Proving that $GL_n(F)$ is non-abelian for $n \geq 2$ and for any field $F$ I'm trying to show that $GL_n(F)$ is non-abelian for any field $F$ and $n \geq 2$. I'm doing so by constructing two $2 \times 2$ matrices that do not commute and "extending" them to $n \times n$ matrices with zeros in every other entry. We define $\displaystyle A = \left[ \begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array}\right]$ and $\displaystyle B = \left[ \begin{array}{cc} 1 & 0 \\ 1 & 1 \end{array}\right]$. A quick calculation reveals that $AB \neq BA$, so the products of the extended $n \times n$ matrices are also not equivalent.
My question is: does this suffice in proving the statement for any field $F$? The only field that I can think of where this might fail is in $\mathbb{Z}/2\mathbb{Z}$, but that turns out to not be the case.
 A: As explained in the comments, the problem is one of extending the two matrices in such a way that 1) the same non-commutativity can be easily verified and 2) the matrices are non-singular.
A standard recipe is to extend by ones along the diagonal and zeros elsewhere.
Why is this "standard"? The way I think about it is in terms of linear transformations. Remeber that matrix multiplication is defined to faithfully represent composition of linear transformations (matrices are with respect to a given basis).
If $n>2$ we can write the vector space $V=F^n$ as a direct sum $V=F^2\oplus F^{n-2}$. The two matrices, $A$ and $B$, can be thought of as two linear transformations, $T_A$ and $T_B$, from $F^2$ to itself. Now we can define two linear transformations from $V$ to itself as follows. Using the above direct sum decomposition we can write any vector
$v\in V$ as $v=(x,y)$ with $x\in F^2$ and $y\in F^{n-1}$. Define $E_A:V\to V$ by declaring $E_A(v)=(T_A(x),y)$ and similarly define $E_B:V\to V$ by $E_B(v)=(T_B(x),y)$.
Because $T_A,T_B$ and the identity mapping of $F^{n-2}$ are all invertible, these linear transformations are both non-singular. They do not commute, because they do not commute on vectors of the form $v=(x,0)$ by your calculation. Thus we are done.
We also see that the matrices of $E_A$ (resp. $E_B$) with respect to the usual basis have the block structure
$$
E_A=\left(\begin{array}{c|c}
A&0\\
\hline
0&I_{n-2}
\end{array}\right)\qquad \text{and}\qquad
E_B=\left(\begin{array}{c|c}
B&0\\
\hline
0&I_{n-2}
\end{array}\right)
$$
respectively.
A: $GL_n(F)\cong Aut(F^n)\,\, n\geq 3.$ where $F$ is an arbitrary field (not necessarily of characteristic zero). Let $f(x_1,x_2\cdots,x_n)=(x_2,x_1,\cdots,x_n)$ and $g(x_1,x_2,\cdots,x_n)=(x_3,x_2,x_1,\cdots,x_n)$ Clearly $f,g\in Aut(F^n)$. Notice that  $f\circ g(1,0,0,\cdots,0)=f(0,0,1,0\cdots,0)=(0,0,1,0,\cdots,0)$ but $g\circ f(1,0,0,\cdots,0)=g(0,1,0,\cdots,0)=(0,1,0,\cdots,0).$ so $f\circ g\not=g\circ f.$ 
A: You are only guaranteed the existence of element $0$ and $1$ in a field.
One possible approach is to define the following matrices:
$A = \left\{ \begin{array}{ll}
         a_{ij} = 1 & \mbox{if $i = j$}\\
         a_{ij} = 1 & \mbox{if $i = 1$ and $j=n$}\\
        a_{ij} = 0 & \mbox{otherwise}.\end{array} \right.$
$B = \left\{ \begin{array}{ll}
         a_{ij} = 1 & \mbox{if $i = -j$}\\
        a_{ij} = 0 & \mbox{otherwise}.\end{array} \right.$
Computing the product of both matrices:
$AB = \left\{ \begin{array}{ll}
         a_{ij} = 1 & \mbox{if $i = 1$ and $j = 1$}\\
         a_{ij} = 1 & \mbox{if $i = - j$}\\
        a_{ij} = 0 & \mbox{otherwise}.\end{array} \right.$
$BA = \left\{ \begin{array}{ll}
         a_{ij} = 1 & \mbox{if $i = n$ and $j = n$}\\
         a_{ij} = 1 & \mbox{if $i = - j$}\\
        a_{ij} = 0 & \mbox{otherwise}.\end{array} \right.$
So $AB \neq BA$
Edited: The original matrices were not invertible
