Question about $4\times4$ matrix representation of a quaternion I have a problem to solve about showing the real quaternion group $\mathbb{H}$ is isomorphic to $M_4(\mathbb{R})$
When trying to define my map I was having trouble coming up with an appropriate map that would work. Then I found these $4\times4$ matrix representation. I can see that it is skew-symmetric. 

Can someone give me an explanation why this is the $4\times4$ matrix you get from a quaternion? 
 A: One way to understand the accuracy of this representation is to investigate the four matrices in terms of the basis elements of H, $\lbrace i,j,k \rbrace$, and the identity element of the field (in this case, the reals), 1. 
Let's go through this process: 
The matrix associated with $a$ is clearly the identity. Thus, it has the properties of 1. 
The matrix associated with $b$, when squared, yields -1.The same goes for the matrix associated with $c$ (a simple calculation confirms both). So we see that these matrices follow some key properties of quaternion bases, namely: 
$$ 
i^2=a \\
 j^2=b \\
 $$
For $a,b$ in the field. 
We can also see that, if we multiply the matrix associated with $b$ by the matrix associated with $c$, we get the matrix associated with $d$ (confirm this for yourself). So, 
$$
ij=k$$. 
Reverse the order in the multiplication and it's easy to see that the final defining factor of quaternion bases holds, i.e.: 
$$
ji=-k$$
So we've shown that the matrices associated with $a,b,c,d$ represent $1,i,j,k$ respectively, and thus that this represents a quaternion algebra. 
