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The first question: what is the proof that LU factorization of matrix is unique? Or am I mistaken?

The second question is, how can theentries of L below the main diagonal be obtained from the matrix $A$ and $A_1$ that results from the row echelon reduction to $U$? ($A=LU$)

Rodrigo de Azevedo
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Tashima Sasaki
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    Note that an $LU$ decomposition does not always exist: for instance among permutation matrices, only the identity matrices have an $LU$ decomposition. If an $LU$ decomposition exists it can be made unique by requiring diagonal entries $1$ for (say) $L$; however if a permutation matrix is thrown into the mix, (e.g. $LUP$) then uniqueness is no longer possible. The general notion is Bruhat decomposition, which at the element level is not unique. – Marc van Leeuwen Aug 21 '12 at 12:34

2 Answers2

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It is unique if you require $diag(L) = (1, \dots 1)$.

Because, assume that the LU-decomposition is not unique. Any other LU-decomp would be of the form: $$A=LU=LI_nU=LDD^{-1}U=(LD)(D^{-1}U)$$ with $D\in\mathbb{R}^{n\times n}$, $D\neq I_n$. For this $(LD)(D^{-1}U)$ to be a LU-decomposition, we need $LD$ lower triangular, $(D^{-1}U)$ upper triangular. In order for $LD$ to stay a lower triangular matrix, $D$ needs to be lower triangular. But $D$ lower triangular $\Rightarrow D^{-1}$ lower triangular matrix $\Rightarrow$ $(D^{-1}U)$ not upper triangular.

Therefore, $D$ can only be a diagonal matrix. Now, because $D\neq I_n \Rightarrow \exists d_{ii} \neq 1$ but in that case $$ (LD)_{ii} = \sum_{j=1} L_{ij}D_{ij} = L_{ii}D_{ii} + \sum_{j=1, j=i} L_{ij}*0 = 1\times D_{ii} +0 \neq 1$$

Resulting in $diag(LD) \neq (1, \dots 1)$ which is a contradiction. So such a $D$ does not exist, meaning the decomposition is unique.

Britta
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    Why "Any other LU-decomp would be of the form..."? What if we have $A = LU$ and $A = MV$ with $M \neq L$ and $V \neq U$? Why we should be able to obtain $M$ from $L$? – Figurinha Sep 09 '20 at 14:53
  • First, It's essential that $M$ is lower unit triangular matrix. This is reasonable since always $L$ is of that type in the $LU$ decomposition. Not allowing $M$ this way would mean, for example, that $M$ does not come from a Gaussian elimination algorithm from which the $LU$ derives from. After that, if we have $A=LU=MV$, we would get $M^{-1} L=VU^{-1}$. This means that $M^{−1}L$ is a upper triangular matrix and a lower triangular matrix, since is a product of two lower triangulars matrices and two upper triangular matrices. Hence, $M^{−1} L=VU^{−1}=D$ is a diagonal matrix. – R. W. Prado Jul 15 '21 at 18:23
  • $U$ is invertible, since $\text{rk}(A)\leq \text{min}{ \text{rk}(L), \text{rk}(U) }$, whenever $A=LU,$ which would mean that $A$ would not be invertible, and this is not compatible with the LU decomposition assumptions. – R. W. Prado Jul 15 '21 at 18:29
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The factorisation is not unique. There are $n^2+n$ coefficients to estimate and only $n^2$ "equations". As such, that is why there are the two "common" methods, Doolittle and Crout see wiki page. For each of these two approaches, you can show that the resulting linear system has a unique solution.

Daryl
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    It seems you're assuming a certain dimension? In any event: one often uses the degrees of freedom available to impose constraints on either the upper or lower triangular factor, e.g. having one of them be unit triangular... – J. M. ain't a mathematician Aug 21 '12 at 10:31
  • Edited! For some reason I had $3\times3$ matrices in my brain... – Daryl Aug 21 '12 at 12:13