[proofplan]
We verify each defining property directly from finite-dimensional matrix algebra. The algebraic $*$-structure comes from the usual matrix product, the identity matrix, and conjugate transpose. The normalized trace is a unital linear functional, positivity and faithfulness follow from the entrywise formula for $\operatorname{tr}_N(A^*A)$, and traciality follows from a finite double-sum computation.
[/proofplan]
[step:Verify the unital $*$-algebra structure on $M_N(\mathbb{C})$]
Let $M_N(\mathbb{C})$ denote the complex [vector space](/page/Vector%20Space) of all functions $A: \{1,\dots,N\} \times \{1,\dots,N\} \to \mathbb{C}$, written as matrices $A = (A_{ij})_{i,j=1}^{N}$. Matrix multiplication is the bilinear operation defined by
\begin{align*}
(AB)_{ij} = \sum_{k=1}^{N} A_{ik}B_{kj}
\end{align*}
for $A,B \in M_N(\mathbb{C})$ and $1 \le i,j \le N$. The identity element is the matrix $I_N \in M_N(\mathbb{C})$ with entries $(I_N)_{ij} = 1$ if $i=j$ and $(I_N)_{ij}=0$ if $i \ne j$.
Define the involution map
\begin{align*}
*: M_N(\mathbb{C}) \to M_N(\mathbb{C}), \quad A \mapsto A^*
\end{align*}
by
\begin{align*}
(A^*)_{ij} = \overline{A_{ji}}.
\end{align*}
For all $A,B \in M_N(\mathbb{C})$ and $\lambda \in \mathbb{C}$, direct entrywise computation gives $(A+B)^* = A^*+B^*$, $(\lambda A)^* = \overline{\lambda}A^*$, $(A^*)^*=A$, and $(AB)^* = B^*A^*$. Thus $M_N(\mathbb{C})$, with this product, unit, and involution, is a unital complex $*$-algebra.
[/step]
[step:Check that the normalized trace is a unital linear functional]
Define the unnormalized matrix trace
\begin{align*}
\operatorname{Tr}: M_N(\mathbb{C}) \to \mathbb{C}, \quad A \mapsto \sum_{i=1}^{N} A_{ii}.
\end{align*}
By definition,
\begin{align*}
\operatorname{tr}_N(A) = \frac{1}{N}\operatorname{Tr}(A)
\end{align*}
for every $A \in M_N(\mathbb{C})$.
Let $A,B \in M_N(\mathbb{C})$ and $\lambda,\mu \in \mathbb{C}$. Since matrix addition and scalar multiplication are entrywise,
\begin{align*}
\operatorname{tr}_N(\lambda A+\mu B)=\frac{1}{N}\sum_{i=1}^{N}(\lambda A_{ii}+\mu B_{ii})=\lambda \operatorname{tr}_N(A)+\mu \operatorname{tr}_N(B).
\end{align*}
Thus $\operatorname{tr}_N$ is complex-linear. Also,
\begin{align*}
\operatorname{tr}_N(I_N)=\frac{1}{N}\sum_{i=1}^{N}1=1.
\end{align*}
Therefore $\operatorname{tr}_N$ is a unital linear functional.
[/step]
[step:Compute $\operatorname{tr}_N(A^*A)$ to prove positivity and faithfulness]
Let $A \in M_N(\mathbb{C})$. For each $1 \le i \le N$, the $i$-th diagonal entry of $A^*A$ is
\begin{align*}
(A^*A)_{ii}=\sum_{k=1}^{N}(A^*)_{ik}A_{ki}=\sum_{k=1}^{N}\overline{A_{ki}}A_{ki}=\sum_{k=1}^{N}|A_{ki}|^2.
\end{align*}
Therefore
\begin{align*}
\operatorname{tr}_N(A^*A)=\frac{1}{N}\sum_{i=1}^{N}\sum_{k=1}^{N}|A_{ki}|^2.
\end{align*}
The right-hand side is a finite sum of non-negative [real numbers](/page/Real%20Numbers), so $\operatorname{tr}_N(A^*A) \ge 0$. Hence $\operatorname{tr}_N$ is positive.
If $\operatorname{tr}_N(A^*A)=0$, then
\begin{align*}
\sum_{i=1}^{N}\sum_{k=1}^{N}|A_{ki}|^2=0.
\end{align*}
Since each term $|A_{ki}|^2$ is non-negative, every term must be zero. Hence $A_{ki}=0$ for all $1 \le k,i \le N$, so $A=0$. Thus $\operatorname{tr}_N$ is faithful.
[guided]
The positivity condition for a $*$-probability space asks us to test elements of the form $A^*A$. This is the matrix analogue of a square norm: its trace should measure the size of $A$.
Fix $A \in M_N(\mathbb{C})$. The involution is conjugate transpose, so for indices $1 \le i,k \le N$ we have
\begin{align*}
(A^*)_{ik}=\overline{A_{ki}}.
\end{align*}
Using the definition of matrix multiplication, the $i$-th diagonal entry of $A^*A$ is
\begin{align*}
(A^*A)_{ii}=\sum_{k=1}^{N}(A^*)_{ik}A_{ki}.
\end{align*}
Substituting the formula for the entries of $A^*$ gives
\begin{align*}
(A^*A)_{ii}=\sum_{k=1}^{N}\overline{A_{ki}}A_{ki}=\sum_{k=1}^{N}|A_{ki}|^2.
\end{align*}
Now apply the normalized trace, which is $N^{-1}$ times the sum of diagonal entries:
\begin{align*}
\operatorname{tr}_N(A^*A)=\frac{1}{N}\sum_{i=1}^{N}(A^*A)_{ii}=\frac{1}{N}\sum_{i=1}^{N}\sum_{k=1}^{N}|A_{ki}|^2.
\end{align*}
Every summand $|A_{ki}|^2$ is a non-negative real number. Since $N \ge 1$, we have
\begin{align*}
\frac{1}{N} > 0.
\end{align*}
Therefore $\operatorname{tr}_N(A^*A)\ge 0$. This proves positivity.
The same formula proves faithfulness. Suppose $\operatorname{tr}_N(A^*A)=0$. Then
\begin{align*}
\frac{1}{N}\sum_{i=1}^{N}\sum_{k=1}^{N}|A_{ki}|^2=0.
\end{align*}
Multiplying by $N$ gives
\begin{align*}
\sum_{i=1}^{N}\sum_{k=1}^{N}|A_{ki}|^2=0.
\end{align*}
A finite sum of non-negative real numbers can equal zero only when each summand equals zero. Therefore $|A_{ki}|^2=0$ for every pair of indices $1 \le k,i \le N$, hence $A_{ki}=0$ for every pair of indices. Thus all entries of $A$ vanish, so $A=0$. This is exactly faithfulness of the positive functional $\operatorname{tr}_N$.
[/guided]
[/step]
[step:Use cyclicity of finite sums to prove traciality]
Let $A,B \in M_N(\mathbb{C})$. By the definition of matrix multiplication and the unnormalized trace,
\begin{align*}
\operatorname{Tr}(AB)=\sum_{i=1}^{N}(AB)_{ii}=\sum_{i=1}^{N}\sum_{j=1}^{N}A_{ij}B_{ji}.
\end{align*}
Since this is a finite double sum and complex multiplication is commutative for scalar entries,
\begin{align*}
\sum_{i=1}^{N}\sum_{j=1}^{N}A_{ij}B_{ji}=\sum_{j=1}^{N}\sum_{i=1}^{N}B_{ji}A_{ij}.
\end{align*}
Renaming the dummy indices in the last expression gives
\begin{align*}
\sum_{j=1}^{N}\sum_{i=1}^{N}B_{ji}A_{ij}=\sum_{j=1}^{N}(BA)_{jj}=\operatorname{Tr}(BA).
\end{align*}
Therefore $\operatorname{Tr}(AB)=\operatorname{Tr}(BA)$, and multiplying by $1/N$ gives
\begin{align*}
\operatorname{tr}_N(AB)=\operatorname{tr}_N(BA).
\end{align*}
Thus $\operatorname{tr}_N$ is tracial.
[/step]
[step:Assemble the defining properties of a faithful tracial $*$-probability space]
A faithful tracial $*$-probability space is a unital complex $*$-algebra $\mathcal{A}$ equipped with a complex-linear functional $\tau: \mathcal{A} \to \mathbb{C}$ such that $\tau(1_{\mathcal{A}})=1$, $\tau(a^*a)\ge 0$ for every $a \in \mathcal{A}$, $\tau(a^*a)=0$ implies $a=0$, and $\tau(ab)=\tau(ba)$ for every $a,b \in \mathcal{A}$.
We have shown that $M_N(\mathbb{C})$ is a unital complex $*$-algebra, that $\operatorname{tr}_N: M_N(\mathbb{C}) \to \mathbb{C}$ is linear and satisfies $\operatorname{tr}_N(I_N)=1$, that $\operatorname{tr}_N(A^*A)\ge 0$ for every $A \in M_N(\mathbb{C})$, that $\operatorname{tr}_N(A^*A)=0$ implies $A=0$, and that $\operatorname{tr}_N(AB)=\operatorname{tr}_N(BA)$ for every $A,B \in M_N(\mathbb{C})$. Applying the definition with $\mathcal{A}=M_N(\mathbb{C})$ and $\tau=\operatorname{tr}_N$, we conclude that $(M_N(\mathbb{C}),\operatorname{tr}_N)$ is a faithful tracial $*$-probability space.
[/step]
