How do I show that every LU matrix A∈K$_n$$_×$$_n$ admits a unique LDU factorisation, that is a triple L∈L$_1$(Kn),D∈D(Kn) and U∈U$_1$(Kn) such that A=LDU? I'm new to these factorisations and I'm just trying to get to grips with the composition of each factorisation.
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If a matrix $A$ has a $L_1U_1$ factorization it also has an $L_2D_2U_2$ factorization with $D_2U_2$ = $U_1$. $U_2$ is just the normalised version of $U_1$ where the diagonal is all one. $L_1$, $L_2$, $U_2$ are all unitriangular matrices.
Adrian AR
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