Let A be equal to this matrix:
Does this matrix has a single LU decomposition?
It is simple to find a decomposition without permutations. However how can I tell if there is more than one?
Basically, I am asking if 2 decompositions with the form A=LU exist for this matrix. Obviously you could find more decompositions but with the form PA=LU.
Because $A$ isn't invertible, there are many possibilities without permutations. Here are two:
$$\left(\begin{matrix} 1 & 2 & 3 \\ 1 & 2 & 4 \\ 1 & 2 & 5 \end{matrix}\right) = \left(\begin{matrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 2 & 1 \end{matrix}\right)\cdot\left(\begin{matrix} 1 & 2 & 3 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{matrix}\right) = \left(\begin{matrix} 1 & 0 & 0 \\ 1 & 1/2 & 0 \\ 1 & 1 & 0 \end{matrix}\right)\cdot\left(\begin{matrix} 1 & 2 & 3 \\ 0 & 0 & 2 \\ 0 & 0 & 0 \end{matrix}\right) $$