector space. This means that it is closed under vector addition and scalar multiplication, and contains the zero vector. In other words, it is a set of vectors in three-dimensional space that can be added and multiplied by scalars to produce other vectors within the same set.

2. How do you prove that a set is a vector subspace of R^3?

To prove that a set is a vector subspace of R^3, you need to show that it satisfies the three properties of a vector space: closure under vector addition, closure under scalar multiplication, and containing the zero vector. This can be done by showing that any two vectors in the set can be added to produce another vector in the set, any vector in the set can be multiplied by a scalar to produce another vector in the set, and the zero vector is included in the set.

3. What are some common examples of vector subspaces of R^3?

Some common examples of vector subspaces of R^3 include the x-y plane, the y-z plane, the x-z plane, and the origin (the point (0,0,0)). These are all subsets of R^3 that satisfy the properties of a vector space, as any two vectors in these sets can be added and multiplied by scalars to produce another vector in the set.

4. Can a vector subspace of R^3 contain more than three dimensions?

No, a vector subspace of R^3 can only contain three dimensions. This is because R^3 refers to three-dimensional space, and any set that satisfies the properties of a vector space in R^3 can only contain three dimensions.

5. How can I use the properties of vector subspaces to solve problems?

The properties of vector subspaces can be used to solve problems involving vectors in three-dimensional space. For example, you can use the closure under vector addition property to add two or more vectors together, or the closure under scalar multiplication property to scale a vector by a certain factor. These properties are also useful in linear algebra, where vector subspaces are used to solve systems of linear equations.

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