Prove that upper triangular matrices are closed under inversion Prove that the group of upper triangle matrices $G\subset GL_n(F)$ is also a subgroup of $GL_n(F)$
Contains identity: $I_n\in G$ since $I_{ij}=0$ when $i>j$
Closed under multi: $C=AB, c_{ij}=\sum_{k=1}^na_{ik}b_{kj},i=1,2..,n,j=1,2..,n$ for all $c_{ij},i>j$ either $i\ge k>j$ or $i>k\ge j$. One factor in each addend is $0$: either $a_{ik}=0$, or $b_{kj}=0$ thus $c_{ij}=0$ if $i>j\rightarrow C\in G$ 
Closed under inversion: We know that a unique inverse exist since G is atleast a subset of $GL_n(F)$. Any tips on how do I proceed form here on? I thought about using my proof above to prove that $A^{-1}\in G$ How do I know there is not a fringe case where $AA^{-1}=I, A^{-1}\not\in G$
 A: If $A$ is an invertible matrix, then its inverse $A^{-1}$ can be expressed as a polynomial in $A$.
Therefore, if $A$ is upper triangular, so is $A^{-1}$, since the product of upper triangular matrices is upper triangular, as you have already proved.
A: Suppose $A^{-1}$ is not upper triangular. Consider any one of the rows, say $i$, of $A^{-1}$ such that it has non - zero entries before it's diagonal entry $a_{ii}$. Let the first of these be in the $j^{th}$ column.
Finally, a hint - consider the element of the product of $A^{-1}A$ in the position $(i,j)$
A: Conceptually, the algebra of upper triangular matrices $UT_n(F)$ is the set of transformations which fixes a complete flag of subspaces of $F^n$.
That is, for all upper triangular matrices, there is a chain of subspaces of $F^n$ written as $0\subseteq V_1\subseteq V_2\subseteq\ldots\subseteq V_n$, where $\dim_F(V_i)=i$, such that for every upper triangular matrix $A$, and each index $i$, $A(V_i)\subseteq V_i$. Conversely given any such flag, you can choose a basis such that the flag-preserving transformations are upper triangular. You should have no problem describing such a flag of subspaces of $F^n$ such that the transformations are the upper triangular matrices.
If $A$ is invertible, then it preserves dimension, so $A(V_i)=V_i$ for all $i$.  The inverse map, if it exists, would map $V_i$ right back into $V_i$, so the inverse is also an upper triangular matrix.
A: Thanks for all the help, I will now attempt to solve this using AnotherJohnDoe's hint. Any pointers or feedback to make this proof more elegant are appreciated.
Let $BA=I,A\in G$ then I choose to look at entries $I_{i1},i=2,3,..$ then $I_{i1}=0=\sum_{k=1}^nb_{ik}a_{k1}$ since $a_{k1}=0,k>1$ the expression reduces to $I_{i1}=0=b_{i1}a_{11}$. Determinant of an upper triangle matrix is the product of the diagonal entries, in our case none of the diagonal entries can be zero. Thus $b_{i1}=0$. 
Moving on to $I_{i2},i=3,4...$ then $I_{i2}=0=\sum_{k=1}^nb_{ik}a_{k2}$ since  $a_{k2}=0,k>2$ the expression reduces to $I_{i2}=0=b_{i1}a_{12}+b_{i2}a_{22}$ since $b_{i1}=0,a_{22}\neq0$ Only solution is $b_{i2}=0$
Now using induction to complete the proof. (Which I might give it a go at a later date)
