Let $L$ be a lower triangular matrix, $D$ diagonal, and $U$ upper triangular.
If $A = LDU$ and also $A = L_1D_1U_1$ with all factors invertible, then $L = L_1$ and
$D = D_1$ and $U = U_1$. 'The three factors are unique!'Hint:
Derive the equation $L_1^{-1}LD = D_1 U_1 U^{-l}.$ Are the two sides triangular or diagonal?
Deduce $L = L^{-1}$ and $U = U^{-1}$ (they all have diagonal $1$'s). Then $D = D_1$.Lemma: The inverse of a (lower/upper) triangular matrix is a (lower/upper) triangular matrix.
Uniqueness Proof: Assume $LDU = L_1D_1U_1$. Objective: Prove $X = X_1$ for each $X = L, D, U$.
In keeping with the hint, $\color{green}{L_1^{-1}}LDU\color{#D555D1}{U^{-1}} = \color{green}{L_1^{-1}}L_1D_1U_1\color{#D555D1}{U^{-1}} \iff \color{green}{L_1^{-1}}LD = D_1U_1\color{#D555D1}{U^{-1}}$ $\Longrightarrow (\text{Lower triangular})D = D_1(\text{Upper triangular}).$
$\color{red}{\bigstar} $ So both sides are diagonal. $ \; L,U,L_1,U_1$ have diagonal $1$’s so $D = D_1$. Then $\color{green}{L_1^{-1}}L = I$ and $U_1\color{#D555D1}{U^{-1}} = I. \qquad \blacksquare$
$\Large{1.}$ I don't apprehend the (gruff) sentences after $\color{red}{\bigstar}$. Would someone please enlarge upon them?
$\Large{2.}$ Without the hint, how would one divine/previse to work with $L_1^{-1}LD = D_1 U_1 U^{-l}$?
This equation looks like the critical one in this proof.
Best Answer
Google this problem and found this page :( but finally I worked out this problem somehow. So I will try to answer this question.
First notice that $L_1^{-1}L$ is a lower triangular matrix and has diagonal 1's, since both $L_1^{-1}$ and $L$ are lower triangular and have diagonal 1's.
Nonzero entries below the diagonal of $L_1^{-1}L$ will cause nonzero entries below the diagonal of $L_1^{-1}LD$ (since $D$ is also diagonal). And if that happens, $L_1^{-1}LD$ cannot equal to $D_1U_1U^{-1}$, thus entries below the diagonal of $L_1^{-1}L$ are all 0's.
Now we can conclude $L_1^{-1}L=I$. Similarly, $U_1U^{-1}=I$.
As for the second question, I can not help.