If $A=LDU$ and also $A=L_1 D_1 U_1$ with all factors invertible, then $L={L}_{1}$ and $D={D}_{1}$ and $U={U}_{1}$. "The three factors are unique."
Derive the equation $L_1^{-1} LD=D_1 U_1 U^{-1}$. Are the two sides triangular or diagonal? Deduce $L=L_1$ and $U=U_1$ (they all have diagonal $1$s). Then $D=D_1$.
This question is very similar to this question; however, some of the assumptions used in that answer are unclear to me.
I can derive that $L_1^{-1} LD= D_1 U_1 U^{-1}$. But I don't understand how to proceed from here. I have a series of questions.
- The question never states that either $L$ is lower triangular nor that $U$ is upper triangular, so how does one go about inferring this?
- I understand that (lower triangular) * (lower triangular)=(lower triangular) but is it also true that (lower triangular) * (arbitrary matrix)=(lower triangular)? If so, can I have some intuition for that? This is necessary for $L_1^{-1} LD$ to be a lower triangular matrix.
I think I understand everything else. Thanks for the insight!
The point of an $LDU$ factorization is that $L$ is lower-triangular, $D$ is diagonal and $U$ is upper triangular. Hence, when you want to show that such a factorization is unique, you assume there are 2 such factorizations, $LDU = A = L_1 D_1 U_1$ and establish that $L=L_1, D=D_1$ and $U=U_1$. Hence, $L_1, D_1, U_1$ are lower-triangular, diagonal, upper-triangular by assumption, just as $L,D,U$ are.
Note that $D$ is not any matrix, so even though $L X$ is not lower triangular for an arbitrary matrix $X$, you will have $LD$ as lower triangular since $D$ is special. So too, $DL$... Can you complete the reason why?