What is the Unique Linear Functional in $M_n(\Bbb F)$ Satisfying Two Conditions?

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Here is this week's POTW:

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Let $\Bbb F$ be a field. Prove that there is a unique linear functional $T : M_n(\Bbb F) \to \Bbb F$ such that

1. $T(I_n) = n$
2. $T(XY) = T(YX)$ for all $X, Y\in M_n(\Bbb F)$

What is the name for $T$?
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No one answered this week's problem. You can read my solution below.

Spoiler

Let $T$ be a linear functional on $M_n(\Bbb F)$ satisfying properties $1$ and $2$. For each $I,j\in \{1,2,\ldots, n\}$, let $E_{ij}$ be the $n\times n$ matrix whose $(I,j)$-entry is $1$ and every other entry is zero. Since $E_{ab}E_{cd} = \delta_{bc}E_{ad}$, property 2 and linearity give $\delta_{bc}T(E_{ad}) = \delta_{da} T(E_{cb})$. It follows that $T(E_{ad}) = 0$ whenever $a \neq d$, and $T(E_{aa}) = T(E_{bb})$. Since $T(I_n) = n$, then $n T(E_{11}) = T(E_{11}) + T(E_{22}) + \cdots + T(E_{nn}) = T(I_n) = n$, so that $T(E_{11}) = 1$.

An arbitrary $X\in M_n(\Bbb F)$ admits an expression $X = \sum_{i,j} x_{ij} E_{ij}$ where $x_{ij}\in \Bbb R$. Linearity of $T$ and the latter observations yield $$T(X) = \sum_{i,j} x_{ij}T(E_{ij}) = \sum_i x_{ii} T(E_{ii}) = T(E_{11})\sum_i x_{ii} = T(E_{11}) \operatorname{trace}(X) = \operatorname{trace}(X)$$

Thus $T$ is the trace map, which is uniquely determined.