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.