I'm studying econometrics but I have some problems with linear algebra.
Consider the following vector $\nu_i = \alpha_i \iota_T + u_i$, with $\alpha_i$ being a scalar, $\iota_T$ is a $T \times 1$ vector of ones, i.e. $\iota_T= (1,1,\ldots,1)^\intercal$, and $u_i=(u_{i1},\ldots,u_{iT})^\intercal$ is still a column vector of size $T$.
Then, I assume that $\alpha_i$ and $u_i$ are uncorrelated, and $Var(\alpha_i)=\sigma^2_\alpha$ and $Var(u_i)=\sigma^2_u$.
Then, the covariance matrix of this vector is:
\begin{equation} Var(\nu_i)= Var(\alpha_i \iota_T + u_i) = Var(\alpha_i \iota_T )+ Var(u_i) = \Omega= \sigma^2_\alpha \iota_T \iota_T^{\intercal} + \sigma^2_u I_T \end{equation} where $I_T$ is a $T$-dimensional identity matrix and $\iota_T \iota_T^{\intercal}$ is a $T \times T$ matrix full of ones.
Can you show me the mathematical steps to get the inverse of $\Omega$?
The book says:
\begin{equation} \Omega^{-1} = \sigma^{-2}_u \left( I_T - \frac{\sigma^2_{\alpha}}{\sigma^2_u + T \sigma^2_{\alpha}} \iota_T \iota^{\intercal}_T \right) \end{equation}
You can derive $\Omega^{-1}$ using the Sherman–Morrison formula. If $A$ is a $T\times T$ invertible matrix and $B$ is a $T\times T$ matrix with rank one, let $g$ be the trace ($tr$) of $BA^{-1}$. Then, for $g \ne -1$: \begin{equation*} (A+B)^{-1} = A^{-1} - \frac{1}{1+g} A^{-1} B A^{-1} \end{equation*} In our case, let $\Omega= A+B$, where $A \equiv \sigma^2_u I_T$ and $A^{-1}$ exists (i.e., $A^{-1}=\sigma^{-2}_u I_T$), and $B=\sigma^2_\alpha \iota_T \iota_T^{\prime}$. Since $B$ is a full matrix of $\sigma^2_\alpha$ and has rank one, $g=tr(BA^{-1})=T \sigma^2_{\alpha} \sigma^{-2}_u$, and $A^{-1} B A^{-1} =\sigma^{-2}_u \sigma^2_\alpha \iota_T \iota_T^{\prime} \sigma^{-2}_u$. Thus, using our original notation for $\Omega$: \begin{equation} \begin{aligned} \Omega^{-1} & = \sigma^{-2}_u I_T - \frac{ 1 }{ 1+ T \sigma^2_{\alpha} \sigma^{-2}_u } \sigma^{-2}_u \sigma^2_\alpha \iota_T \iota_T^{\prime} \sigma^{-2}_u \\ & = \sigma^{-2}_u \left( I_T - \frac{\sigma^{-2}_u \sigma^2_\alpha }{1+ T \sigma^2_{\alpha} \sigma^{-2}_u} \iota_T \iota_T^{\prime} \right) \end{aligned} \end{equation} Then, dividing the numerator and denominator of the second term in parentheses by $\sigma^{-2}_u$: \begin{equation} \Omega^{-1} = \sigma^{-2}_u \left( I_T - \frac{ \sigma^2_\alpha }{\sigma^{2}_u + T \sigma^2_{\alpha}} \iota_T \iota_T^{\prime} \right) \end{equation}