Suppose $L$ is a linear operator on a finite dimensional vector space $V$ over a field of characteristic $2$. If $L\circ T=T\circ L$ for all isomorphisms $T$ on $V$, does this imply $L=\lambda I$ for some $\lambda$?
I am interested because I was able to prove it is true when the ground field is not of characteristic $2$.
I chose $\{x_1,\dots,x_n\}$ to be a basis of $V$. I defined the isomorphism $S_{ij}$ as the isomorphism swapping $x_i$ and $x_j$, and fixing all other basis vectors. I express $L(x_i)=a_{i1}x_1+\cdots+a_{in}x_n$ for each $i$. Evaluating $$ L(S_{ij}(x_i))=S_{ij}(L(x_i)) $$ and equating coefficients shows $a_{ii}=a_{jj}$ for all $i,j$, and $a_{ij}=a_{ji}$ for all $i,j$. I then defined for $i\neq j$, $N_{ij}$ as the isomorphism which sends $x_i\mapsto x_j$, and $x_j\mapsto -x_i$, and fixes all other basis elements, which is an isomorphism as it sends a basis to a basis. Evaluating $$ L(N_{ij}(x_i))=N_{ij}(L(x_i)) $$ and equating coefficients shows $$ -a_{ij}=a_{ji}=a_{ij} $$ so $a_{ij}=0$ when $i\neq j$. It then follows that $L=aI$ where $a=a_{11}=\cdots=a_{nn}$. The last part hinges on the fact that $2x=0\implies x=0$. Does the result still hold in the case where the characteristic is $2$?
