<s>[INST] Can $\mathbb{R}^3$ with Hadamard product be represented as matrices? 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?
 [/INST] 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.
</s>