Given a triadiagonal Matrix A with
$$A = \begin{bmatrix} d_1 &e1\\ c_2 & d_2 & e_2 \\ & c_3 & \ddots & \ddots \\ & & \ddots & \ddots & e_{n-1} \\ & & & c_n & d_n \\ \end{bmatrix} \quad $$
Assume that A is invertible and has a LU factorization.
Proof that L and U have this structure:
$$ L = \begin{bmatrix} 1 \\ l_2 & 1 \\ & l_3 & \ddots \\ & & \ddots & \ddots \\ & & & l_n & 1 \\ \end{bmatrix} \quad U = \begin{bmatrix} u_1 & e_1 \\ & u_2 & e_2 \\ & & \ddots & \ddots \\ & & & \ddots & e_{n-1} \\ & & & & u_n \\ \end{bmatrix} $$
I've found out, that an invertible matrix has diagonal elements $\neq 0$. I tried to start by A and do a LU factorization, but I couldn't get to the needed structure. Than I've read that this could be shown by induction.
So the simplest case should be $n = 2$. Then I would have $$ A = \begin{bmatrix} d_1 & e_1 \\ c_2 & d_2 \\ \end{bmatrix} \quad L = \begin{bmatrix} 1 & 0 \\ l_2 & 1 \\ \end{bmatrix} \quad U = \begin{bmatrix} u_1 & e_1 \\ 0 & u_2 \\ \end{bmatrix}$$ Now, I don't know how to show that this case is true. In all literature I just found the comment, that this is trivial, but I don't understand why. I also don't know how to show for further cases by induction that the structure must hold. Any help would be much appreciated.
The statement is not true in general.
$$L = \begin{bmatrix} 1 \\ l_2 & 1 \\& & \ddots & \ddots \\ & & & l_n & 1 \\ \end{bmatrix} \quad U = \begin{bmatrix} u_1 & e_1 \\ & u_2 & e_2 \\ & & \ddots & \ddots \\ & & & \ddots & e_{n-1} \\ & & & & u_n \\ \end{bmatrix}$$ will generate the same product $LU$ as $2L$ and $(1/2) U$