10.2 Determine if the set of vectors form a vector space

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SUMMARY

The discussion centers on determining whether the set of vectors of the form $\begin{bmatrix} x\\y\\5 \end{bmatrix}$ in $\Bbb{R}^3$ forms a vector space. It is concluded that this set does not form a vector space because it is not closed under addition, as demonstrated by the sum resulting in $\begin{bmatrix} x_1+x_2+x_3\\y_1+y_2+y_3\\15 \end{bmatrix}$. The third component remains constant at 5, violating the closure property required for vector spaces. The conversation also touches on the relationship between vector spaces and subspaces, affirming that subspaces are indeed vector spaces in their own right.

PREREQUISITES
  • Understanding of vector space properties
  • Familiarity with linear algebra concepts
  • Knowledge of closure properties in vector addition
  • Basic comprehension of subspaces
NEXT STEPS
  • Study the definition and properties of vector spaces in linear algebra
  • Learn about closure properties and their implications for vector spaces
  • Explore examples of vector spaces and subspaces in $\Bbb{R}^n$
  • Investigate the effects of varying components in vector definitions on their vector space status
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Students of linear algebra, educators teaching vector space concepts, and anyone interested in the foundational principles of vector spaces and subspaces.

karush
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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.
 
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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
 

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