TITLE: How can LU factorization be used in non-square matrix?
QUESTION [18 upvotes]: In my textbook, there is some information about LU factorization of square matrix $A$, but not about non-square matrix.
How can LU factorization be used to factorize non-square matrix?

REPLY [27 votes]: I'll illustrate how to understand the LU-decomposition of a particular $3 \times 4$ matrix below. The method works just as well for other sizes since the LU-decomposition arises naturally from the study of Gaussian elimination via multiplication by elementary matrices.
$$
\begin{array}{ll}
A  \ = &\left[ 
 \begin{array}{cccc}
   1  & 2 & -3 & 1\\
   2  & 4 & 0  & 7 \\
   -1 & 3 & 2  & 0 
 \end{array} 
\right]
\ \underrightarrow{r_2-2r_1 \rightarrow r_2} \
\left[ 
 \begin{array}{cccc}
   1  & 2 & -3 & 1\\
   0  & 0 & 6  & 5 \\
   -1 & 3 & 2  & 0 
 \end{array} 
\right]
\ \underrightarrow{r_3+r_1 \rightarrow r_3} \\ & \\
&\left[ 
 \begin{array}{cccc}
   1  & 2 & -3 & 1\\
   0  & 0 & 6  & 5 \\
   0  & 5 & -1  & 1 
 \end{array} 
\right]
\ \underrightarrow{r_2 \leftrightarrow r_3} \
\left[ 
 \begin{array}{cccc}
   1  & 2 & -3 & 1\\
   0  & 5 & -1  & 1 \\
   0  & 0 & 6  & 5 
\end{array} 
\right] = \ U
\end{array}
$$
We have $U = E_3E_2E_1A$ hence $A = E_1^{-1}E_2^{-1}E_3^{-1}U$ and we can calculate the product $E_1^{-1}E_2^{-1}E_3^{-1}$ as follows:
$$ 
\begin{array}{ll}
I  \ = &\left[ 
 \begin{array}{ccc}
  1  & 0 & 0   \\
  0  & 1 & 0   \\
  0  & 0 & 1 
 \end{array} 
\right]
\ \underrightarrow{r_2 \leftrightarrow r_3} \
\left[ 
 \begin{array}{ccc}
  1  & 0 & 0   \\
  0  & 0 & 1   \\
  0  & 1 & 0 
 \end{array} 
\right] 
\ \underrightarrow{r_3-r_1 \rightarrow r_3}  \\ & \\
&\left[ 
 \begin{array}{ccc}
  1  & 0 & 0   \\
  0  & 0 & 1   \\
  -1  & 1 & 0 
 \end{array} 
\right]
\ 
\underrightarrow{r_2+2r_1 \rightarrow r_2} \
\left[ 
 \begin{array}{ccc}
  1  & 0 & 0   \\
  2  & 0 & 1   \\
  -1  & 1 & 0 
 \end{array} 
\right] = PL
\end{array}
$$
I have inserted a "$P$" in front of the $L$ since the matrix above is not lower triangular. However, if we go one step further and let $r_2 \leftrightarrow r_3$ then we will obtain a lower triangular matrix:
$$ 
\begin{array}{ll}
PL  \ = &\left[ 
 \begin{array}{ccc}
  1  & 0 & 0   \\
  2  & 0 & 1   \\
  -1  & 1 & 0 
 \end{array} 
\right]
\ \underrightarrow{r_2 \leftrightarrow r_3} \
\left[ 
 \begin{array}{ccc}
  1  & 0 & 0   \\
  -1  & 1 & 0 \\
  2  & 0 & 1   \\
 \end{array} 
\right] =L
\end{array}
$$
Therefore, we find that $E_1^{-1}E_2^{-1}E_3^{-1}=PL$ where $L$ is as above and $P = E_{2 \leftrightarrow 3}$. This means that $A$ has a modified $LU$-decomposition. Some mathemticians call it a $PLU$-decomposition,
$$ A = \underbrace{\left[ 
 \begin{array}{ccc}
  1  & 0 & 0   \\
  0  & 0 & 1   \\
  0  & 1 & 0 
 \end{array} 
\right]}_{P}
\underbrace{\left[ 
 \begin{array}{ccc}
  1  & 0 & 0   \\
  -1  & 1 & 0 \\
  2  & 0 & 1   \\
 \end{array} 
\right]}_{L}\underbrace{\left[ 
 \begin{array}{cccc}
   1  & 2 & -3 & 1\\
   0  & 5 & -1  & 1 \\
   0  & 0 & 6  & 5 
\end{array} 
\right]}_{U} = \underbrace{\left[ 
 \begin{array}{ccc}
  1  & 0 & 0   \\
  2  & 0 & 1   \\
  -1  & 1 & 0 
 \end{array} 
\right]}_{PL}\underbrace{\left[ 
 \begin{array}{cccc}
   1  & 2 & -3 & 1\\
   0  & 5 & -1  & 1 \\
   0  & 0 & 6  & 5 
\end{array} 
\right]}_{U}. $$
Since permutation matrices all satisfy the condition $P^k=I$ (for some $k$) the existence of a $PLU$-decomposition for $A$ naturally suggests that $P^{k-1}A = LU$. Therefore, even when a $LU$ decomposition is not available we can just flip a few rows to find a $LU$-decomposable matrix. This is a useful observation because it means that the slick algorithms developed for $LU$-decompositions apply to all matrices with just a little extra fine print.
Much of the writing above can be spared if we adopt the notational scheme illustrated below. 
$$ 
\begin{array}{ll}
A  \ = &\left[ 
 \begin{array}{cccc}
   1  & 2 & -3 & 1\\
   2  & 4 & 0  & 7 \\
   -1 & 3 & 2  & 0 
 \end{array} 
\right]
\ \underrightarrow{r_2-2r_1 \rightarrow r_2} \
\left[ 
 \begin{array}{cccc}
   1  & 2 & -3 & 1\\
   (2)  & 0 & 6  & 5 \\
   -1 & 3 & 2  & 0 
 \end{array} 
\right]
\ \underrightarrow{r_3+r_1 \rightarrow r_3} \\ & \\
&\left[ 
 \begin{array}{cccc}
   1  & 2 & -3 & 1\\
   (2)  & 0 & 6  & 5 \\
   (-1)  & 5 & -1  & 1 
 \end{array} 
\right]
\ \underrightarrow{r_2 \leftrightarrow r_3} \
\left[ 
 \begin{array}{cccc}
   1  & 2 & -3 & 1\\
   (-1)  & 5 & -1  & 1 \\
   (2)  & 0 & 6  & 5 
\end{array} 
\right] = \ U
\end{array}
$$
We find if we remove the parenthetical entries from $U$ and ajoing them to $I$ then it gives back the matrix $L$ we found previously:
$$
U = \left[ 
 \begin{array}{cccc}
   1  & 2 & -3 & 1\\
   0  & 5 & -1  & 1 \\
   0  & 0 & 6  & 5 
\end{array} 
\right] 
\qquad 
L=\left[ 
 \begin{array}{ccc}
   1  & 0 & 0 \\
   -1  & 1 & 0 \\
   2  & 0 & 1  
\end{array} 
\right].
$$
Hope this helps.