Showing that the noncentered model can be written in centered form

Question

The model for each $y_i$ can be written in terms of centered $x$ variables as

$$\begin{align}y_i &= \beta_0+\beta_1x_{i1}+ \cdots+\beta_kx_{ik}+\epsilon_i \\&=\alpha +\beta_1(x_{i1}-\bar{x}_1)+\beta_2(x_{i2}-\bar{x}_2+\cdots+\beta_k(x_{ik}-\bar{x})+\epsilon_i \end{align}$$

$i = 1, 2, ..., n$ where

$$\alpha = \beta_0+\beta_1\bar{x}_1+ \cdots + \beta_k\bar{x}_k$$

I have to write this but in noncentered form, here is what I have tried:

$$\begin{align}\mathbf{x\hat{\beta}}&=\mathbf{x(x^Tx)^{-1}x^Ty} \\&\implies \mathbf{\hat{\beta}(x-\bar{x})=y[x(x^Tx)^{-1}x^T-\bar{x}(\bar{x}^T\bar{x})^{-1}\bar{x}^T]} \end{align}$$

Then to find $\alpha$ we have something like

$$\alpha = \mathbf{\bar{x}\beta}=\mathbf{\bar{x}(\bar{x}^T\bar{x})^{-1}\bar{x}y} \\ \implies \mathbf{\alpha-\beta(x-\bar{x})}=\mathbf{\bar{x}(\bar{x}^T\bar{x})^{-1}\bar{x}y+y[x(x^Tx)^{-1}x^T-\bar{x}(\bar{x}^T\bar{x})^{-1}\bar{x}^T]}$$

Answer 1

We can show the noncentered model by the following:

$$y_i=\begin{pmatrix}j & X_C \end{pmatrix}\pmatrix{\alpha_0 \\ \beta_1} = \alpha_0j+\beta_1X_C+\epsilon$$

We have that $$X_C = (I - \frac{J}{n})X_1 = \pmatrix{x_{11}-\bar{x}_1 & \cdots & x_{1k}-\bar{x}_k \\ \vdots & \cdots & \vdots \\ x_{n1}-\bar{x}_1 & \cdots & x_{nk}-\bar{x}_k }$$