Find a linear functional on $M_n(\mathbb{C})$ which preserves matrix multiplication I know that a linear functional on $M_n(\mathbb{C})$ is $f(A)={\rm tr}(XA)$.
It is obvious that I have to find the matrix X such that $f$ preserves matrix multiplication. I thought I could replace $X$ with a Jordan normal form, since similar matrices have the same trace.
However, it does not work, it is difficult to find the Jordan normal form.
 A: There is only one multiplicative functional $f: M_n(\mathbb{C}) \to \mathbb{C}$ when $n > 1$, which is $f=0$.
Indeed, if $f: M_n(\mathbb{C})\to \mathbb{C}$ is a multiplicative functional, then $\ker(f)$ is a two-sided ideal in $M_n(\mathbb{C})$. But $M_n(\mathbb{C})$ is a simple algebra, so either $\ker(f) = M_n(\mathbb{C})$ which implies $f=0$ or $\ker(f)=0$ which implies $f$ is injective, which is impossible by dimension reasons.
A: Another proof which doesn't need simpleness of $M_n(\mathbb{C})$:
Let $E_i$ be the identity matrix with all columns set to zero except for the $i$th. Then $f(E_i) f(E_j) = 0$ whenever $i \not = j$, which means there is at most one $k$ with $f(E_k)$ nonzero. But also $f(Q X Q^{-1}) = f(X)$ for all $X$ and invertible $Q$, and all of the $E_i$s are similar, so  $f(E_k) = 0$ too (if $n > 1$). Finally then for any $X$ we have $f(X) = f(X I) = f(X) \sum_i f(E_i) = 0$, as desired.
A: If $U$ is a strictly upper-triangular matrix (strictly meaning it has zeros on the diagonal) then it is nilpotent: $U^n = 0$. Therefore $f(U)^n = f(U^n) = f(0) = 0$ and hence $f(U) = 0$. So $f$ is zero on strictly upper triangular matrices, and by similar logic strictly lower triangular matrices: so $f$ is some linear function of the diagonal only.
Now suppose that $f \neq 0$. Since $f(A) \neq 0$ for some $A$, and then $f(I) = f(IA) / f(A) = 1$, and we can see that for any invertible matrix $B$ we have $1 = f(I) = f(B) f(B^{-1})$, and so $f$ is nonzero on the set of invertible matrices. But provided that $n \geq 2$, there exist invertible matrices with zeros along the diagonal (take any permutation with no fixed points for example, like a cycle), which should be mapped to zero by the first observation. Hence we have a contradiction, and $f = 0$.
I think a key observation here is that if $f$ is nonzero anywhere then it is nonzero on the set of invertible matrices: you should be able to get from there to any number of contradictions.
