Let $A$ be a ring and $B:=(\lambda_{ij})_{ij}$ an upper triangular $n\times n$ matrix over $A$ whose diagonal elements are invertible. I want to show that $B^{-1}$ exists and is upper triangular.
Let $(e_i)_i$ be the canonical basis of $A^{\oplus n}_d$. Since the diagonal elements of $B$ are invertible, there exists $u\in\mathbb{GL}(A^{\oplus n}_d)$ such that $u(e_i)=\sum_{j=1}^ne_j\lambda_{ji}$. This shows that $B^{-1}=M(u^{-1})=\left(\langle e^*_k,u^{-1}(e_i)\rangle\right)_{ki}$.
Now, we have to finish by showing that $B^{-1}$ is in fact upper triangular: i.e. that for $i<k$ we have $\langle e^*_k,u^{-1}(e_i)\rangle=0$; this means that the $k$-th component of $u^{-1}(e_i)$ w.r.t. the basis $(e_i)_i$ is equal to zero.
Take $i<k$ in the interval $[1,n]$. Then $u^{-1}(e_i)=\sum_{l=1}^ne_la_l$ and
$$e_i=\sum_{l=1}^n\sum_{j=1}^ne_j\lambda_{jl}a_l=\sum_{j=1}^ne_j\left(\sum_{l=1}^n\lambda_{jl}a_l\right).$$
This implies that $\sum_{l=1}^n\lambda_{kl}a_l=0$. I am having difficulty showing that $a_k=0$. Any suggestions?
Best Answer
Doesn't a simple induction argument work? We know that $e_{n} = a_{nn}^{-1} u_n$ and now $e_{n-1} = u_{n-1} - a_{n-1, n} e_n = u_{n-1} - a_{n-1, n} a_{n, n}^{-1} u_n,$ and so on.