# If $n > 1$, there are no non-zero $*$-homomorphisms $M_n(\Bbb{C}) \to \Bbb{C}$

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?

• Yes, or just use the algebra homomorphism condition on $E_{ii}=E_{ij}E_{ji}$, $j\neq i$. – user10354138 Jul 29 at 8:13
• @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? – user745578 Jul 29 at 8:21
• 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. – MaoWao Jul 29 at 8:39
• @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. – MaoWao Jul 29 at 8:43
• @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. – 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 $$\phi(A)=\phi(P)\left(\phi(L)+\phi(U)\right)\phi(Q)=0$$.