$\DeclareMathOperator{\End}{End}$ $\DeclareMathOperator{\Aut}{Aut}$ $\DeclareMathOperator{\id}{id}$ Let $V$ be a vector space over a field $\mathbb{K}$. Let $\End(V)\overset{*}{\longrightarrow}\End(V)$ be a ring automorphism. I am trying to show that there is a $\theta\in\Aut(V)$ so that $$T^*=\theta T\theta^{-1}\qquad \forall T\in\End(V)$$ I have figured out that if such a $\theta$ exists then $\theta^*=\theta$ and $\theta^2=\lambda\id_V$ for some $\lambda\in\mathbb{K}$.
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How did you conclude that $\theta$? – Ben Grossmann May 10 '21 at 17:03
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Set $T=\theta$ to get $\theta^=\theta$. To get $\theta^2=\lambda id_V$ use that commutator $[T,\theta^\theta]=0 \quad T\in End(V)$ implies $\theta^*\theta=\lambda id_V$. – Rasmus May 10 '21 at 17:05
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I see, thanks for clarifying – Ben Grossmann May 10 '21 at 17:06
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@Troposphere $\DeclareMathOperator{\End}{End}$ $\End(V)\overset{}{\longrightarrow}\End(V)$ is another way to write $:\End(V)\to\End(V)$ – Ben Grossmann May 10 '21 at 17:08
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OK, I see. $ {} $ – Troposphere May 10 '21 at 17:08
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2Possible duplicate of Automorphisms of matrix rings. Does this answer your question? – Christoph May 10 '21 at 17:11
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Yes. Thanks. It is an answer for the finite dimensional case. – Rasmus May 10 '21 at 17:19
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$\DeclareMathOperator{\End}{End} \DeclareMathOperator{\Aut}{Aut} \DeclareMathOperator{\id}{id}$
$\theta$ exists if $V$ is finite-dimensional.
Fix a non-zero $u\in V$ and $\phi\in V'$ where $V'$ denotes the dual space. Then $u\otimes \phi\in\End(V)$ and is non-zero. Since $*$ is injective $\exists z\in V$ so that $(u\otimes\phi)^*z\neq0$. Define $\theta\in\End(V)$ by $\theta x\equiv (x\otimes\phi)^*z$. Let $T\in\End(V)$. $$\begin{align*} \theta T x &=\left((Tx)\otimes\phi\right)^*z \\ &=\left(T(x\otimes\phi)\right)^*z \\ &=T^*(x\otimes\phi)^*z \\ &=T^*\theta x \end{align*}$$ So $\theta T=T^*\theta$. Let $w\in V$. Since $\theta u\neq0$, by axiom of choice there is a $S\in \End(V)$ so that $S\theta u=w$. Since $*$ is surjective, $S=U^*$ for a $U\in \End(V)$. $U^*\theta u=w$. Since $\theta U=U^*\theta$, $\theta Uu=w$, so $\theta$ is surjective. If $V$ is finite-dimensional, this means $\theta$ is invertible and $T^*=\theta T\theta^{-1}$.
Rasmus
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