The OLS estimator can be written has:
$(X'X)^{-1}X'Y$
I am interested in analyzing this expression by having $X$ has a block matrix:
$X = [\begin{matrix} X_a & X_b \end{matrix}]$
So that the OLS estimator can be written has:
$([\begin{matrix} X_a & X_b \end{matrix}]'[\begin{matrix} X_a & X_b \end{matrix}])^{-1}[\begin{matrix} X_a & X_b \end{matrix}]'Y$
I begin by working out $(\begin{bmatrix} X_a & X_b \end{bmatrix}'\begin{bmatrix} X_a & X_b \end{bmatrix})^{-1}$ :
$\begin{bmatrix} X_a & X_b \end{bmatrix}'\begin{bmatrix} X_a & X_b \end{bmatrix}^{-1} = \begin{bmatrix} X_a'X_a & X_a'X_b \\ X_b'X_a & X_b'X_b \end{bmatrix}^{-1}$
We know that the inverse of a block matrix can be written has (I based myself on https://en.wikipedia.org/wiki/Block_matrix) :
$$ \begin{bmatrix} A & B \\ C & D \end{bmatrix}^{-1} =\begin{bmatrix} (A - BD^{-1}C)^{-1} & 0 \\ 0 & (D-CA^{-1}B)^{-1} \end{bmatrix}\begin{bmatrix} I & -BD^{-1} \\ -CA^{-1} & I \end{bmatrix} $$
The problem is that the first term of this multiplication with our $X = [\begin{matrix} X_a & X_b \end{matrix}]$ collapses to the null matrix.
For instance:
$$ -BD^{-1} = -X_a'X_b'^{-1} $$
So
$$ A-BD^{-1}C = X_a'X_a-X_a'X_b'^{-1}X_b'X_a = 0 $$
And an equivalent result is obtained for $(D - CA^{-1}B)^{-1}$
What am I doing wrong ?