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?

It can be shown by direct computation. First, we need the following lemma.


Lemma.  Let $g\in GL(n,F)$. Let $h:=g^{-1}$. If there are entries of $g$ and $h$ such that 
  $$g_{ij}\neq 0,h_{kj}\neq 0,i>j,k\geq j,$$
  then $g$ is not in $N(B)$, the normalizer of $B$.

(Proof)Let $b=I+E_{jk}$, where $E_{jk}$ is the matrix whose entries are all zero except for the $(j,k)$-entry. Since $k\geq j$, $b$ is an element of $B$. Notice that $(gbg^{-1})_{ij}=(g)_{ij}(h)_{kj}\neq 0$. So $gbg^{-1}$ is not upper triangular(, for $i>j$). Hence $gbg^{-1}\not \in B$. Therefore $g\not \in N(B)$. $\square$

Now we prove that $N(B)=B$ by induction on $n$. 
Let us denote by $B_n$ the set of invertible upper triangular matrices of size n and by $N(B_n)$ its normalizer in $GL(n,F)$. Assuming that $N(B_{n-1})=B_{n-1}$, we prove $N(B_{n})=B_{n}$.
Suppose that $g_0\in N(B)$. Let $h_0:=g_0^{-1}$. Since $g_0$ and $h_0$ are invertible, neither of them has a column of zero. If we have $(g_0)_{i,1}\neq 0, (h_0)_{k,1}\neq 0$ for some $i,k$, then the above lemma implies that $i=1$. On the other hand, Since $g_0$ and $h_0$ are invertible, neither of them has a column of zero. Hence $g_0$ has the form$$g_0=\left[\begin{array}{cccc}
d_1 & \ast & \cdots & \ast\\
0 & \ast & \cdots & \ast\\
\vdots &  & \vdots\\
0 & \ast & \cdots & \ast
\end{array}\right].$$
We then multiply a product $c_1$ of elementary matrices that corresponds to adding multiples of the first column to other columns, which is in $B$, to obtain an element $g_1:=g_0c_1$ of the form
 $$g_{1}=\left[\begin{array}{cccc}
d_1 & 0 & \cdots & 0\\
0 & \ast & \cdots & \ast\\
\vdots &  & \vdots\\
0 & \ast & \cdots & \ast
\end{array}\right].$$
The $(n-1)\times (n-1)$ matrix $g_1'$ that is obtained by deleting the first row and the first column of $g_1$ is in $N(B_{n-1})$; thus by induction hypothesis, $g_1'\in B_{n-1}$. It follows that $g_1\in B_{n}$. Hence $N(B_{n})\subset B_{n}$.The converse inclusion holds trivially.
