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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 $2$?

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marked as duplicate by Marc van Leeuwen, Dennis Gulko, Nick Peterson, ncmathsadist, Dan Rust Oct 30 '13 at 13:47

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

You're looking for the center of the group $GL(n,\mathbb{F}_{2})$, which is indeed scalar multiplies of the identity:… – Jackson Jun 10 '13 at 1:20
@Kally What is the definition of $\text{GL}(n,\Bbb{F}_2)$? – user38268 Jun 10 '13 at 2:03
up vote 2 down vote accepted

As mentioned by Jackson Walters in his comment to your question, the answer is yes. His comment links to a proof that is tailor-made for your problem. However, since the set of all linear isomorphisms spans the set of all linear endomorphisms, $L$ must commute with every linear map $T$ on $V$. So, you don't actually need a special proof; any proof for the statement that "every linear map on $V$ that commutes with all linear maps on $V$ is a scalar multiple of the identity" can be reused here.

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