Let $n\geq 1$. Let $V$ be the vector space of all $n\times n$ matrcies over the field $F$. Let $B$ be a fixed element of $V$, and let $T_B$ be the linear operator on $V$ defined by $T_B(A)=AB-BA$. Show that det $T_B=0$.
Proof:For $B\ne O$. Det$T_B$ is the determinant of any matrix represantation of T.The columns of the matrix of $T_B$ relative to the ordered base B consisting of the matrix'units'$E_{ij}$ i.e matrices with exactly one zero entry in the ijth position,are$[T_B(E_{ij})]_B$, $1\le i,j \le n$. For any i,j $T_B(E_{i,j})=E_{i,j}B-BE_{i,j}$ is but since B is an element of V $$B=\sum_{s,k}^n B_{sk}E_{sk}$$ and from the properties of unit matrices $E_{ij}E_{sk}=E_{ik} for j=s$ and O for $j\ne s$. Then $T_B(E_{i,j})=E_{i,j}B-BE_{i,j}=\sum_{s,k}^n B_{sk}E_{ij}E_{sk}-\sum_{sk}^n B_{sk}E_{sk}E_{ij}=\sum_k^n B_{jk}E_{ik}-\sum_s^n B_{si}E_{sj}=\sum_k B_{jk}E_{ik}-B_{ki}E_{kj}$.For i=j the $E_{ii}$ coordinate of the $T_B(E_{i,i})$vector is zero and for $i\ne j$ the $E_{ji}$ coordinate is zero becuse if not then $E_{ji}=B_xE_{ik}-B_yE_{kj}$ for some $B_x,B_y$wich can not be since $E_{ji},E_{ik},E_{kj}$are independent. From the above we conclude that each $E_{ji}$ coordinate position of the transformation matrix $ T_B(E_{i,j})$column is equal to zero and also if we row-reduce the matrix since these zero positions are distinct we will have at least one zero row ,meaning that det$T_B=0$. For B=O $T_B$ is the zero transformation wich is represented by the zero matrix wich have zero determinant.
I wonder if my proof is right if it is not any hints will be helpful, if it is any alternative proofs will be much appreciated.