TITLE: Demonstrate that a matrix has no LU factorization
QUESTION [5 upvotes]: Have to show that 
$$\begin{bmatrix}0 & 1\\1 & 1\end{bmatrix}$$
has no LU factorization.
It seems trivial just to say that this cannot have an LU decomposition because it is a lower triangular matrix already.
What is a more formal way of suggesting that this cannot be furth decomposed into LU?

REPLY [10 votes]: Suppose $A = \pmatrix{0 & 1 \\ 1 & 1}$ has an $LU$ decomposition.
Then $A=LU$ for some lower triangular matrix $L$ and upper triangular matrix $U$.
Let $L= \pmatrix{ l_{11} & 0 \\ l_{21} & l_{22}}$ and $U = \pmatrix{ u_{11} & u_{12} \\ 0 & u_{22}}$
By matrix multiplication, we conclude the following:
$a_{11} = l_{11} \cdot u_{11} = 0$, so it must be either $l_{11}=0$ or $u_{11}=0$.
Suppose $l_{11}=0$, since $a_{12} = l_{11} \cdot u_{21}$ this would imply $a_{12} = 0 \neq 1$. A contradiction.
Suppose $u_{111}=0$, since $a_{21} = l_{21} \cdot u_{11}$ this would imply $a_{21} = 0 \neq 1$. A contradiction.
So we conclude that $A$ cannot have an $LU$ decomposition. $\square$