Normalised LU and LDU factorisations with nonzero pivots are unique

For claim 1, the first row of $A$ equals the first row of both $U$ and $U^{\mathrm{\prime }}$, because the first rows of $L$ and $L^{\mathrm{\prime }}$ are $(1,0,\dots ,0)$. Hence the first rows of $U$ and $U^{\mathrm{\prime }}$ agree, in particular $u_{00}=u_{00}^{\mathrm{\prime }}\mathrm{\ne }0$. Then the first-column equations are $a_{i0}={\mathrm{\ell }}_{i0}{u}_{00}={\mathrm{\ell }}_{i0}^{\mathrm{\prime }}{u}_{00}^{\mathrm{\prime }}$ for $i>0$, so ${\mathrm{\ell }}_{i0}={\mathrm{\ell }}_{i0}^{\mathrm{\prime }}$. Subtracting the common first-column outer product leaves two factorizations of the same trailing principal block by unit lower and upper triangular matrices of size $n-1$ with nonzero diagonal. Induction gives equality of the trailing blocks, hence $L=L^{\mathrm{\prime }}$ and $U=U^{\mathrm{\prime }}$.

For claim 2, put $V:=DU$ and $V^{\mathrm{\prime }}:=D^{\mathrm{\prime }}{U}^{\mathrm{\prime }}$. Because $D,D^{\mathrm{\prime }}$ are diagonal and $U,U^{\mathrm{\prime }}$ are unit upper triangular, $V,V^{\mathrm{\prime }}$ are upper triangular and their diagonals are exactly the diagonals of $D,D^{\mathrm{\prime }}$, so they are nonzero. The factorisations $A=LV=L^{\mathrm{\prime }}{V}^{\mathrm{\prime }}$ are therefore of the kind handled in step 1.1, which gives $L=L^{\mathrm{\prime }}$ and $V=V^{\mathrm{\prime }}$. Taking diagonals of $V=V^{\mathrm{\prime }}$ yields $D=D^{\mathrm{\prime }}$, and then $DU=D^{\mathrm{\prime }}{U}^{\mathrm{\prime }}$ forces $U=U^{\mathrm{\prime }}$.

Steps 1.1 and 2.1 prove claims 1 and 2. The nonzero-pivot hypothesis is essential: without it, a zero row of the upper factor leaves lower-triangular multipliers undetermined.