Granger's representation theorem: Johansen's version

Question

In his book 'Likelihood based inference in cointegrated Var', in order to get the expression for the Granger's representation theorem,, Johansen claims that:

(1)

$$\beta \bot(\alpha' \bot \beta \bot )^{-1} \alpha' \bot + \alpha (\beta' \alpha)^{-1} \beta' = I $$

Where:

$\alpha$ is NxR $\text{rank}(\alpha) =R$

$\beta$ is NxR $\text{rank}(\beta) =R$

$\beta \bot $ is NxN-R $\text{rank}(\beta \bot) =N-R$

$\alpha \bot $ is NxN-R $\text{rank}(\alpha \bot) =N-R$

$\alpha' \alpha \bot =0$

$\beta' \beta \bot =0$

I am not able to prove (1). Can you help me, please?

Answer 1

For ease of typing, let $$\eqalign{ A &= \alpha,\quad B=\beta,\quad X = \alpha\!\perp,\quad Y=\beta\!\perp \\ A^TX &= 0 \quad\implies X^TA = 0 \\ B^TY &= 0 \quad\implies Y^TB = 0 \\ }$$ Now consider the matrix $$\eqalign{ M &= Y(X^TY)^{-1}X^T + A(B^TA)^{-1}B^T \\ }$$ Straightforward calculations yield. $$\eqalign{ B^TM &= B^T \\ X^TM &= X^T \\ MA &= A \\ MY &= Y \\ M^2 &= M \\ }$$ The only idempotent $M$ which acts as an identity for all of those linearly independent row/column vectors -- recall that $(A,B,X,Y)$ are full rank -- is the identity matrix.