Elegant proof for nonsingular upper triangular matrix has an upper triangular inverse I am looking for (an elegant proof or a proof that does not use so many results) that a nonsingular upper triangular matrix $A$ has an upper triangular inverse. Here is what I have:
Nonsingular $\iff \det(A) \neq 0 \iff$ the rows of $A$ form a basis for $\mathbb{R}^n$
This is where my proof is "awkward/not-elegant/I'm not so sure what to do": From here we can just apply the reduced-row-reduction algorithm. And we will get $I$ the identity matrix. Thus $A^{-1}$ is the product of elementary matrices multiplied by $A$.\


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*I guess a good proof would show that we will indeed get the identity matrix.

 A: Hint
We have
$$A^{-1}=\frac{1}{\det A}\mathrm{Adj}(A)$$
Do you know how we calculate the adjugate of $A$?
Added: Another proof: Let $T_n$ denote the vector  space of $n$ by $n$ upper triangular matrices and define the linear transformation:
$$f_A: T_n\rightarrow T_n,\quad M\mapsto AM$$
so $f_A$ is well defined since the product of two upper triangular matrices is upper triangular and if $M\in\ker f_A$ then $f_A(M)=AM=0$ so $A^{-1}AM=M=0$ and then $f_A$ is injective so it's also bijective (since $T_n$ is a finite dimensional vector space).
Now since $I_n\in T_n$ then there's a unique matrix $X\in T_n$ such that $f_A(X)=AX=I_n$ so $X=A^{-1}$ is an upper triangular matrix.
A: Let $u$ be an inversible endomorphism of some vector space $E$. For every subspace $F \subset E$, it is clear that $F$ is stable by $u$ if, and only if, it is stable by $u^{-1}$. Apply this with the family of subspaces $F_i = \mathrm{span}\,(e_1,\dots,e_i)$ for $1 \leq i \leq n$ and the canonical endomorphism of $\Bbb R^n$ induced by $A$.
