Full-matrix Jordan algebra M_3(C)^+

The full-matrix Jordan algebra $M_{3}(\mathbb C)^+$ has the same vector space as $M_{3}(\mathbb C)$ , but the product is

X\circ Y=\tfrac12(XY+YX).

The unit is $I_{3}$ . This object includes all square matrices; self-adjointness is not required.

Why it is Jordan

The new product is commutative. Expanding both sides of $X^2\circ(X\circ Y)=X\circ(X^2\circ Y)$ in the original associative product gives the same four terms, which proves the Jordan identity. The matrix-entry coordinates still give dimension $3^2$ over $\mathbb C$ .

Category distinction

The associative matrix object and this Jordan object retain separate records because their multiplications differ. The symmetrization is a construction, not an assertion that the identity map is a multiplicative homomorphism between the two products. A Jordan map need preserve only the symmetrized product. At size one both products coincide.