If $n > 1$, there are no non-zero $*$-homomorphisms $M_n(\Bbb{C}) \to \Bbb{C}$. A $*$-homomorphism is an algebra morphism $\varphi: M_n(\Bbb{C}) \to \Bbb{C}$ with $\varphi(\overline{A}^T) = \overline{\varphi(A)}$

I tried to show that every $*$-morphism must be zero: if $\varphi$ is such a $*$-morphism, then $E_{ij}^2 = 0$ for $i \neq j$ (here $E_{ij}$ is the matrix that is $1$ on position $(ij)$, $0$ elsewhere). Thus $\varphi(E_{ij}) = 0$ for $i \neq j$. Hence, $$\varphi(A) = \sum_i a_{ii} \varphi(E_{ii})$$ so if I can show that $\varphi(E_{ii}) = 0$ I will be done. Unfortunately, I can't see why this should be true. I did not yet fully use the fact that it is a $*$-morphism. The only thing I can see is that $\varphi(A) \in \Bbb{R}$ for all $A$ since $\varphi(E_{ii})= \varphi(\overline{E_{ii}}^T) = \overline{\varphi(E_{ii}})$. If $\varphi$ is non-zero, then there is $A$ with $\varphi(A) \in \Bbb{R}\setminus \{0\}$. But then $\varphi(iA) = i\varphi(A) \notin \Bbb{R}$, a contradiction. Thus $\varphi=0$. Is this correct?

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    $\begingroup$ Yes, or just use the algebra homomorphism condition on $E_{ii}=E_{ij}E_{ji}$, $j\neq i$. $\endgroup$ – user10354138 Jul 29 at 8:13
  • $\begingroup$ @user10354138 So we actually have that every algebra morphism $M_n(\Bbb{C}) \to \Bbb{C}$ is zero? We don't need that it preserves the $*$-operations? $\endgroup$ – user745578 Jul 29 at 8:21
  • $\begingroup$ Why would you have $\phi(A)\in\mathbb{R}$? The diagonal entries are not necessarily real. The trace is a map that has all the properties you used, and is certainly not zero. $\endgroup$ – MaoWao Jul 29 at 8:39
  • $\begingroup$ @user1551 That is true, but I wrote "all the properties you used", and the OP did not use the multiplicativity of $\phi$ for anything but $\phi(E_{ij})=0$ for $i\neq j$, which is also true for the trace. $\endgroup$ – MaoWao Jul 29 at 8:43
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    $\begingroup$ @user1551 Once again, I know that the trace is not multiplicative. But the trace also has the property that $\phi(E_{ij})=0$ for $i\neq j$ and $\phi(E_{ii})\in \mathbb{R}$. These two facts are not sufficient to show that $\phi=0$, and that is what the OP tries to do. $\endgroup$ – MaoWao Jul 29 at 8:48

We can prove that when $n\ge2$ and $\mathbb F$ is a field, the only mapping $\phi:M_n(\mathbb F)\to\mathbb F$ that is additive and multiplicative is zero. (So, we don't need $\phi$ to be an algebra morphism, not to say a $\ast$-homomorphism over $\mathbb C$.)

For any $A\in M_n(\mathbb F)$, we have $A=P_1DQ$ for some nonsingular matrices $P_1,Q$ and some diagonal matrix $D$. Let $P_2$ be any permutation matrix with a zero diagonal. Then $P_2^TD$ has a zero diagonal too. It follows that $A=(P_1P_2)(P_2^TD)Q$ can be written as $A=P(L+U)Q$ where $P,Q$ are nonsingular, $L$ is strictly lower triangular and $U$ is strictly upper triangular.

Since $L^n=0$, we have $\phi(L)^n=\phi(L^n)=0$. Therefore $\phi(L)=0$ and similarly $\phi(U)=0$. Hence $\phi(A)=\phi(P)\left(\phi(L)+\phi(U)\right)\phi(Q)=0$.

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  • $\begingroup$ The above comment shows an easier way imo. Thanks though. $\endgroup$ – user745578 Jul 29 at 10:16

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