How to prove that the following isn't true (using 3 by 3 matrix):
Given A is a square and a singular matrix (which means non invertible), if LU decomposition is possible without the use of permutation, then the LU decomposition is not unique.
Any help?
Consider $A=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0\end{bmatrix}$
Suppose you adopt the convention that $L$ is unit lower triangular.
$$L=\begin{bmatrix} 1 & 0 & 0 \\ l_{21} & 1 & 0 \\ l_{31} & l_{32} & 1\end{bmatrix}$$ and
$$U=\begin{bmatrix} u_{11} & u_{12} & u_{13} \\ 0 & u_{22}& u_{23} \\ 0 & 0 & u_{33}\end{bmatrix}$$
Direct computation show that it it unique.
From the first row of $A$, We can deduce that $u_{11}=1, u_{12}=0=u_{13}$.
From the second row of $A$, we can deduce that $l_{21}=0$, and hence $u_{22}=1$, $u_{23}=0$.
From the third row, $l_{31}=0, l_{32}=0, u_{33}=0$.