Proving Solutions of Linear Transformations Using Kernel and Fixed Vectors

Click For Summary

Discussion Overview

The discussion revolves around proving a property of linear transformations, specifically the relationship between solutions of the equation T(x) = b and the kernel of the transformation. Participants explore the conditions under which a vector x_1 is a solution based on the fixed solution x_0 and elements from the kernel of T.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant proposes that a vector x_1 is a solution of T(x) = b if and only if it can be expressed as x_1 = x_h + x_0, where x_h is in the kernel of T.
  • Another participant emphasizes the importance of logical conviction in arguments, suggesting that one should critically assess their own reasoning.
  • A later reply expresses personal doubt about the correctness of one's own arguments, highlighting a common struggle with self-assurance in mathematical reasoning.
  • Another participant encourages the development of clear and simple logical arguments to build confidence in one's reasoning.

Areas of Agreement / Disagreement

Participants express varying levels of confidence in their arguments, with some acknowledging self-doubt. There is no consensus on the correctness of the initial proof or the method of self-assessment.

Contextual Notes

Participants do not fully resolve the logical steps involved in the proof, and there are indications of uncertainty regarding the clarity and correctness of their arguments.

laminatedevildoll
Messages
211
Reaction score
0
[tex]\ Let T: V \rightarrow W[/tex] be a linear transformation, let [tex]b \in W[/tex]be a fixed vector, and let [tex]x_0 \in V[/tex] be a fixed solution of
[tex]T(x)=b.[/tex] Prove that a vector [tex]x_1 \in V[/tex]is a solution of [tex]T(x)=b,[/tex] if and only if [tex]x_1[/tex]is of the form [tex]x_1=x_h +x_0[/tex]where [tex]x_h \in kerT[/tex]

I started out by saying that

[tex]x_i \in X_i[/tex]

[tex](x_1... x_n) \in \prod[/tex] (where i=1 and h is at the top) [tex]X_0[/tex]

[tex](x_1... x_n) \in \prod X_i[/tex]

[tex]x_i \in X_i \rightarrow x_1[/tex] is not equal to the empty set for all i.

I am not sure if I am doing this right. I'd appreciate any feedback.
 
Last edited:
Physics news on Phys.org
You're looking for these:

[tex]\neq \emptyset[/tex]

[tex]\prod_{i=1}^{h}[/tex]

[itex]\ker T[/itex]
 
Last edited:
to see if you are right, ask yourself if your argument is logically convincing. it is crucial to be able to decide for yourself, if such arguments are correct. i.e. practice playing both roles, argue it then ask if it could possibly be wrong, then answer yourself.
 
I think that it's convincing, but sometimes I find it hard to convince myself that it's right because I doubt myself all the time.
 
well that's the goal to achieve. to reduce your arguments to logic so simple and clear that you can persuade yourself that you must be right. keep practicing.
 

Similar threads

  • · Replies 25 ·
Replies
25
Views
5K
  • · Replies 40 ·
2
Replies
40
Views
5K
  • · Replies 27 ·
Replies
27
Views
3K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 6 ·
Replies
6
Views
3K
Replies
22
Views
4K
  • · Replies 52 ·
2
Replies
52
Views
4K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 12 ·
Replies
12
Views
5K