1
$\begingroup$

I have this confusion about the time complexity of least squares problem. Suppose

minimize $||Ax-b||^2$

Analytical solution = $x^* = (A^TA)^{-1}A^Tb$

computational time proportional to $n^2k$, $A \in \mathbb{R}^{k\times n}$

I didn't get how the computation time is $n^2k$

Here is the source http://classes.soe.ucsc.edu/cmps290c/Spring05/lectures/intro.pdf

$\endgroup$
  • 2
    $\begingroup$ $\epsilon$ (\epsilon) is a variable name whereas your $A$ is in the set $\mathbb{R}^{kn}$. So you should use $\in$ (\in). $\endgroup$ – xavierm02 Mar 8 '13 at 12:19
  • $\begingroup$ Also to get $\mathbb{R}$, you can use \mathbb{R}. And I think you should simply use $kn$ instead of $kxn$ but if you really want the multiplication operator to appear, use $\times$ (\times). $\endgroup$ – xavierm02 Mar 8 '13 at 12:21
  • $\begingroup$ In fact you should keep the $\times$ because otherwise, we wouldn't be able to interpret it as a matrix. $\endgroup$ – xavierm02 Mar 8 '13 at 12:25
  • $\begingroup$ check out this link for the reasoning of $n^2k$... math.stackexchange.com/questions/84495/… $\endgroup$ – Learner Mar 8 '13 at 13:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.