10.2 Determine if the set of vectors form a vector space

Click For Summary

Discussion Overview

The discussion revolves around determining whether a specific set of vectors, represented as $\begin{bmatrix} x\\y\\5 \end{bmatrix}$ in $\Bbb{R}^3$, forms a vector space. The conversation includes considerations of vector space properties, particularly closure under addition and scalar multiplication, and touches on related concepts from linear algebra.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested
  • Homework-related

Main Points Raised

  • One participant argues that the set of vectors is not closed under addition because the third component remains constant at 5, leading to a resultant vector with a third component of 15.
  • Another participant confirms that subspaces are vector spaces and questions the scenario if the third component were zero, implying that closure might hold in that case.
  • A different participant suggests that if the third component is zero, all vectors would lie on the same plane, which could imply closure under addition and scalar multiplication.
  • One participant reflects on their previous experience with linear algebra and mentions a potential overlap with another question they might have posted, indicating a possible familiarity with the topic.
  • Another participant clarifies that their previous comment was not meant as criticism but rather as an inquiry about the question's history.

Areas of Agreement / Disagreement

Participants express differing views on the closure properties of the vector set in question. There is no consensus on whether the set forms a vector space, particularly regarding the implications of the third component being fixed at 5 versus being zero.

Contextual Notes

Participants reference definitions and properties of vector spaces and subspaces, but there are unresolved assumptions regarding the implications of the third component's value on closure properties.

karush
Gold Member
MHB
Messages
3,240
Reaction score
5
Determine if the set of vectors
$\begin{bmatrix}
x\\y\\5
\end{bmatrix}\in \Bbb{R}^3$
form a vector space
ok if I follow the book example I think this is what is done
$\begin{bmatrix} x_1\\y_2\\5 \end{bmatrix}
+\begin{bmatrix} x_2\\y_2\\5 \end{bmatrix}
+\begin{bmatrix} x_2\\y_2\\5 \end{bmatrix}
=\begin{bmatrix} x_1+x_2+x_3\\y_1+y_2+y_3\\15 \end{bmatrix}$
since the third entry is 15, the set of such vectors is not closed under addition and hence is not a subspaceI assume in this case a vector space and sub space are the same.
 
Physics news on Phys.org
Right: sub spaces are by definition vector spaces in their own right. Your proof looks good to me! What if the third component was zero?
 
well if the 3rd conponent is zero then everying is just on the same plane
so multiplication or addition would close

well i think anyway?
 
Didn't you post this question somewhere else many moons ago? I could swear that I responded to this one at some point in the past.

-Dan
 
I don't think so

But I took linear Algebra a year ago it might be very similar..

They combined the LA and De class
Not sure why.

I mark the homework probs I go to MHB for help with the MHB logo to avoid dbb.

But I post a lot since I'm very deaf and classroom is nil to me.

Sorry I'm probably overload here
 
It wasn't a criticism. I was just wondering.

-Dan
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 11 ·
Replies
11
Views
2K
Replies
8
Views
2K
  • · Replies 27 ·
Replies
27
Views
2K
  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 1 ·
Replies
1
Views
5K
  • · Replies 7 ·
Replies
7
Views
2K