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Let $\mathfrak{M}$ be the algebra of matrices $n\times n$ over $\mathbb{C}$, and let $\mathfrak{O}$ be the algebra of linear operators over $\mathbb{C}^n$.

It is well-known that every basis $\mathcal{B}$ in $\mathbb{C}^n$ defines an isomorphism (which I will also denote $\mathcal{B}$) between $\mathfrak{M}$ and $\mathfrak{O}$. I have two questions about it:

First: is every isomorphism between $\mathfrak{M}$ and $\mathfrak{O}$ defined by a special basis?

Second: Can we induce an addition on the set of the set of all basises?

The most obvious way is that if $\mathcal{A}\in\mathfrak{O}$ and $\mathcal{B}_1(\mathcal{A}) = A_1\in\mathfrak{M}$, while $\mathcal{B}_2(\mathcal{A}) = A_2\in\mathfrak{M}$, we can define $(\mathcal{B}_1+\mathcal{B}_2)(\mathcal{A})=A_1+A_2$. But is this correct? And if it is, what is the geometric representation of the sum of two basises?

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  • $\begingroup$ If a basis defines an isomorphism you must mean the algebra of matrices and the algebra of linear operators. (Saying "an algebra of" means it's a subalgebra of the algebra of all matrices or operators...) $\endgroup$ – David C. Ullrich Jun 4 '18 at 16:54
  • $\begingroup$ @DavidC.Ullrich thank you for correction, I edited the post. I mean the algebra of all operators and all matrices $\endgroup$ – Michael Freimann Jun 4 '18 at 16:56
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First: This is true because every automorphism of $\mathfrak{M}$ is inner: Let $\mathcal{B}$ be any basis of $\mathbb{C}^n$ and let $\varphi \colon \mathfrak{D} \to \mathfrak{M}$ be the induced isomorphism of $\mathbb{C}$-algebras. Then for every other isomorphism of $\mathbb{C}$-algebras $\psi \colon \mathfrak{D} \to \mathfrak{M}$ it follows that there exists an invertible matrix $S \in \operatorname{GL}_n(\mathbb{C})$ such that $\psi \varphi^{-1} \colon \mathfrak{M} \to \mathfrak{M}$ is given by conjugation with $S$, i.e. we have that $$ \psi(\varphi^{-1}(A)) = S A S^{-1} $$ for every matrix $A \in \mathfrak{M}$. In other words, we have that $$ \psi(f) = S \varphi(f) S^{-1} $$ for all $f \in \varphi^{-1}(\mathfrak{M}) = \mathfrak{D}$. If $\mathcal{B}'$ is the basis of $\mathbb{C}^n$ such that $S$ is the basis change matrix from $\mathcal{B}$ to $\mathcal{B}'$ (i.e. the $j$-th column of $S$ represents the $j$-th basis vector of $\mathcal{B}$ with respect to the basis $\mathcal{B}'$) then it follow that $\psi$ is precisely the $\mathbb{C}$-algebra isomorphism $\mathfrak{D} \to \mathfrak{M}$ which is induced by $\mathcal{B}'$.


Second: Note that the pointwise sum of two $\mathbb{C}$-algebra homomorphisms $\varphi_1, \varphi_2 \colon \mathfrak{D} \to \mathfrak{M}$ is itself never an algebra homomorphism (unless $n = 0$) because $$ (\varphi_1 + \varphi_2)(\operatorname{id}) = \varphi_1(\operatorname{id}) + \varphi_2(\operatorname{id}) = I + I = 2I \neq I. $$ Because of this, the proposed addition on basis isn’t well-defined.

We can, however, give the set $\operatorname{Iso}_{\text{$\mathbb{C}$-$\mathbf{Alg}$}}(\mathfrak{D}, \mathfrak{M})$ of $\mathbb{C}$-algebra automorphisms $\mathfrak{D} \to \mathfrak{M}$ the structure of a heap via $$ [f,g,h] = f g^{-1} h \,. $$ This can also be expressed as follows: By fixing a $\mathbb{C}$-algebra isomorphim $\varphi \colon \mathfrak{D} \to \mathfrak{M}$, i.e. an element $\varphi \in \operatorname{Iso}_{\text{$\mathbb{C}$-$\mathbf{Alg}$}}(\mathfrak{D}, \mathfrak{M})$, we get a bijection $$ \operatorname{Iso}_{\text{$\mathbb{C}$-$\mathbf{Alg}$}}(\mathfrak{D}, \mathfrak{M}) \to \operatorname{Aut}_{\text{$\mathbb{C}$-$\mathbf{Alg}$}}(\mathfrak{M}, \mathfrak{M}), \quad \psi \mapsto \varphi \psi^{-1}. $$ By using this bijection we can pull back the group structure of $\operatorname{Aut}_{\text{$\mathbb{C}$-$\mathbf{Alg}$}}(\mathfrak{M}, \mathfrak{M})$ to a group structure on $\operatorname{Iso}_{\text{$\mathbb{C}$-$\mathbf{Alg}$}}(\mathfrak{D}, \mathfrak{M})$, which has $\varphi$ as the identity. But I don’t know how useful this construction is.

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