deletions | additions       diff --git a/Yggm Matrix Properties.tex b/Yggm Matrix Properties.tex  index 5a2d504..0ab4cb0 100644  --- a/Yggm Matrix Properties.tex  +++ b/Yggm Matrix Properties.tex 
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%[Work your genius here Ioannis!]  We can now state the following Theorem.\\\\  \textbf{Theorem 3.1.} If every row of the $Y_{bus}$ matrix sums to zero and det$Y_{LL}$=0, then the rows every row  of $Y_{GGM}$ sum sums  to zero. If every row of the $Y_{bus}$ matrix sums approximately to zero, or if it sums to zero but det$Y_{LL}$=0, then every row of the matrix $Y_{GGM}$ sums approximately to zero.\\\\ \textbf{Proof.} The matrix $Y_{GGM}$can be written as  \begin{equation}\label{eq1}  Y_{GGM}=Y_{GG}+Y_{GL}F_{LG}, 
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\begin{equation}\label{eq4}  \sum^m_{j=1}a_{ij}+\sum^n_{j=1}b_{ij}=0,\quad \forall i=1,2,...,m.  \end{equation}  By replacing \eqref{eq3} into \eqref{eq2} and by using \eqref{eq4} \eqref{eq2}  we get \[  \sum_{j=1}^mg_{ij}=\sum_{j=1}^ma_{ij}+\sum_{j=1}^m\sum_{k=1}^n b_{ik}c_{kj}=-\sum_{j=1}^mb_{ij}+\sum_{j=1}^m\sum_{k=1}^n b_{ik}c_{kj}=-\sum_{j=1}^nb_{ij}+\sum_{j=1}^m\sum_{k=1}^n  b_{ik}c_{kj}, \]  or, equivalently,   \[  \sum_{j=1}^mg_{ij}=-\sum_{k=1}^mb_{ik}+\sum_{j=1}^m\sum_{k=1}^n b_{ik}c_{kj}=-\sum_{k=1}^mb_{ik}+\sum_{k=1}^nb_{ik}(\sum_{j=1}^mc_{kj}), \sum_{j=1}^mg_{ij}=-\sum_{k=1}^nb_{ik}+\sum_{j=1}^m\sum_{k=1}^n b_{ik}c_{kj}=-\sum_{k=1}^nb_{ik}+\sum_{k=1}^nb_{ik}(\sum_{j=1}^mc_{kj})=\sum_{k=1}^n[-b_{ik}+b_{ik}(\sum_{j=1}^mc_{kj})],  \] or, equivalently,  \begin{equation}\label{eq5}  \sum_{j=1}^mg_{ij}=\sum_{k=1}^n[(-1+\sum_{j=1}^mc_{kj})b_{ik}].  \end{equation}  From [9], if $det(Y_{LL})\neq 0$, then every row of $F_{LG}$ sums to one, i.e. $\sum_{j=1}^mc_{kj}=1$, $\forall k=1,2,...,n$ and thus from \eqref{eq5} $\sum_{j=1}^mg_{ij}=0$, i.e. every row of $Y_{GGM}$ sums to zero.  If $det(Y_{LL})=0$, then every row of $F_{LG}$ sums approximately to one, i.e. $\sum_{j=1}^mc_{kj}\cong 1$, $\forall k=1,2,...,n$ and thus from \eqref{eq5} $\sum_{j=1}^mg_{ij}\cong 0$, i.e. every row of $Y_{GGM}$ sums approximately to zero.  With similar steps if every row of $Y_{bus}$ sums approximately to zero then every row of $Y_{GGM}$ sums approximately to zero. The proof is completed.