Least squares problem

  • Thread starter chuy52506
  • Start date
  • #1
77
0

Homework Statement


Let
A=
|2 -1 -1|
|-1 2 -1|
|-1 -1 2|
and
B=
|1|
|2|
|3|


Homework Equations



Find the x in which minimizes ||Ax-b||2


The Attempt at a Solution


I tried to solve it by using this formula (A**A)-1A**b=x but i get the inverse of A*A equal 0
 

Answers and Replies

  • #2
Greetings! Right, since A does not have linearly independent columns, [tex]A^TA[/tex] is not invertible. Call [tex]B = A^TA[/tex] and [tex]\vec{y} = A^T\vec{b}[/tex] and try using row reduction to solve the matrix equation [tex]B\vec{x} = \vec{y}[/tex] for [tex]\vec{x}[/tex].
 
  • #3
77
0
When i try to solve it i get the last row in rref of B=A*A to be a row of 0's equal to 3=/
Is there any other way to solve this??
 
  • #4
I got that

[tex] B = A^TA = \[ \left[ \begin{array}{ccc}
6 & -3 & -3 \\
-3 & 6 & -3 \\
-3 & -3 & 6 \end{array} \right]\][/tex]

and

[tex] \vec{y} = A^T\vec{b} = \[ \left[ \begin{array}{c}-3 \\
0 \\3 \end{array} \right]\][/tex].

Is this what you got? Then we can row reduce

[tex] [ B\ \vec{y} ] = \[ \left[ \begin{array}{cccc}
6 & -3 & -3 & -3 \\
-3 & 6 & -3 & 0 \\
-3 & -3 & 6 & 3 \end{array} \right]\][/tex]

In fact, reducing this shows that there is a free variable, meaning that there is a whole line worth of solutions that give the best approximation.
 
  • #5
77
0
thanks!!
 

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