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

$$ (1) \hspace{1.4 cm}\Omega=\sigma^2_{c}\boldsymbol{1}\boldsymbol{1'} + \sigma^2_{\varepsilon}I_T=(\sigma^2_{\varepsilon}+T\sigma^2_{c})\boldsymbol{1}(\boldsymbol{1}'\boldsymbol{1})^{-1}\boldsymbol{1}'+\sigma^2_{\varepsilon}(I_T- \boldsymbol{1}(\boldsymbol{1}'\boldsymbol{1})^{-1}\boldsymbol{1}')$$ $$\Omega^{-1}=\frac{1}{\sigma^2_{\varepsilon}+T\sigma^2_{c}}\boldsymbol{1}(\boldsymbol{1}'\boldsymbol{1})^{-1}\boldsymbol{1}'+\frac{1}{\sigma^2_{\varepsilon}}(I_t-\boldsymbol{1}(\boldsymbol{1}'\boldsymbol{1})^{-1}\boldsymbol{1}')$$ $$|\Omega|=(\sigma^2_{\varepsilon}+T\sigma^2_{c})\sigma^{{2(T-1)}}_{\varepsilon}$$

1 is a T-vector of ones.

The Lemma states:

$For \hspace{0.2cm} \lambda_1>0\hspace{0.2cm}and\hspace{0.2cm} \lambda_2>0\hspace{0.2cm}we\hspace{0.2cm}define\hspace{0.2cm}the\hspace{0.2cm}covariance\hspace{0.2cm}\Sigma\hspace{0.2cm} by$ $$\Sigma=\lambda_1 P + \lambda_2 Q,$$ $so\hspace{0.2cm}that$ $$x'\Sigma x=\lambda_1 T \bar{x}^2+\lambda_2\sum\nolimits_{i=1}^T (x_i-\bar{x})^2$$ $Then$ $$\Sigma^{-1}=\lambda_1^{-1} P + \lambda_2^{-1} Q,$$ $so \hspace{0.2cm} that$ $$x'\Sigma^{-1} x=\lambda_1^{-1} T \bar{x}^2+\lambda_2^{-1}\sum\nolimits_{i=1}^T (x_i-\bar{x})^2.$$ $The\hspace{0.2cm} eigenvalues \hspace{0.2cm} of\hspace{0.2cm} \Sigma\hspace{0.2cm} are\hspace{0.2cm}\lambda_1\hspace{0.2cm}with\hspace{0.2cm}multiplicity\hspace{0.2cm}one, \hspace{0.2cm}and\hspace{0.2cm}\lambda_2\hspace{0.2cm}with\hspace{0.2cm} multiplicity\hspace{0.2cm}T-1, \hspace{0.2cm}so\hspace{0.2cm}that $ $$|\Sigma|=\lambda_1\lambda_2^{T-1}$$.

Can anyone explain the second equality in (1)? E.g. 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.