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.

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
