Exact solution of linear least square problem

[Demon117]

Homework Statement

What is the exact solution to the linear least square problem

$$\left(\begin{array}{ccc}1 & 1 & 1 \\\epsilon & 0 & 0 \\0 & \epsilon & 0 \\0 & 0 & \epsilon \\ \end{array}\right)$$$$\left(\begin{array}{c}x_{1}\\x_{2}\\x_{3}\end{array}\right)$$=$$\left(\begin{array}{c}1\\ 0\\ 0\\ 0\end{array}\right)$$

as a function of epsilon.

The Attempt at a Solution

I've tried numerous methods, including Cholesky factorization of $$A^{T}A$$, which is of course

$$A^{T}A$$ = $$\left(\begin{array}{ccc} 1+\epsilon^{2} & 1 & 1 \\1 & 1+\epsilon^{2} & 1 \\1 & 1 & 1+\epsilon^{2}\end{array}\right)$$

The Cholesky factorization is a difficult one but in the end I get a vector that cannot possibly be a solution to this problem, what I get is

$$\vec{x}$$ = $$\left(\begin{array}{c} 1+\epsilon^{2} \\1 \\1 \end{array}\right)$$

Would anyone be willing to give me a few pointers?

 

[Demon117]

Homework Statement

What is the exact solution to the linear least square problem

$$\left(\begin{array}{ccc}1 & 1 & 1 \\\epsilon & 0 & 0 \\0 & \epsilon & 0 \\0 & 0 & \epsilon \\ \end{array}\right)$$$$\left(\begin{array}{c}x_{1}\\x_{2}\\x_{3}\end{array}\right)$$=$$\left(\begin{array}{c}1\\ 0\\ 0\\ 0\end{array}\right)$$

as a function of epsilon.

The Attempt at a Solution

I've tried numerous methods, including Cholesky factorization of $$A^{T}A$$, which is of course

$$A^{T}A$$ = $$\left(\begin{array}{ccc} 1+\epsilon^{2} & 1 & 1 \\1 & 1+\epsilon^{2} & 1 \\1 & 1 & 1+\epsilon^{2}\end{array}\right)$$

The Cholesky factorization is a difficult one but in the end I get a vector that cannot possibly be a solution to this problem, what I get is

$$\vec{x}$$ = $$\left(\begin{array}{c} 1+\epsilon^{2} \\1 \\1 \end{array}\right)$$

Would anyone be willing to give me a few pointers?

Well, I've tried and tried again . . . but I keep ending up with something like this:

$$\vec{x}$$ = $$\left(\begin{array}{c} 1/(3+\epsilon^{2}) \\1/(3+\epsilon^{2}) \\1/(3+\epsilon^{2}) \end{array}\right)$$