LU decomposition of a diagonally dominant matrix Let $A \in M_{n\times n}(\mathbb{R})$ a matrix such that $a_{ij}< 0$ for $i \ne j$ and  $A$ is diagonally (row) dominant, that is $a_{ii}>\sum_{j\ne i} |a_{ij}|$. I know that all the leading minors of $A$ are $>0$, and so $A$ has an LU decomposition 
$$A= L \cdot D \cdot U$$
where $L$ is lower triangular with $1$ on the diagonal, $D$ is a diagonal matrix with positive diagonal elements, $U$ is upper triangular with $1$ on the diagonal.
I would like to get a confirmation whether the off-diagonal elements of $L$ and $U$ are negative. 
 A: After the first step of the elimination (introduce the ${\bf D}$ if you want), we have
$$
{\bf A}:=
\begin{bmatrix}
a_{11} & {\bf a}_{12}^T \\
{\bf a}_{21} & {\bf A}_{22}
\end{bmatrix}
=
\begin{bmatrix}
1 & 0^T \\ {\bf l}_{21} & {\bf I}
\end{bmatrix}
\underbrace{\begin{bmatrix}
u_{11} & {\bf u}_{12}^T \\
0 & \tilde{\bf A}_{22}
\end{bmatrix}}_{\tilde{\bf A}}
$$
where
$$
u_{11} = a_{11}>0, \quad {\bf u}_{12}={\bf a}_{12}<0, \quad {\bf l}_{21}=\frac{1}{a_{11}}{\bf a}_{21}<0, \quad \tilde{\bf A}_{22}={\bf A}_{22}-\frac{1}{a_{11}}{\bf a}_{21}{\bf a}_{12}^T.
$$
The LU factorization continues by applying this step to the submatrix $\tilde{\bf A}_{22}$, which is again diagonally dominant with negative off-diagonal entries.
Using the row diagonal dominance of ${\bf A}$ and the inequality $$|\alpha|\geq|\alpha-\beta|-|\beta|\tag{*}$$ we show that, for all $i \geq 2, $
$$
\begin{split}
\tilde{a}_{ii}
&=
a_{ii}-\frac{a_{i1}a_{1i}}{a_{11}}
\\&>
\sum_{j\neq i}|a_{ij}|-\frac{a_{i1}a_{1i}}{a_{11}}
\\&=
|a_{i1}|+\sum_{j\geq 2\\j\neq i}|a_{ij}|-\frac{a_{i1}a_{1i}}{a_{11}}
\\&\overset{(*)}{\geq} 
|a_{i1}|+\sum_{j\geq 2\\j\neq i}\left|a_{ij}-\frac{a_{i1}a_{1j}}{a_{11}}\right|-\frac{|a_{i1}|}{a_{11}}\sum_{j\neq i}|a_{1j}|
\\&=
\sum_{j\geq 2\\j\neq i}|\tilde{a}_{ij}|
+|a_{i1}|\left(1-\frac{1}{a_{11}}\sum_{j\neq i}|a_{1j}|\right)
\\&>
\sum_{j\geq 2\\j\neq i}|\tilde{a}_{ij}|, \end{split}
$$
so $\tilde{\bf A}_{22}$ is also row diagonally dominant. It is easy to see that $\tilde{\bf A}_{22}$ has negative off-diagonal elements because $\frac{1}{a_{11}}{\bf a}_{21}{\bf a}_{12}^T>0$.
