Zero-product Jordan algebra C^3

The vector space has a nonzero vector $x$, whereas $e\circ x=0$ for every $e$. Thus no element acts as a multiplicative unit. This is an object of $\mathbf{Jord}$ over the stated field, but is not an object of either $\mathbf{UJord}$ or $\mathbf{Jord}_1$. Positivity of the dimension matters: the zero-dimensional algebra is outside this family.

Every $\mathbb C$-linear map $T$ preserves the product, since

Conversely, a Jordan homomorphism over $\mathbb C$ is required to be $\mathbb C$-linear. Therefore, in $\mathbf{Jord}_{C}$,

These identifications use the standard coordinate basis. On endomorphisms, composition becomes ordinary matrix multiplication and pointwise addition gives the matrix algebra. On automorphisms, composition gives the general linear group $\operatorname{GL}_{3}(\mathbb C)$. Invertibility is sufficient because the inverse of an invertible linear map also preserves the zero product.

The resulting matrix algebra has its ordinary associative product, and is a different object from this zero-product Jordan algebra.

As a real Jordan algebra this object has dimension $6$, still with zero product. The same proof gives real-linear endomorphism algebra $M_{6}(\mathbb R)$ and automorphism group $\operatorname{GL}_{6}(\mathbb R)$, after choosing the real basis obtained from the complex coordinate basis and its multiples by $i$. These are larger collections of maps than the complex-linear ones. For example, coordinatewise complex conjugation is a real-linear automorphism but is not complex-linear.