# Constrained fixed effects

I need to show that constraining the sum of group level fixed effects (in this case, zero) has no effect on the coefficients of the regressors. My intuition is that each constrained d_i is a perfect linear transformation of the non constrained constants. I looked in Greene, and Cameron and Trivedi, but I did not find a formal treatment of it.

Using Greene 2012, theorem 3.8. Construct the transforming matrix, $P$, s.t:
$\mathbf{P}^{-1}=\begin{bmatrix} 1 & 0 & \cdots & 0 & \cdots & \cdots & 0 \\ 0 & \ddots & 0 & \vdots & \vdots & \vdots & \vdots \\ \vdots & 0 & 1_K & 0 & \vdots & \vdots & \vdots \\ & \cdots & 0 & 1_{\alpha} & 0 & \vdots & \vdots \\ & \cdots & \vdots & -1_{\alpha} & 1_{D_2} & 0 & \vdots \\ \vdots & \cdots & 0 & \vdots & 0 & \ddots & 0 \\ 0 & \cdots & 0 & -1_{\alpha}& 0 & 0 & 1_{D_N} \\ \end{bmatrix}$
Since $\mathbf{P}$ does not transform the regressors (that part of the matrix is idempotent), only the constant are transformed to be distance from the mean of the omitted group (or any other constrain on them, such as to sum to 0). And according to theorem 3.8, $R^2$ is unchanged.