Proving Non-Trivial Solutions of Ax = b

  • Context: Undergrad 
  • Thread starter Thread starter HaLAA
  • Start date Start date
  • Tags Tags
    Linear algebra
Click For Summary

Discussion Overview

The discussion revolves around the conditions under which the equation Ax = b has non-trivial solutions, specifically focusing on the relationship between vector b and the span of matrix A. Participants are exploring definitions and implications related to linear algebra concepts.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant states that if Ax = b has a non-trivial solution, then b is in the span of A.
  • Another participant seeks clarification on the phrase "b span in A" and requests the source of the claim.
  • A different participant suggests that b should be in the span of the column vectors of A and mentions that the proof involves the definition of span and matrix multiplication.
  • Another contribution explains that the "range" of A, which consists of vectors of the form Av, is sometimes referred to as the "span" of A, and illustrates how the columns of A can be shown to span this range.

Areas of Agreement / Disagreement

Participants have not reached a consensus on the exact phrasing or implications of the claim regarding b and the span of A. There are differing interpretations and clarifications being sought.

Contextual Notes

Some assumptions about the definitions of span and range are not explicitly stated, and there may be unresolved details regarding the proof of the relationship between Ax = b and the span of A.

HaLAA
Messages
85
Reaction score
0
As I study today, I read through my textbook that says if Ax=b has non-trival solution, then b span in A.

I want to know how can I prove it.
 
Physics news on Phys.org
I don't understand what you mean by "b span in A". Can you post the exact claim you came across? You should also tell us where you found it. What book? What page number?
 
Sounds like it should be b is in THE span OF (the column vectors of) A. The proof would just be knowing the definition of span, plus multiplying A times x and writing it out in terms of the components of x.
 
The "range" of A, the set of all vector of the form Av for some v in the domain of A, is sometimes called the "span" of A. Since we can take, in succession, v= <1, 0, 0, ..., 0>, v= <0, 1, 0, ..., 0>, v= <0, 0, 1, ..., 0>, to v= <0, 0, 0, ..., 1> and applying A to each of those gives a column of A, it can be shown that the columns of A span the range of A. And, if A is invertible, form a basis for it.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K