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.

$GL_n(F)\cong Aut(F^n)\,\, n\geq 3.$ 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 $f\circ g\not=g\circ f.$ 
