I'm trying to understand the derivation of a ML-estimator and more specifically the rewriting of the covariance matrix $\Sigma$. In this rewriting, a lemma is used to show that:
$$
\tag{1} \Omega=\sigma^2\_{c}\boldsymbol{1}\boldsymbol{1'} + \sigma^2\_{\varepsilon}I\_T=(\sigma^2\_{\varepsilon}+T\sigma^2\_{c})\boldsymbol{1}(\boldsymbol{1}'\boldsymbol{1})^{-1}\boldsymbol{1}'+\sigma^2\_{\varepsilon}(I\_T- \boldsymbol{1}(\boldsymbol{1}'\boldsymbol{1})^{-1}\boldsymbol{1}')$$
$$\Omega^{-1}=\frac{1}{\sigma^2\_{\varepsilon}+T\sigma^2\_{c}}\boldsymbol{1}(\boldsymbol{1}'\boldsymbol{1})^{-1}\boldsymbol{1}'+\frac{1}{\sigma^2\_{\varepsilon}}(I\_t-\boldsymbol{1}(\boldsymbol{1}'\boldsymbol{1})^{-1}\boldsymbol{1}')$$
$$|\Omega|=(\sigma^2\_{\varepsilon}+T\sigma^2\_{c})\sigma^{{2(T-1)}}\_{\varepsilon}$$
Here, $\boldsymbol{1}$ is a T-vector of ones.


The Lemma states:



> 
> For $\lambda\_1 > 0$ and $\lambda\_2 > 0$ we define the covariance $\Sigma$ by:
>  $$\Sigma = \lambda\_1 P + \lambda\_2 Q ,$$
>  so that
>  $$x'\Sigma x = \lambda\_1 T \bar{x}^2+\lambda\_2\sum\nolimits\_{i=1}^T (x\_i-\bar{x})^2$$
>  Then
>  $$\Sigma^{-1} = \lambda\_1^{-1} P + \lambda\_2^{-1} Q ,$$
>  so that
>  $$x'\Sigma^{-1} x = \lambda\_1^{-1} T \bar{x}^2+\lambda\_2^{-1}\sum\_{i=1}^T (x\_i-\bar{x})^2 .$$
>  The eigenvalues of $\Sigma$ are $\lambda\_1$ with multiplicity one, and $\lambda\_2$ with multiplicity $T-1$, so that
>  $$ |\Sigma| = \lambda\_1\lambda\_2^{T-1} .$$
> 
> 
> 


Can anyone explain the second equality in (1)? I.e. this equality
$$\sigma^2\_{c} + \sigma^2\_{\varepsilon}I\_T=(\sigma^2\_{\varepsilon}+T\sigma^2\_{c})\boldsymbol{1}(\boldsymbol{1}'\boldsymbol{1})^{-1}\boldsymbol{1}'+\sigma^2\_{\varepsilon}(I\_T- \boldsymbol{1}(\boldsymbol{1}'\boldsymbol{1})^{-1}\boldsymbol{1}')$$


Please let me know if the question is off topic.



Understanding the derivation of a ML-estimator (statistics)
You can simply multiply $\Omega^{-1}$ as given by the author times $\Omega$ and see that the product is $I$. This confirms that the formula for $\Omega^{-1}$ is correct. 


The key here is that


$(I-1(1'1)^{-1}1')(1(1'1)^{-1}1')=1(1'1)^{-1}1'-1(1'1)^{-1}1'=0$


so the cross terms vanish. 


