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I've used the following idea as a black box for some time now, but it occurred to me I don't fully understand why it's true.

Suppose $A=M_n(R)$ is the algebra of square matrices over some division ring $R$. Then for any $\phi\in\operatorname{Aut}(A)$, we can actually write $\phi$ as the composition of an automorphism induced by an automorphism $\psi$ of $R$ and the conjugation by some unit of $A$. More explicitly, for $\psi\in\operatorname{Aut}(R)$, this induces an automorphism $\tilde{\psi}$ of $A$ by applying $\psi$ to each of the entries in the matrix, for example, $$ M=\begin{pmatrix} a_{11} & a_{12}\\ a_{21} & a_{22} \end{pmatrix} \mapsto \tilde{\psi}(M)\begin{pmatrix} \psi(a_{11}) & \psi(a_{12})\\ \psi(a_{21}) & \psi(a_{22}) \end{pmatrix} $$ and then we can conjugate by an invertible matrix in $A$, say $N$, to get $N\tilde{\psi}(M)N^{-1}$. I don't think the order of applying $\tilde{\psi}$ or conjugating matters, since if I conjugate first, then I could apply a different $\tilde{\psi}$. So the composition would be something like $\phi=\varphi_N\circ\tilde{\psi}$ where $\varphi_N$ is the conjugation by $N$ map.

My question is, why can any automorphism $\phi$ of $A$ actually be decomposed in this way?

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    I'm still trying to figure out what your second paragraph means... can you formalize it a bit more please :)2012-04-29
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    @Nastassja: It think it should say "conjugation by" instead of "conjugation of"? By the automorphism induced by some $\psi\in\operatorname{Aut}(R)$ I presume you mean the componentwise application of $\psi$?2012-04-29
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    @rschwieb Sorry, I admit I'm having a somewhat hard time expressing what I mean :(. I will try to fix it.2012-04-30
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    In your statement about your "gut feeling" you never use $\rho$. What did you intend to say?2012-04-30
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    @Nastassja: if $\rho$ is an automorphism of $A$ then it isn't an element of $A$, and I'm not sure what you mean by conjugation here. Perhaps you mean the following: giving $N$ an $A$-module structure means specifying a ring map $A \to \text{End}(N)$, and any _endomorphism_ $\rho : A \to A$ (no invertibility necessary) defines a new ring map $A \to A \to \text{End}(N)$ by composition.2012-04-30
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    @QiaochuYuan Thanks, I think you've phrased it much better and more accurately than I did.2012-04-30
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    @Nastassja I'm curious: do you happen to remember what you were studying when you started using this "black box"? Or some problems you applied it in? I just wonder if it's something else in disguise.2012-04-30
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    @rschwieb This was from the later portions of a second course in linear algebra I took last fall. I've tried to formalize and better explain the second paragraph. It was too vague this first time around.2012-05-02

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