Matrix representation of the Quaternions? Can anyone explain how why the matrix representation of the quaternions using real matrices is constructed as such?
 A: It is a simple consequence of the multiplication rules of the quaternions.   
So
$$
\eqalign{
  & \left( {a + bi + cj + dk} \right) \cdot 1 = \left( {\matrix{
   a & b & c & d  \cr 
   { - b} & a & { - d} & c  \cr 
   { - c} & d & a & { - b}  \cr 
   { - d} & { - c} & b & a  \cr 
 } } \right)^{\,T} \left( {\matrix{
   1  \cr 
   0  \cr 
   0  \cr 
   0  \cr 
 } } \right) = \left( {\matrix{
   1  \cr 
   0  \cr 
   0  \cr 
   0  \cr 
 } } \right)^{\,T} \left( {\matrix{
   a & b & c & d  \cr 
   { - b} & a & { - d} & c  \cr 
   { - c} & d & a & { - b}  \cr 
   { - d} & { - c} & b & a  \cr 
 } } \right) = \left( {\matrix{
   a  \cr 
   b  \cr 
   c  \cr 
   d  \cr 
 } } \right)  \cr 
  & \left( {a + bi + cj + dk} \right) \cdot i = \left( {\matrix{
   a & b & c & d  \cr 
   { - b} & a & { - d} & c  \cr 
   { - c} & d & a & { - b}  \cr 
   { - d} & { - c} & b & a  \cr 
 } } \right)^{\,T} \left( {\matrix{
   0  \cr 
   1  \cr 
   0  \cr 
   0  \cr 
 } } \right) = \left( {\matrix{
   { - b}  \cr 
   a  \cr 
   { - d}  \cr 
   c  \cr 
 } } \right) \cr} 
$$
and so on
A: I am not sure exactly what you mean by asking "... why the ...". "Why" questions can be hard to answer satisfactory in math.
The claim is that the Quaternions $\mathbb{H}$ are isomorphic (as $\mathbb{R}$-algebras) to the given set of matrices. The isomorphism looks like this:
$$
\phi: a + bi + cj + dk \longmapsto \begin{pmatrix}a & b & c & d  \\  -b & a & -d & c \\ -c &d &a& - b\\ -d& -c & b& a\end{pmatrix}.
$$
To "understand" why this is true, you "simply" check that this is an isomorphism. 
You check for example that $\phi$ is bijective, which is clear from the construction.
Then you check that $\phi$ is an algebra homomorphism, so you need for $x,y\in \mathbb{H}$ and $\lambda \in \mathbb{R}$:


*

*$\phi(xy) = \phi(x)\phi(y)$ for $x,y\in\mathbb{H}$

*$\phi(x+y) = \phi(x) + \phi(y)$

*$\phi(\lambda x) = \lambda\phi(x)$


The last two are not difficult to check. The first one requires a bit of work.
Even though this does not answer the minus signs are where there are in the matrix, I highly recommend that you try to prove that $\phi$ is a homomorphism. This exercise will make you more familiar with the Quaternions.
But note if you check property $3$ above you would need (as a special case)
$$
\phi((bi)(bi))  = \phi(ib)\phi(ib).
$$
That is you would need
$$
\begin{pmatrix} -b^2 & 0 & 0& 0  \\  0 & -b^2 & 0 & 0 \\ 0 &0 &-b^2& 0\\ 0& 0 & 0& -b^2\end{pmatrix} = \begin{pmatrix} 0 & b & 0& 0  \\  -b & 0 & 0 & 0 \\ 0 &0 &0& -b\\ 0& 0 & b& 0\end{pmatrix}\begin{pmatrix} 0 & b & 0& 0  \\  -b & 0 & 0 & 0 \\ 0 &0 &0& -b\\ 0& 0 & b& 0\end{pmatrix}.
$$
So here you can see that you "need" the minus on all the $b$'s. In this case it comes down to the fact that $i^2 = -1$.
A: If It is not a misunderstanding, the question of "why as such (possible others?)" is equivalent to the problem of finding out ALL 4x4 real matrix representations of the Quaternions. The fact is: all those matrix representations are conjugated to each other. In other words, this is the only 4x4 real matrix representation of the Quaternions up to equivalent.  
