I was trying to prove this -
Let $A$ be a lower triangular $n\times n$ matrix with nonzero entries on the diagonal. Show that $A$ is invertible and the inverse of $A$ is lower triangular.
[Hint: Explain why $A$ can be changed into $I$ using only row replacements and scaling. (Where are the pivots?) Also, explain why the row operations that reduce $A$ to $I$ change $I$ into a lower triangular matrix.]
I believe we can prove this by considering the matrix $[A|I]$, now we since $A$ is lower triangular we can first scale the pivot in the first row to 1 and then use it to remove all subsequent entries in first column below it then scale the pivot in column 2 and so on.
Clearly we can reduce the matrix to the identity matrix but at the same time I will have the same row operations so we'll be multiplying and adding multiples of the pivots to rows below it and generating a lower triangular matrix only since there will be no instance when we add a row to a row above it we'll only use the pivots to remove the entries in the column below it. So the inverse generated will be a lower triangular matrix only using the theorem which states that if a set of row operations reduce a matrix to I then they reduce I to the its inverse
Now this is logically correct as far as I know but this doesn't look much like a proof rather an intuitive argument to me and there is where I always get stuck. I get to a point where I know why it's true but don't know how to write it formally. So using this example can someone give tips and techniques to write it in a more "proof like" manner if I may say so
I guess it's been asked here but I'm not looking for a proof per say but a way to convert my intuition into a set of rigorous statements