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
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:
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top