Embedding from a rational Hamilton quaternion algebra into $M(2,\mathbb{R})$

Question

We knew that there are some embeddings as $\mathbb{Q}$-algebras from $H=\{a+bi+cj+dk|a,b,c,d\in \mathbb{Q}, i^2=j^2=-1,k=ij=-ji\}$ into $M_2(\mathbb{C})$, for instance: $a+bi+cj+dk \mapsto \begin{bmatrix} a+bi&c+id\\-c+id&a-ib \end{bmatrix}$

My question: Is there an embedding into $M(2,\mathbb{R})$ ?

Answer 1

On dimension grounds, such an embedding would be a bijection, and so an isomorphism. But $H$ and $M(2,\Bbb R)$ aren't isomorphic: the latter has zero-divisors.

There is an embedding into $M(4,\Bbb R)$.

ADDED IN EDIT

I see the question has been edited to make it about the quaternion algebra over $\Bbb Q$ not over $\Bbb R$.

Actually this makes no difference. The homomorphisms from your new $H$ to a given $\Bbb R$-algebra $A$ correspond to the $\Bbb R$-algebra homomorphisms from $H\otimes_{\Bbb Q}\Bbb R$ to $A$. Of course $H\otimes_{\Bbb Q}\Bbb R$ is the classical Hamiltonian quaternions $\Bbb H$.

Answer 2

Let me try to have a short proof of non-existence of an embedding of $H$ (even over $\mathbb Z$) into $M_2(\mathbb R)$. If such an enbedding exists, then its restriction to quaternion group $Q_8$ is a faithful representation of the group. The real faithful reprsentation of $Q_8$ is well-known of having minimal degree 4, not 2. You may easily find this argument on internet.