show a invertible matrix has unique LU factorization

Question

I have a n by n invertible matrix and want to show that it has a unique LU decomposition when the diagonal of the lower triangular matrix are ones.

I prove this by contradiction by saying there exist a diagonal matrix D such that A=LDD'U is another LU decomposition and showing that D actually does not exist.

How do I prove by using the inverse of the triangular matrix? with the fact that its diagonal entries are ones.

Thanks!

Answer 1

Suppose that $A$ invertible has a $LU$ decomposition where the diagonal of $L$ has only ones on the diagonal. This implies that $L$ is also invertible and therefore $U$.

Now supposing that

$$A=L_1 U_1 = L_2 U_2$$ we get

$$L_2^{-1} L_1 = U_2 U_1^{-1}$$ and therefore that both $L_2^{-1} L_1, U_2 U_1^{-1}$ are diagonal. As $L_1, L_2$ are supposed to only have ones on the diagonal, $L_2^{-1} L_1$ is the identity matrix, as well as $U_2 U_1^{-1}$. This implies the desired conclusion $L_1=L_2$ and $U_1=U_2$.