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if we are given a set of points $\{(x_i,y_i)\}$ and we are looking to fit a straight line $ax+b$ "as close" to the points as possible we are building a set of equations.

One way is to take the partial derivatives with respect to $a,b$ and find the min.

On the matrix form we multiply by $A^T$ when $A$ is entries of $1,x,x^2,..$ why is it smilier to finding the min using derivative?

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  • $\begingroup$ Smilier ? What does it mean? $\endgroup$ – mathreadler Jan 14 at 10:52
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    $\begingroup$ Similar, I guess ? $\endgroup$ – nicomezi Jan 14 at 10:55
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There is some cookbook about matrix and vector calculus on the internet where we can derive $${\bf x_o}=\min_{\bf x}\{\|{\bf Ax-b}\|_2^2\}$$ by

  1. Differentiating the expression $$\|{\bf Ax-b}\|_2^2=({\bf Ax-b})^T({\bf Ax-b})$$ (With respect to $\bf x$)
  2. Setting this derivative to zero.
  3. Solving for $\bf x$

Doing these steps is a good exercise.

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  • $\begingroup$ Wow this great, must be bookmarked. $\endgroup$ – newhere Jan 14 at 15:38
  • $\begingroup$ I need to derive $(Ax-b)^T(Ax-b)$ is deriving matrices-by-vectors? $\endgroup$ – newhere Jan 14 at 15:54
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    $\begingroup$ You can multiply it together and differentiate each resulting term with help of the book. $\endgroup$ – mathreadler Jan 14 at 17:01

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