How is the x,y,z part a vector space?

Click For Summary
SUMMARY

The discussion focuses on the verification of the vector space properties of a specific set defined by vectors in the form of (x, y, z). The key point is that for a set to be a vector space, it must be closed under addition and scalar multiplication. The solution demonstrates that if u and v are vectors in the space, then u + dv is also in the space, confirming closure. The vectors are expressed in terms of real numbers, ensuring that all components remain real, thus fulfilling the requirements of a vector space.

PREREQUISITES
  • Understanding of vector space properties
  • Familiarity with vector addition and scalar multiplication
  • Knowledge of real numbers and their properties
  • Basic linear algebra concepts
NEXT STEPS
  • Study the definition and properties of vector spaces in linear algebra
  • Learn about closure properties in vector spaces
  • Explore examples of vector spaces in R^n
  • Investigate the role of scalar multiplication in vector spaces
USEFUL FOR

Students of linear algebra, educators teaching vector space concepts, and anyone seeking to deepen their understanding of vector space properties and their applications.

SpiffyEh
Messages
191
Reaction score
0

Homework Statement


I uploaded the problem because its easier to see.


Homework Equations





The Attempt at a Solution



The solution is there from a practice exam. I don't understand the x,y,z part and how it fulfills the multiplication requirement of a vector space. Could someone please explain to me why it does?
 

Attachments

  • problem.png
    problem.png
    20.1 KB · Views: 485
Physics news on Phys.org
In the solution they are handling the addition and multiplication at the same time.

They define u and v as vectors in the space, then show that u + dv (where d is any real) is also in the space.

So after multiplying and adding the vectors, you have a new vector:

[tex]\left(\begin{array}{cc} 4(a + da') + 3(b + db')\\0\\(c + dc') - 2\\(a + da') + (b + db')\end{array}\right)[/tex]

So now all we have to do is show that all three (a+da'), (b+db'), (c+dc') are real numbers. Since a,b,c,d are all real, this is obviously the case, and hence the space is closed under addition and multiplication.

In the solution, they converted these expressions to (x,y,z), but that's unnecessary.
 
oh! That makes sense, i didn't see where they came from. Thank you
 

Similar threads

Replies
9
Views
2K
  • · Replies 18 ·
Replies
18
Views
3K
  • · Replies 15 ·
Replies
15
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
4
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K