Why Isn't the Least Square Solution Unique with Linearly Dependent Columns?

  • Thread starter Thread starter td21
  • Start date Start date
  • Tags Tags
    Matrix Square
Click For Summary

Homework Help Overview

The discussion revolves around the uniqueness of the least squares solution when dealing with a matrix that has linearly dependent columns. Participants are exploring the implications of linear dependence in relation to least squares solutions.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are questioning why the least squares solution is not unique in the presence of linearly dependent columns. Some are attempting to relate this to the concept of projection onto the column space of the matrix. Others are drawing analogies to geometric interpretations, such as describing a point with a line.

Discussion Status

The discussion is ongoing, with participants raising questions and offering tentative reasoning. There is a mix of attempts to clarify concepts and explore the implications of linear dependence, but no consensus has been reached yet.

Contextual Notes

Some participants reference the need for linear independence in the context of calculating inverses, indicating a potential gap in understanding related to the mathematical properties of matrices involved in least squares solutions.

td21
Gold Member
Messages
177
Reaction score
8

Homework Statement


If a matrix A has linearly dependent columns and b is a vector, then the least square solution is not unique.


Homework Equations





The Attempt at a Solution


I know that the "projection" onto column space of A is unique, but why the least square solution isn't?
 
Physics news on Phys.org
help?
 
If you have linearly dependent columns, you have too few independent equations to get a unique? Its like trying to describe only one point with a line.
Or am I wrong? It's two years since I had the course
 
http://en.wikipedia.org/wiki/Linear_least_squares_(mathematics )

B = (X'X)^-1 X'y

(X'X)^-1 doesn't it have to have linearly independent columns if you want to calculate the inverse?

Again I'm only guessing
 
Last edited by a moderator:

Similar threads

  • · Replies 30 ·
2
Replies
30
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
4
Views
2K
Replies
15
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K