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Let $V$ be a vector space with dimension greater than 1 over the field $F$ and $Sim = \{(f\in \operatorname{Lin}(V))\mapsto gf(g^{-1}) : g\in \operatorname{GL}(V)\}$, ie $Sim$ is the set of all similarity transforms on $\operatorname{Lin}(V)$. Let $Aut$ be the set of all automorphisms of the $F$-algebra $\operatorname{Lin}(V)$.

What is $Aut\setminus Sim$? In particular, is it empty?

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Let $M$ denote the $F$-algebra $\hom(V, V)$. An $F$-algebra homomorphism $M \to \hom(V, V)$ is tantamount to an $M$-module structure on $V$ in the category of $F$-vector spaces. But the category of $M$-modules is equivalent to $Vect_F$ (there is a Morita equivalence between the categories given by the functor $- \otimes_F V: Vect_F \to Mod_M$). That is to say, every $M$-module is isomorphic to a direct sum of copies of $V$, and in particular, any module structure on $V$ is isomorphic to the standard $M$-module structure on $V$.

So, if $\mu': M \otimes_F V \to V$ is any module structure, and $\mu: M \otimes_F V \to V$ denotes the standard structure, there is an isomorphism $g: V \to V$ such that $g \circ \mu' = \mu \circ (M \otimes_F g)$. Putting all this together, any $F$-algebra homomorphism $M \to \hom(V, V)$ is obtained by conjugating the standard homomorphism (the identity) by an isomorphism $g: V \to V$.

Edit: If this proof seems too categorical, there is a proof in slightly alternative language given on page 401 of Bilinear algebra: An Introduction to the algebraic theory of quadratic forms by Szymiczek (Google books here.

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  • $\begingroup$ Am I missing a reason for you writing $M \to \hom(V, V)$ instead of $M \to M$? $\endgroup$
    – user5810
    Commented Oct 16, 2010 at 2:12
  • $\begingroup$ I wrote it that way in remembrance of the canonical bijection between linear maps $M \to \hom_F(V, V)$ and linear maps $M \otimes_F V \to V$, and (by specializing) between algebra maps and module structures. I was trying to segue from a question about algebras to a module-theoretic question. $\endgroup$ Commented Oct 16, 2010 at 2:19
  • $\begingroup$ Okay, I'm lost in the category theory, but how could we possibly get the zero homomorphism by conjugating the identity with an isomorphism? $\endgroup$
    – user5810
    Commented Oct 16, 2010 at 3:07
  • $\begingroup$ That's not an algebra homomorphism -- it doesn't preserve the identity. $\endgroup$ Commented Oct 16, 2010 at 3:16
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    $\begingroup$ I am confused: are you assuming that $V$ is finite-dimensional? $\endgroup$
    – t3suji
    Commented Oct 20, 2010 at 14:34
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If $V$ is finite-dimensional, this fact is a simple consequence of Skolem-Noether Theorem

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  • $\begingroup$ which, of course, links to an error page. $\endgroup$
    – user5810
    Commented Oct 18, 2010 at 16:25
  • $\begingroup$ Of course, you know, this means war! $\endgroup$
    – Bugs Bunny
    Commented Oct 20, 2010 at 11:02
  • $\begingroup$ Bugs was won :-) $\endgroup$ Commented Oct 20, 2010 at 12:23

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