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Let $D\neq 0$ be a linear transformation from $Mat_{n\times n}(K)$ to $Mat_{n\times n}(K)$ that preserves multiplication: $D(AB)=D(A)D(B)$. Show that there exists nonsingular $C$, such that $D(A)=CAC^{-1}$.

I think we may first show $D(E)=E$ from the fact that $D(A)D(E)=D(A)$ for all $A$, but I have no idea how to show this. Can we show there exist some $A$ such that $D(A)$ is nonsingular?

Any hints are welcomed.

Dreamworld2001
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  • False if $D \equiv 0$. Otherwise, what can you about $D(I)$? – geetha290krm Mar 14 '23 at 11:36
  • @geetha290krm I think it is idempotent and D(A)D(I)=D(A), but I cannot deduce further. Could you give me more hints? Thank you :) – Dreamworld2001 Mar 14 '23 at 13:19
  • Note that $\text{Mat}_{n \times n}(K)$ is a simple ring. I believe it follows that if $D$ is a non-zero, linear and multiplication-preserving map, then $\ker(D) = {0}$. As a result, it follows that $D(E)$ is invertible (I assume you're using $E$ to refer to the identity matrix) and thus $D(E) = E$. – Ben Grossmann Mar 14 '23 at 14:55
  • From there, we could use the fact that every matrix-ring automorphism is inner. – Ben Grossmann Mar 14 '23 at 14:57
  • If you are ok with characteristic $0$ (for now) you can do this with the trace. Recognize composing trace with your linear map $T$ (out of habit I use T instead of D) is a class function and $T(\mathbf e_k\mathbf e_k^T)$ is an idempotent matrix, so its trace is equal to the same $ r \in \mathbb N$ for all k. Then $T(I)$ is an idempotent matrix with trace $=n \cdot r \leq n \implies r$ is zero or one and $T(I)$ is $0$ or $I$ and the former implies $T=0$, so it must be the latter. – user8675309 Mar 15 '23 at 00:20
  • @BenGrossmann Thank you! Your argument is clear! – Dreamworld2001 Mar 15 '23 at 04:21
  • @user8675309 Why is $\mathrm{trace}\circ T$ a class function? I think to see this we need $T(A)^{-1}=T(A^{-1})$ first? – Dreamworld2001 Mar 15 '23 at 04:29
  • regarding class function: notice that $\text{trace}\big(T(SAS^{-1})\big)=\text{trace}\big(T(S)T(A)T(S^{-1})\big)=\text{trace}\big(T(S^{-1})T(S)T(A)\big)=\text{trace}\big(T(S^{-1}S)T(A)\big)=\text{trace}\big(T(I)T(A)\big)=\text{trace}\big(T(IA)\big)=\text{trace}\big(T(A)\big)$ – user8675309 Mar 15 '23 at 05:06
  • My original argument shows $\text{trace}\big(T(\mathbf e_k\mathbf e_k^T)\big)=1$ and now consider the other standard basis vectors for $V$: $\text{trace}\big(T(\mathbf e_k\mathbf e_j^T)\big)=0$ for $j\neq k$ since $\big(T(\mathbf e_k\mathbf e_j^T)\big)^2= \mathbf 0$ i.e. $T(\mathbf e_k\mathbf e_j^T)$ is a nilpotent matrix. Conclude $\text{trace}\big(A\big)= \text{trace}(T\big(A\big))$ for arbitrary $A\in M_n(\mathbb F)$. And by non-degeneracy of bilinear (Killing) form $b(A,B)=\text{trace}\big(AB\big)$ conclude $\dim \ker T =0$. – user8675309 Mar 15 '23 at 05:14
  • @user8675309 Thank you! I understand your argument, but it seems that the trace-method does not work when $\mathrm{char}(K)\mid n$. – Dreamworld2001 Mar 15 '23 at 06:26

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