# How to obtain the inverse of the variance covariance matrix of GLS (Random Effects Model)

In the standard GLS set up how do you find the inverse of the variance covariance matrix?

$$y _ { i t } = \beta _ { 0 } + x _ { i t } ^ { \prime } \beta + \alpha _ { i } + u _ { i t } \hspace{35pt} u _ { i t } \sim I I D ( 0 , \sigma _ { u } ^ { 2 })\hspace{35pt} \alpha _ { i } \sim I I D ( 0 , \sigma _ { \alpha } ^ { 2 } )$$

$$\alpha _ { i }$$ is the random error, $$u_{it}$$ is the error term, $$x _ { i t }$$ are the control variables, $$\Omega$$ is the variance covariance matrix of GLS estimator it is symmetric and positive deﬁnite. $$l _ { T } l _ { T } ^ { \prime }$$ is a t x t matrix of ones $$I_{T}$$ is the t x t identity matrix. $$\sigma _ { \alpha } ^ { 2 }$$ is the variance of the random effect and $$\sigma _ { u } ^ { 2 }$$ is the variance of the error term.

$$V \{ \alpha _ { i } l _ { T } + u _ { i } \} = \Omega = \sigma _ { \alpha } ^ { 2 } l _ { T } l _ { T } ^ { \prime } + \sigma _ { u } ^ { 2 } I _ { T } \hspace{35pt} (a)$$

$$\Omega ^ { - 1 } = \sigma _ { u } ^ { - 2 } [ I _ { T } - \frac { \sigma _ { \alpha } ^ { 2 } } { \sigma _ { u } ^ { 2 } + T \sigma _ { \alpha } ^ { 2 } } l _ { T } l _ { T } ^ { \prime } ] \hspace{35pt} (b)$$

$$\Omega ^ { - 1 } = \sigma _ { u } ^ { - 2 } [ ( I _ { T } - \frac { 1 } { T } l _ { T } l _ { T } ^ { \prime } ) + \psi \frac { 1 } { T } l _ { T } l _ { T } ^ { \prime } ]$$

$$\psi = \frac { \sigma _ { u } ^ { 2 } } { \sigma _ { u } ^ { 2 } + T \sigma _ { \alpha } ^ { 2 } }$$

Does anyone know the algebraic steps required to obtain the inverse of the variance covariance matrix $$\Omega ^ {-1}$$ from the variance covariance matrix i.e. go from $$(a)$$ to $$(b)$$. All of the papers I have looked at skip this step so I suspect its straight forward but my matrix algebra is not great so I can't figure it out.

Once you have the solution, you can of course check that it is correct by verifying that $$(a)\cdot (b)$$ is equal to the identify matrix, which is standard matrix algebra.