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 $A$.\
- I guess a good proof would show that we will indeed get the identity matrix.
Best Answer
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.