Automorphism of $\mathcal{M}_n(\mathbb{C})^r$

Question

I'm stucked with this question, I have no clue. We denote $\mathcal{M}_n(\mathbb{C})^r:=\mathcal{M}_n(\mathbb{C})\times\ldots\times\mathcal{M}_n(\mathbb{C})$ $r$ times, I have to show that if $f$ is an automorphism of $\mathcal{M}_n(\mathbb{C})^r$, then there exists a permutation $\sigma$ of $\{1,\ldots,r\}$ such that $f(e_i)=e_{\sigma(i)}$ for all $i\in\{1,\ldots,r\}$ where $e_i=(0,\ldots,0,\underbrace{I_n}_{\text{index }i},0,\ldots,0)$. The only idea that came to my mind is that the matrices of $f(e_i)$ are projections, thus their trace is equal to their rank, I believe one can show that all except one are of rank $0$ with this reasoning... Thank you for your help.

Answer 1

The set $S = \{e_1,\dots,e_r\}$ can be characterized as the set of elements $M$ of $\mathcal M_n(\Bbb C)^r$ that satisfy the following:

• $M$ has rank $n$,
• $M$ is in the center of $\mathcal M_n(\Bbb C)$ (i.e. $MX = XM$ for all $X \in \mathcal M_n(\Bbb C)^r$),
• $M$ is a projection (i.e. $M^2 = M$).

Because $f$ is an automorphism, it must preserve the above properties. Thus, $f(S) \subset S$. Because $f$ is invertible, $f|_S : S \to S$ must be a an invertible map, which means that there exists a permutation $\sigma$ for which $f(e_i) = e_{\sigma_i}$, as was desired.

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Proof that $f$ preserves rank:

Note that for $M \in \mathcal M_n(\Bbb C)$, the set $$ S_M = \{MX : X \in \mathcal M_n(\Bbb C)\} $$ is a linear subspace of $\mathcal M_{n}(\Bbb C)^r$ that satisfies $\dim(S_M) = \operatorname{rank}(M)\cdot n$. Note that $f(S_M) = S_{f(M)}$, which means that $\dim S_M = \dim S_{f(M)}$, which means that $\operatorname{rank}(M) = \operatorname{rank}(f(M))$.