Doolittle transformation is non-unique for singular matrices Decomposing the singular matrix $$A = \begin{bmatrix} 
1 & 2 \\ 
1 & 2
\end{bmatrix} = \begin{bmatrix}1 & 0 \\ 1 & 1\end{bmatrix}\begin{bmatrix}1 & 2 \\ 0 & 0\end{bmatrix}=LU$$
by Doolittle decomposition seems to be unique for this case. But how to proove that this is not necessarily unique?
 A: The row of zeroes in your $U $ allows you to play with the second column of $L $. You have 
$$
 \begin{bmatrix} 
1 & 2 \\ 
1 & 2
\end{bmatrix} = \begin{bmatrix}1 & 0 \\ 1 & x\end{bmatrix}\begin{bmatrix}1 & 2 \\ 0 & 0\end{bmatrix}
$$ for any choice of $x $.
If you need $x=1$, there is no other choice and the decomposition is unique for that matrix.  Such is the case for any singular $2\times2$ matrix: if
$$
A=\begin{bmatrix}r&s\\ tr&ts\end{bmatrix}
=
\begin{bmatrix}1&0\\
 x&1\end{bmatrix}
 \begin{bmatrix}a&b\\
 0&c\end{bmatrix},
$$ it follows immediately that $a=r $, $b=s $, $x=t $, $c=0$.
For $3\times3$, here is an example where you are free to choose $z $:
 $$
A=\begin{bmatrix}1&1&1\\1&1&1\\1&1&1\end{bmatrix}
=
\begin{bmatrix}1&0&0\\
 1&1&0\\
1&z&1\end{bmatrix}
 \begin{bmatrix}1&1&1\\
 0&0&0\\ 0&0&0\end{bmatrix}.
$$
On the other hand, the decomposition is always unique when $A$ is non-singular and we require one of the two triangular matrices to have all ones in the diagonal, like
 $$
A=\begin{bmatrix}1&2&3\\4&6&9\\5&8&11\end{bmatrix}
=
\begin{bmatrix}1&0&0\\
 4&1&0\\
5&1&1\end{bmatrix}
 \begin{bmatrix}1&2&3\\
 0&-2&-3\\ 0&0&-1\end{bmatrix}
$$
is unique.
