Normalizer of upper triangular group in ${\rm GL}(n,F)$ The following question has already appeared on mathstack:

If $B$ is the subgroup of ${\rm GL}(n,F)$ consisting of upper triangular matrices then normalizer of $B$ in ${\rm GL}(n,F)$ is $B$ itself.

I know a proof of this using Bruhat decomposition of ${\rm GL}(n,F)$. 
Question: Can we prove above theorem without using Bruhat decomposition?

Why came to this question: Consider the general linear Lie algebra $L=\mathfrak{gl}(n,F)$; in it, let $T=\mathfrak{t}(n,F)$ be the upper triangular sub-algebra. Then normalizer of $T$ in $L$ is $T$ itslef, and this can be proved just by considering a very simple decomposition of ${\mathfrak gl}(n,F)$: write any element as sum of upper triangular matrix and lower triangular matrix whose diagonal is $0$. 
But then for problem above, is it necessary to use Bruhat decomposition?

$\DeclareMathOperator{\GL}{GL}$$\newcommand{\Span}[1]{\left\langle #1 \right\rangle}$$\newcommand{\Set}[1]{\left\{ #1 \right\}}$Let $e_0, e_1, \dots, e_{n-1}$ be a basis with respect to which $B$ is upper-triangular, and write
$$
V_i = \Span{ e_j : j \ge i}.
$$
Allow me to use row vectors, so that the group $G = \GL(n, F)$ acts on the right.
Then 
$$
B = \Set{b \in G : V_i b \subseteq V_i \text{ for each $i$}}.
$$
Moreover, 

$V_{n-i}$ is the unique subspace $W$ of dimension $i$ such that $W B \subseteq W$. 

This is proved by induction on $i$. For $i = 1$ we have that $V_{n-1}$ is the unique common eigenspace for the elements of $B$ (just consider the element of $B$ which is a single Jordan block of size $n \times n$ and eigenvalue $1$, say), then pass to $V / V_{n-1}$ and use induction.
Let $g \in N_{G}(B)$. Then for each $b \in B$ we have $g b g^{-1} \in B$, that is for all $i$
$$
V_{i} g b g^{-1} \subseteq V_i
$$
or
$$
(V_{i} g) b \subseteq V_i g.
$$
It follows from the above that $V_{i} g = V_{i}$, so that $g \in B$.
