How is A= UL different from A = LU
Using Guassian elimination introducing zeros into the columns of $A$ in the order, $n: - 1:2$ and producing the factorization $A = UL$ where $U$ is unit upper triangular and $L$ is lower triangular. How is this different from $A = LU.$ Can I relate the two procedures?
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$\begingroup$$$ \pmatrix{ 1 & 2 \\ 1 & 1 } = \pmatrix{1 & 2 \\ 0 & 1} \pmatrix{-1 & 0 \\ 1 & 1} \\ \pmatrix{ 1 & 2 \\ 1 & 1 } = \pmatrix{1 & 0 \\ 1 & 1} \pmatrix{1 & 2 \\ 0 & -1} $$So evidently the two approaches produce different results.
One (weak) relation is that if you start with $$ A = L U $$and transpose everything, you get $$ A^t = U^t L^t $$and because $U^t$ is lower-triangular, and $L^t$ is upper triangular, you get$$ A^t = L' U', $$but this time $L'$ has the ones on the diagonal rather than $U$ having ones on the diagonal. I guess that's really a relationship between two different versions of LU factorization, rather than one being UL factorization.
$\endgroup$ 1 $\begingroup$Observe that if $$A = LU$$
then (assuming $A^{-1}$ exists.) :
$$A^{-1} = U^{-1}L^{-1}$$ We can verify this by hand if you don't believe it.
What remains is to show that $U^{-1}$ is also upper triangular and $L^{-1}$ is also lower triangular if $U,L$ are, respectively.
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