I know how LDU decomposition works and moreover, I know that these closed formulas are more difficult to compute and hence it is not done in practice, but I want to understand the following cite from Wikipedia:
Closed formula When an LDU factorization exists and is unique, there is a closed (explicit) formula for the elements of L, D, and U in terms of ratios of determinants of certain submatrices of the original matrix A.In particular, $D_{1}=A_{1,1}$ and for i = 2 , … , n $D_i$ is the ratio of the i -th principal submatrix to the ( i − 1 )-th principal submatrix. Computation of the determinants is computationally expensive, so this explicit formula is not used in practice.
Unfortunately I do not understand how this formula looks. Especially what is the i-th principal submatrix. I am not familiar with this term and it is not explained in more detail. I tried reformulating the recursive formulas that are usually used to explicit ones but this gets very fast very messy and I did not find an easy technique for doing so.
Here is the full wikipedia article: https://en.wikipedia.org/wiki/LU_decomposition
Thank you very much in advance for your help.
EDIT: Hello,
I was thinking about it since yesterday and as soon as I post the question, I realize what part of the answer must be. The definition of principal submatrix that is meant, it also called leading principal submatrix and obtained by keeping the first i rows and column of a matrix and eliminating the others. Hence $D_i = \frac{det A_i}{det A_{i-1}}$. The part that is still missing is the formula for $L_i$ and $U_i$.
Use slice notation $m:n=[m,m+1,..,n]$, so that for example $A_i=A_{1:i,1:i}$. Then we want to achieve the decomposition in the step $i\to i+1$ $$ A_{i+1}=\pmatrix{A_i&A_{1:i,i+1}\\A_{i+1,1:i}&A_{i+1,i+1}} = \pmatrix{L_i&0\\L_{i+1,1:i}&1} \pmatrix{D_i&0\\0&d_{i+1}} \pmatrix{U_i&U_{1:i,i+1}\\0&1} $$ Of course, $A_i=L_iD_iU_i$ by iteration assumption. Also from the determinant laws, $\det(A_i)=\det(D_i)$ so that $d_{i+1}=\det(A_{i+1})/\det(A_i)$.
For the other parts multiply out the matrix product $$ =\pmatrix{L_iD_i&0\\ L_{i+1,1:i}D_i&d_{i+1}} \pmatrix{U_i&U_{1:i,i+1}\\0&1} = \pmatrix{A_i&L_iD_iU_{1:i,i+1}\\L_{i+1,1:i}D_iU_i&L_{i+1,1:i}D_iU_{1:i,i+1}+d_{i+1}} $$ to then get $$ L_iD_iU_{1:i,i+1}=A_{:i,i+1}=(L_iD_iU_i)A_i^{-1}A_{1:i,i+1}\implies U_{1:i,i+1}=U_iA_i^{-1}A_{1:i,i+1} \\ L_{i+1,1:i}D_iU_i=A_{i+1,1:i}=A_{i+1,1:i}A_i^{-1}(L_iD_iU_i)\implies L_{i+1,1:i}=A_{i+1,1:i}A_i^{-1}L_i \\ d_{i+1}=A_{i+1,i+1}-L_{i+1,1:i}D_iU_{1:i,i+1} $$ Remember that the entries of $A_i^{-1}=\frac{A_i^\#}{\det(A_i)}$ (using the adjoint/adjunct/adjugate which is the transpose of the co-factor matrix) are fractions of determinants, and for further consideration, that constructs like $A_i^\#A_{1:i,i+1}$ can be interpreted as consisting of Lagrange determinant formulas, giving $-\bigl(A_{i+1}^\#\bigr)_{1:i,i+1}$. Thus in total $$ U_{1:i,i+1} = -\frac{U_i\bigl(A_{i+1}^\#\bigr)_{1:i,i+1}}{\det(A_i)}\\ L_{i+1,1:i} = -\frac{\bigl(A_{i+1}^\#\bigr)_{i+1,1:i}L_i}{\det(A_i)} $$