TITLE: LU factorization of a non-symmetric matrix
QUESTION [6 upvotes]: I want to find $L$ and $U$ for the following non-symmetric matrix $A$:
$
 A = \begin{bmatrix}
a & r & r & r \\
a & b & s & s \\ 
a & b & c & t \\
a & b & c & d\\
     \end{bmatrix}
$
I'm not sure how the matrix being non-symmetric helps me. Does $A$ have to be a square matrix which is symmetric and invertible? Do I need to calculate its determinant to be nonzero in the first place? 
Since I don't know the answers to the above questions, I calculated $U$ anyway:
$
 U = \begin{bmatrix}
a&r&r&r\\
0&b-r&s-r&s-r\\ 
0&0&c-s&b-s\\
0&0&0&d-b
     \end{bmatrix}
$
But I don't know how to proceed about $L$. 
I'd also know what are the four conditions to get $A=LU$ with four pivots. I think the diagonal elements of both $L$ and $U$ must be nonzero to get four pivot elements, right? So $a \neq 0, b \neq r, c \neq s, d \neq b$, am I right?

REPLY [2 votes]: If $b\ne r$, $c\ne s$, you indeed get the $LU$ decomposition
$$
A=LU=\begin{bmatrix}
1 & 0 & 0 & 0 \\
-a & 1 & 0 & 0 \\
-a & 0 & 1 & 0 \\
-a & 0 & 0 & 1
\end{bmatrix}\,
\begin{bmatrix}
a&r&r&r\\
0&b-r&s-r&s-r\\ 
0&0&c-s&b-s\\
0&0&0&d-b
\end{bmatrix}
$$
The conditions $a\ne0$ and $d\ne b$ are irrelevant. If $a=0$ and $b=d$ the rank is $2$; if $a=0$ and $d\ne b$ the rank is $3$; if $a\ne0$ and $d=b$ the rank is $3$; if $a\ne0$ and $d\ne b$ the rank is $4$.
However, the $LU$ decomposition doesn't always exist. Let's see the case $b=s=r$, so that the first step in the elimination produces
$$
U = \begin{bmatrix}
a&r&r&r\\
0&0&0&0\\ 
0&0&c-s&0\\
0&0&0&d-b
\end{bmatrix}
$$
If $c\ne s$ or $d\ne b$, there's no way to write $A=LU$ with $L$ lower triangular and $U$ upper triangular.