Delving into GL(n,F)

We should still show that any other matrix won't commute with all matrix of $GL_n(F)$ .

Without much details, for $n\geq 2$ :

Suppose $A$ always commutes. Take matrix $B_{i,j}=I_n+\Delta_{i,j}$ , where $\Delta_{i,j}=(\delta_{i,j})$ with $\delta_{i,j}=1_F$ if $i=j$ and $0$ otherwise. So $B_{i,j}=(b_{i,j})$ is the identity matrix where we added $1_F$ to $b_{i,j}$ . We have $\det(B)=\det(I_n)+\det(\Delta_{i,j})=1+0=1$ , so the matrix is in $GL_n(F)$ .

$AB_{i,j}=B_{i,j}A \Leftrightarrow A(I_n+\Delta_{i,j})=(I_n+\Delta{i,j})A \Leftrightarrow A\Delta_{i,j}=\Delta_{i,j}A$ . $A$ must commute with $\Delta_{i,j}$ (we can generalize this and show that $A$ must commute with any matrix, even outside of $GL_n{F}$ ).

$A\Delta_{i,j}$ is just a matrix with the $i^{\mathrm{th}}$ column of $A$ in the $j^{\mathrm{th}}$ column (and zeroes everywhere else). $\Delta_{i,j}A$ is just a matrix with the $j^{\mathrm{th}}$ line of $A$ in the $i^{\mathrm{th}}$ line.

If both are the same, that means that $a_{i,i}=a_{j,j}$ and that $a_{j,i}$ must be zero when $i\neq j$ . Then, $A$ is scalar.