Show that if matrix $A \in \mbox{GL}_n(\mathbb{R})$ and $A$ has decomposition $A=LR$, where $L$ is a normed triangular matrix $(a_{ii}=1$, for all $i$.) and $R$ is an invertible upper triangular matrix then the decomposition is unique.
I was trying to prove this and thought about using the fact that if a matrix has an inverse it is unique so my understanding was if $A=LR \implies A^{-1}=(LR)^{-1}=R^{-1}L^{-1}$ has to be unique but I'm not sure if this statement implies that the original decomposition is unique. Is there a simpler way to prove this?
I will assume that $L$ is lower triangular. Suppose $A$ has two such decompositions $A = L_1R_1 = L_2R_2$. Then $L_{2}^{-1}L_1 = R_2R_{1}^{-1}$. It is easy to check that the LHS is lower triangular and has $1$ on the diagonal and that the RHS is upper triangular. It follows that both sides are the identity matrix.