TITLE: Involutions of full matrix ring $M_n(R)$
QUESTION [6 upvotes]: Hellow, I want to describe all involutions of full matrix ring over field and all involutions of matrix polynomial ring.
Is it true or false that every involution of the full matrix ring $T = M_n(R)$ over field $R$ has the follwing form
$$
A \to C^{-1}A^TC,
$$
for all $A\in M_n(R)$ and some fixed matrix $C$?
What can we say about involutions of the matrix-polynomial ring $T[x]$?

REPLY [3 votes]: Every involution of $T$ is of the form $\phi(A)=C^{-1}\sigma(A^T)C$ for an invertible matrix $C$, and a field automorphism $\sigma\colon R\to R$ satisfying $\sigma^2=\iota$.
Note that, for an involution $\phi$, then $\theta\colon T\to T$ defined by $\theta(A)=\phi(A^T)$ is a ring automorphism. So $\theta$ preserves the center of $T$. As the center of $T$ is the diagonal elements $\lambda I$ for $\lambda\in R$, we have $\theta(\lambda I)=\sigma(\lambda)I$ for a field automorphism $\sigma$. As $\phi\phi(\lambda I)=\lambda I$, we have $\sigma^2(\lambda)=\lambda$. Then, $\chi(A)=\phi(\sigma(A^T))$ is an automorphism of $T$, which preserves $\lambda I$ for each $\lambda\in R$, so is an automorphism of $R$-algebras. So, $\chi$ is an inner automorphism, meaning it is of the form $\chi(A)=C^{-1}AC$ for an invertible $C$. Therefore, $\phi(A)=C^{-1}\sigma(A^T)C$.
In fact, $\phi\phi(A)=A$, so $C^{-1}\sigma(C)^TA\sigma(C)^{-T}C=A$, showing that $C^{-1}\sigma(C)^T$ is in the center of $T$, so is equal to $\lambda I$ for some $\lambda\in R$. So, $C^T=\lambda \sigma(C)$. Taking the transpose $C=C^{TT}=\lambda\sigma(C^T)=\lambda^2 C$. So, $\lambda=\pm1$ and either $C^T=\sigma(C)$ or $C^T=-\sigma(C)$.