Let $R$ be an unitary ring, and let $M_{n}(R)$ be the set of all $n\times n$ upper triangular Matrices. Prove that $A\in M_{n}(R)$ is invertible if and only if $a_{i,i}$ is invertible for all $i\in\{ 1,\ldots ,n\}$.
If $A$ is invertible, $\exists B\in T_{n}(R)$ such that $AB=I$, so I can write $$1_{R}=\sum_{k=1}^{n}a_{1,k}b_{k,1}$$ but $A, B$ are upper triangular matrices, so $a_{i,j}, b_{i,j}$ are $0_{R}$ if $i>j$ so $1_{R}=a_{11}b_{11}$, hence $a_{11}$ is invertible and I can do the same thing for the other elements.
The problem, for me, is to prove the other implication. For $n=2$ it's easy to compute an inverse for $A$, but I don't know how to do an induction argument over the order of the Matrix. Can someone give me an advice? Thanks before!
If $A$ is upper triangular with invertible diagonal entries, it can be written $A=DU$ where $D$ is diagonal with the same diagonal entries as $A$ and $U$ is upper triangular with diagonal entries equal to $1$.
The matrix $D$ is invertible by construction. The matrix $U$ can be written $U=I+N$, where $N$ is strictly upper triangular, hence $N^n=0$. Therefore, $U$ is invertible with inverse $\sum_{k=0}^{n-1} (-N)^k$. Hence $A=DU$ is invertible.