# Granger's representation theorem: Johansen's version

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?

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.