Ldu decomposition
From linear algebra:
LDU decomposition states that for every square matrix A, there exists:
 a permutation matrix P
 a lowertriangular matrix L with diagonals of 1
 a diagonal matrix D
 and an upper triangular matrix U with diagonals of 1
such that PA = LDU.
The permutation matrix can possibly be equal to the unit matrix I. If it is, than LDU is a unique decomposition, that is, for any other decomposition L!=L*, D!=D*, and U!=U*.
We can show this by showing a contradiction if it's not true:

If we solve the equation to:
If you multiply a lowertriangular matrix by another, you get a lower triangular matrix. Therefore the inverse of a lowertriangular matrix must also be a lowertriangular matrix. If you multiply a lowertriangular matrix by a diagonal matrix, you still have a lowertriangular matrix. Therefore, the left side of the above equation is a lowertriangular matrix. The same arguments show that the right side must be an uppertriangular matrix. Therefore, the only way the above equation could hold (besides with null matrices) is if the terms (L with the inverse of L*, and U with the inverse of U*) cancel, which shows that they are equal because a term would only cancel if it was its own inverse.