Find optimal least square solution to the normal equation What is the optimal solution for $\beta_1$ and $\beta_2$ in the following normal equation:
$$\beta _{ 1 }\sum _{ i=1 }^{ n }{ { x }_{ i } } +\beta _{ 0 }=\sum _{ i=1 }^{ n }{ { y }_{ i } } $$
EDIT
Suppose you are given a set of data $(x_i,y_i)$ with y = $\beta_1x+\beta_0$ and $\beta_1,\beta_0 \in \Bbb R$ are the parameters we want to determine.
A good criteria to find parameter values is to find $\beta_1$ and $\beta_0$ such that the residual sum of squares is minimum. In other words, we find the
values $\beta_0$, $\beta_1$ that minimize:
$$\sum _{ i=1 }^{ n }{ ({ \beta  }_{ 1 }{ x }_{ i }+{ \beta  }_{ 0 }-y_{ i })^{ 2 } } $$ which I wrote as 
$$\left \|\ A\begin{pmatrix} \beta _{ 1 } \\ \beta _{ 0 } \end{pmatrix}-b\   \right \|^2$$
with $A=\begin{pmatrix} { x }_{ 1 } & 1 \\ { x }_{ 2 } & 1 \\ ... & ... \\ { x }_{ n } & 1 \end{pmatrix}$ and $b = \begin{pmatrix} y_{ 1 } \\ { y }_{ 2 } \\ ... \\ y_{ n } \end{pmatrix}$
From there, you can extract the normal equation
$$\beta _{ 1 }\sum _{ i=1 }^{ n }{ { x }_{ i } } +\beta _{ 0 }=\sum _{ i=1 }^{ n }{ { y }_{ i } } $$
Now equation is, how do you find the optimal solution for $\beta_1$ and $\beta_0$?
 A: The normal equation $(A^TA)\vec{\beta} = A^T\vec{y}$ actually gives you two equations:
$\displaystyle\underbrace{\begin{pmatrix}\sum_{i=1}^{n}x_i^2 & \sum_{i=1}^{n}x_i \\ \sum_{i=1}^{n}x_i & n\end{pmatrix}}_{A^TA} \underbrace{\begin{pmatrix} \beta_1 \\ \beta_0 \end{pmatrix}}_{\vec{\beta}} = \underbrace{\begin{pmatrix} \sum_{i=1}^{n}x_iy_i \\ \sum_{i=1}^{n}y_i \end{pmatrix}}_{A^T\vec{y}}$
$\displaystyle\beta_1\sum_{i = 1}^{n}x_i^2 + \beta_0\sum_{i=1}^{n}x_i = \sum_{i=1}^{n}x_iy_i$ (1)
$\displaystyle\beta_1\sum_{i = 1}^{n}x_i + \beta_0n = \sum_{i=1}^{n}y_i$ (2)
Now, you have a two variable, two unknown system. You should be able to solve this. 
A: Given a sequence of measurements, $\left\{ x_{k}, y_{k} \right\}_{k=1}^{m}$, and a trial function
$$
  y(x) = \beta_{0} + \beta_{1} x,
$$
the linear system is
$$
  \begin{align}
    \mathbf{A} \beta &= y \\
\left[
  \begin{array}{cc}
    1 & x_{1} \\
    1 & x_{2} \\
    \vdots & \vdots \\
    1 & x_{m}
  \end{array}
\right]
%
\left[
  \begin{array}{c}
    \beta_{0} \\
    \beta_{1}
  \end{array}
\right]
%
 &=
%
\left[
  \begin{array}{c}
    y_{1} \\
    y_{2} \\
   \vdots \\
    y_{m}
  \end{array}
\right].
%
  \end{align}
%
$$
Here are the normal equations expressed in terms of column vectors:
$$
  \begin{align}
   \mathbf{A}^{*} \mathbf{A} \beta &= \mathbf{A}^{*} y \\
\left[
  \begin{array}{cc}
    \mathbf{1} \cdot \mathbf{1} & \mathbf{1} \cdot x \\
    x \cdot \mathbf{1}          & x \cdot x
  \end{array}
\right]
%
\left[
  \begin{array}{c}
    \beta_{0} \\
    \beta_{1}
  \end{array}
\right]
%
 &=
%
\left[
  \begin{array}{c}
    \mathbf{1} \cdot y \\
    x \cdot y
  \end{array}
\right].
%
  \end{align}
%
$$
The solution is
$$
  \beta 
  = \left( \mathbf{A}^{*}\mathbf{A} \right)^{-1} \mathbf{A}^{*} b
  = \frac{1}{
\left( \mathbf{1} \cdot \mathbf{1} \right)
\left( x \cdot x \right) - 
\left( \mathbf{1} \cdot x \right)^{2}}
%
\left[
  \begin{array}{rr}
    x \cdot x & -\mathbf{1} \cdot x \\
    -\mathbf{1} \cdot x & \mathbf{1} \cdot \mathbf{1}
  \end{array}
\right]
%
\left[
  \begin{array}{c}
    \mathbf{1} \cdot y \\
    x \cdot y
  \end{array}
\right].
$$
