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\textbf{L\"owdin's theorem (?):} Even though it is so simple,
I couldn't find any good proof. Therefore I put it here.

Let:

\begin{equation}
  \matr{H}\matr{U} = \matr{S}\matr{U}\matr{E}
\end{equation}

then it is possible to transform this equation to
standard eigenvalue problem

\begin{equation}
  \matr{H}^\mathrm{eff} \matr{U}^\mathrm{eff} =
  \matr{U}^\mathrm{eff} \matr{E} 
\end{equation}

with 

\begin{equation}
  \matr{H^\mathrm{eff}} = \matr{S}^{\left. {-1} \middle/ {2} \right.}
  \matr{H}\matr{S}^{\left. {-1} \middle/ {2} \right.}
\end{equation}

$\blacktriangleleft$ 

First, some obvious remarks:

\begin{equation}
  \begin{aligned}
 \matr{1} &= \matr{S}^{\left. {-1} \middle/ {2} \right.}
 \matr{S}^{\left. {1} \middle/ {2} \right.} = 
 \matr{S}^{\left. {1} \middle/ {2} \right.}
 \matr{S}^{\left. {-1} \middle/ {2} \right.} \\
 \matr{S} &=  \matr{S}^{\left. {1} \middle/ {2} \right.}
 \matr{S}^{\left. {1} \middle/ {2} \right.}
  \end{aligned}
 \label{eq:eq10}
\end{equation}

And second, we use them for some obvious transformations.
We start with the initial equation

\begin{equation}
 \matr{H}\matr{U} = \matr{S}\matr{U}\matr{E}
\end{equation}

expand $\matr{S}$ matrix using square roots

\begin{equation}
 \matr{H}\matr{U} = 
 \matr{S}^{\left. {1} \middle/ {2} \right.}
 \matr{S}^{\left. {1} \middle/ {2} \right.} 
 \matr{U}\matr{E}
  \label{ eq:eq11 }
\end{equation}

then we introduce unity matrix, and expand it with square roots of
$\matr{S}$ as follows

\begin{equation}
 \matr{H}\matr{1}\matr{U} = 
 \matr{S}^{\left. {1} \middle/ {2} \right.}
 \matr{S}^{\left. {1} \middle/ {2} \right.} 
 \matr{U}\matr{E} 
   \label{ eq:eq12 }
\end{equation}
\begin{equation}
 \matr{H}\matr{S}^{\left. {-1} \middle/ {2} \right.}
 \matr{S}^{\left. {1} \middle/ {2} \right.}  
 \matr{U} = \matr{S}^{\left. {1} \middle/ {2} \right.}
 \matr{S}^{\left. {1} \middle/ {2} \right.} 
 \matr{U}\matr{E}
   \label{ eq:eq13 }
\end{equation}

then multiply to the left with 
$\matr{S}^{\left. {-1} \middle/ {2} \right.}$

\begin{equation}
 \matr{S}^{\left. {-1} \middle/ {2} \right.} \times \Bigg|
 \matr{H}
 \matr{S}^{\left. {-1} \middle/ {2} \right.}
 \matr{S}^{\left. {1} \middle/ {2} \right.}  
 \matr{U} = 
 \matr{S}^{\left. {1} \middle/ {2} \right.}
 \matr{S}^{\left. {1} \middle/ {2} \right.} 
 \matr{U}\matr{E}
   \label{ eq:eq15 }
\end{equation}

so we get

\begin{equation}
 \matr{S}^{\left. {-1} \middle/ {2} \right.}
 \matr{H}
 \matr{S}^{\left. {-1} \middle/ {2} \right.}
 \matr{S}^{\left. {1} \middle/ {2} \right.}  
 \matr{U} = 
 \matr{S}^{\left. {1} \middle/ {2} \right.} 
 \matr{U}\matr{E}  
\end{equation}

and then we introduce new notation and get the result

\begin{equation}
 \begin{aligned}
 \matr{U}^{\mathrm{eff}} &= 
 \matr{S}^{\left. {1} \middle/ {2} \right.} 
 \matr{U} \\
 \matr{H}^{\mathrm{eff}} &=
 \matr{S}^{\left. {-1} \middle/ {2} \right.}
 \matr{H}
 \matr{S}^{\left. {-1} \middle/ {2} \right.}
 \end{aligned}
\end{equation}

End of proof. $\blacktriangleright$