MHB Problem about equivalent conditions for a basis of a free module

Click For Summary
The discussion centers on the conditions for a basis of a free module, specifically the implications between statements labeled a, b, c, and d. The main issue raised is the assumption that there are only finitely many nonzero elements in b, which is linked to the uniqueness of the basis. The participant demonstrates that b implies d and c, and that c implies a, but questions the reasoning behind the finite nature of a_z in b. The conclusion drawn is that the finiteness of a_z in b stems from Z being a generating set for the module M, confirming its status as a basis. The conversation highlights the intricacies of understanding basis conditions in free modules.
Ganitadnya
Messages
1
Reaction score
0
Here's the problem:
View attachment 7922

I don't see why there should be only finitely many nonzero a_z in b. I was able to prove uniqueness assuming that there only finitely many nonzero. I was able to show b implies d and b implies c, c implies a.
 

Attachments

  • uGyGDN2.png
    uGyGDN2.png
    16.9 KB · Views: 121
Last edited by a moderator:
Physics news on Phys.org
So you're referring to the implication (a) $\implies$ (b). There are only finitely many $a_z$ in (b) simply because $Z$ is a generating set for $M$ (since it is a basis by (a)).
 
Thread 'How to define a vector field?'
Hello! In one book I saw that function ##V## of 3 variables ##V_x, V_y, V_z## (vector field in 3D) can be decomposed in a Taylor series without higher-order terms (partial derivative of second power and higher) at point ##(0,0,0)## such way: I think so: higher-order terms can be neglected because partial derivative of second power and higher are equal to 0. Is this true? And how to define vector field correctly for this case? (In the book I found nothing and my attempt was wrong...

Similar threads

Replies
48
Views
4K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 15 ·
Replies
15
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
Replies
4
Views
2K
Replies
3
Views
2K