It is known that $\mathbb{R}^2$ with Hadamard product, represented as pairs or numbers $(a,b)$ with element-wise operations is isomorphic to real matrices of the form $\left( \begin{array}{cc} \frac{a+b}{2} & \frac{a-b}{2} \\ \frac{a-b}{2} & \frac{a+b}{2} \\ \end{array} \right)$, and also to the split-complex numbers.
But I wonder, whether it is possible to represent $\mathbb{R}^3$ with Hadamard product and element-wise operations as real matrices in a similar way?
The following works for complex matrices. Let $\omega = \exp(i 2\pi/3)$ be a primitive cube root of $1$. Define $$ H(a,b,c) := \begin{bmatrix} \frac{a+b+c}{3} & \frac{a+\omega b+\omega^2 c}{3} & \frac{a+\omega^2 b+\omega c}{3} \\ \frac{a+\omega^2 b+\omega c}{3} & \frac{a+b+c}{3} & \frac{a+\omega b+\omega^2 c}{3} \\ \frac{a+\omega b+\omega^2 c}{3} & \frac{a+\omega^2 b+\omega c}{3} & \frac{a+b+c}{3} \end{bmatrix} $$
Then $$ H(a,b,c)+ H(d,e,f) = H(a+d,b+e,c+f) \\ H(a,b,c) H(d,e,f) = H(ad,be,cf) $$
Probably we can adapt this to get real $6 \times 6$ matrices, replacing each complex number by a real $2\times 2$ matrix in the usual way.