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I really don't know how to start/solve the following problem, any help is appreciated

Here is the question:

It consists of a and b, I already solved a

a) Let $A∈ℝ^{m*n}$ and $b∈ℝ^{m}$ with $rank(A)=n$. Suppose x is the solution for the least squares problem given by $A$ and $b$.

Show that: $||Ax-b||^2=b^T(I_m-A(A^TA)^{-1})b$

I proved this is the case so now it can be used as a given

b)By using least squares method, find the best circle of the form $x^2+y^2=r^2$ representing the pairs $(x_i,y_i)$ for $i=1,2,...,n$

2 Answers 2

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The only freedom is given by the radius of the circle and in the perfect case you would have that all points lie on it, or that $x_i^2+y_i^2=r^2=:R$ for all data points. Consequently we would like to have \begin{align*} \begin{bmatrix} 1\\1\\ \vdots \\ 1 \end{bmatrix} R=\begin{bmatrix} x_1^2+y_1^2\\ x_2^2+y_2^2\\ \vdots\\ x_n^2+y_n^2\\ \end{bmatrix}. \end{align*} This does not have a solution generally, so we take the least squares approach. Multiplying both sides by $\begin{bmatrix}1&1& \cdots & 1 \end{bmatrix}$ for the normal equation(s) yields \begin{align*} nR=\sum_{i=1}^{n}\left(x_i^2+y_i^2\right). \end{align*} So the desired circle has radius equal to $r=\sqrt{R}=\sqrt{\dfrac{1}{n}\sum_{i=1}^{n}\left(x_i^2+y_i^2\right)}$.

N.B.: I think you should eventually start solving your homework yourself, especially considering your question history. I will not be the one responsible, but you might get in trouble some day if you keep doing this.

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@Brayan Shali has posted the least squares solution $R=\mathbf{A}^{\dagger}b$. The pseudoinverse for a vector $$ v^{\dagger} = \frac{v^{*}}{\lVert v \rVert}. $$ That is, the average value is the least squares solution.