Tracial states on $M_n$ not being Trace Background: Let $A$ be a unital *-algebra. A state $\varphi$ is a positive (i. e. $\forall x \in A: \varphi(x^*x) \geq 0$) unital linear functional, w. l. o. g. normalized. It is said to be tracial if $\forall x, y \in A: \varphi(xy) = \varphi(yx)$.
Is there for some $n \in \mathbb N$ a tracial state on the matrix algebra $M_n$, which is neither trivial nor the trace operator?
And how to find such a counterexample without wasting much time?
 A: There is a single tracial state in $M_n(\mathbb C)$. Suppose $\varphi:M_n(\mathbb C)\to\mathbb C$ is a tracial state. Looking at the matrix units, we have
$$
\varphi(E_{kj})=\varphi(E_{ k1}E_{1j})=\varphi(E_{1j}E_{k1})=\delta_{k,j}\,\varphi(E_{11}). 
$$
So, for any $A\in M_n(\mathbb C)$ we have
$$
\varphi(A)=\varphi\Big(\sum_{k,j}a_{kj}\,E_{kj}\Big)=\sum_{k,j}a_{kj}\,\delta_{k,j}\varphi(E_{11})=\varphi(E_{11})\,\sum_ka_{kk}=\varphi(E_{11})\,\operatorname{Tr}(A). 
$$
As $\varphi$ is a state,
$$
1=\varphi(I)=\varphi\Big(\sum_kE_{kk}\Big)=\sum_k\varphi(E_{kk})=n\varphi(E_{11}). 
$$
So $\varphi(E_{11})=\frac1n$ and so
$$
\varphi(A)=\frac1n\,\operatorname{Tr}(A). 
$$
A: No. The matrix algebra $\mathbb{M}_n$ has a unique tracial state namely $\text{tr}_n$ given by $\text{tr}_n[a_{i,j}]=\frac{1}{n}\sum_{i=1}^na_{ii}$.
We will solve the following exercise from Rordam's blue book to prove this:
Let $A$ be a unital C*-algebra with a trace $\tau:A\to\mathbb{C}$.

*

*Verify that the maps $\tau_n:M_n(A)\to\mathbb{C}$ given by $\tau_n[a_{i,j}]_{i,j}=\sum_{j=1}^n\tau(a_{jj})$ are traces on $M_n(A)$ for each $n\geq1$.

*Assume that $\rho:M_n(A)\to\mathbb{C}$ is a trace and that $\rho(\text{diag}(a,0,\dots,0))=\tau(a)$ for all $a\in A$. Show that $\rho=\tau_n$.

*Using 2, show that there exists only on trace $\text{Tr}$ on $M_n(\mathbb{C})$ such that $\text{Tr}(e)=1$, where $e=\text{diag}(1,0,\dots,0)$.

Proof: Indeed, we have
$$\tau_n([a_{i,j}]\cdot[b_{i,j}])=\tau_n\big(\big[\sum_{k=1}^na_{ik}b_{kj}\big]\big)=\sum_{j=1}^n\sum_{k=1}^n\tau(a_{jk}b_{kj})=\sum_{k=1}^n\sum_{j=1}^n\tau(b_{kj}a_{jk})=\sum_{k=1}^n\tau\big(\sum_{j=1}^nb_{kj}a_{jk}\big)=$$
$$=\tau_n\big(\big[\sum_{j=1}^nb_{ij}a_{jk}\big]_{i,k}\big)=\tau_n([b_{i,j}]\cdot[a_{i,j}])$$
for all matrices $[a_{i,j}]$ and $[b_{i,j}]$. Boundedness is immediate: $$|\tau_n[a_{i,j}|\leq\sum_{i=1}^n|\tau(a_{ii})|\leq n\cdot\|\tau\|\cdot\|[a_{i,j}]\|_{M_n(A)}$$
for any matrix $[a_{i,j}]\in M_n(A)$.


*We will use tensor language, so the matrix $[a_{i,j}]$ is written as $\sum_{i,j=1}^na_{i,j}\otimes e_{i,j}$ where $e_{i,j}$ are the matrix units of $M_n(\mathbb{C})$. Recall that the matrix units behave as $e_{i,j}\cdot e_{k,l}=e_{i,l}$ iff $j=k$ and the product is $0$ otherwise. So, using the linearity of the trace we have
$$\rho[a_{i,j}]=\sum_{i,j=1}^n\rho(a_{i,j}\otimes e_{i,j}).$$
We distinguish two cases. If $i=j$, then we have that $$\rho(a_{i,i}\otimes e_{i,i})=\rho\big((1_A\cdot a_{i,i}\cdot 1_A)\otimes(e_{i,1}\cdot e_{1,1}\cdot e_{1,i})\big)=\rho((1_A\otimes e_{i,1})\cdot(a_{i,i}\otimes e_{1,1})\cdot(1_A\otimes e_{1,i}))=$$
$$\rho((1_A\otimes e_{1,i})\cdot (1_A\otimes e_{i,1})\cdot (a_{i,i}\otimes e_{1,1}))=\rho(1_A\otimes e_{1,1})\cdot(a_{i,i}\otimes e_{1,1}))=\rho(a_{i,i}\otimes e_{1,1})=$$ $$=\rho(\text{diag}(a_{i,i},0,\dots,0))=\tau(a_{i,i}).$$
If $i\neq j$, then
$$\rho(a_{i,j}\otimes e_{i,j})=\rho\big((1_A\cdot a_{i,j}\cdot 1_A)\otimes(e_{i,1}\cdot e_{1,1}\cdot e_{1,j})\big)=\rho((1_A\otimes e_{i,1})\cdot(a_{i,j}\otimes e_{1,1})\cdot(1_A\otimes e_{1,j}))=$$
$$=\rho((1_A\otimes e_{1,j})\cdot(1_A\otimes e_{i,1})\cdot(a_{i,j}\otimes e_{1,1}))=0$$
and thus $\rho[a_{i,j}]=\sum_{i=1}^n\tau(a_{i,i})=\tau_n[a_{i,j}]$ and we are done.


*An obvious application of 2 for $A=\mathbb{C}$.
Now if $\tau$ is a tracial state on $\mathbb{M}_n$, then $\tau(1)=1$. Set $e_i$ for the diagonal matrix with 0 everywhere except the $i$ slot on the main diagonal where the entry is $1$. Note that we can always find a unitary $u_{i,j}$ such that $u_{i,j}e_iu_{i,j}^*=e_j$, since these are all rank one projections. But then, by the tracial property we have that $\tau(e_j)=\tau(u_{i,j}e_iu_{i,j}^*)=\tau(u_{i,j}^*u_{i,j}e_i)=\tau(e_i)$. Therefore, since $1_{\mathbb{M}_n}=e_1+\dots+e_n$, we have that $1=\tau(e_1)+\dots+\tau(e_n)=n\tau(e_1)$. Now $n\cdot\tau$ is a trace on $\mathbb{M}_n$ and satisfies $n\tau(e_1)=1$, so by (3) of the exercise we have that $\tau=\text{tr}_n$.
