Show the automorphism $\phi={}^t(A^{-1})$ is not inner for $|F|>3$

Asked Mar 01 '17 at 08:50

Active Mar 01 '17 at 09:42

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I have a question that asks to show the automorphism $\phi= {}^t(A^{-1})$ is not inner for $|F|>3$ (The trace of the inverse of $A$).

For $\phi$ is be inner, it would have to be the case that there exists some $B\in GL_n(F)$ such that

\begin{equation}\phi_B(A)=BAB^{-1}= {}^t(A^{-1})\quad\quad \forall A\in GL_n(F)\end{equation}

The key to this problem lies in the fact that our field $F$ has more than 3 elements. I tried to use the above equation to show a contradiction, but I ended up with pages upon pages of calculations without anything becoming obvious to me about where the contradiction was, so then I thought perhaps I'm approaching this the wrong way. Can anyone offer a hint?

edited Mar 01 '17 at 09:42

Andreas Caranti

68,873

asked Mar 01 '17 at 08:50

ai.jennetta

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1

Transpose, not trace. – darij grinberg Mar 01 '17 at 09:01
Consider how the automorphism acts on the matrix $a I_n $ for different nonzero $a $. Could a linear map act like that? – darij grinberg Mar 01 '17 at 09:05
See also http://math.stackexchange.com/questions/1040202/prove-the-automorphism-given-by-phi-leftg-right-leftg-1-rightt-is-n?rq=1 (this is only about the case $F=\Bbb R$, but it could help). – Arnaud D. Mar 01 '17 at 09:06

Show the automorphism $\phi={}^t(A^{-1})$ is not inner for $|F|>3$

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