Can anyone explain how why the matrix representation of the quaternions using real matrices is constructed as such?
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asked Jan 26, 2013 at 21:06
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3$\begingroup$ Maybe helpful: Think of the Quaternions as a four dimensional vector space over $\mathbb{R}$ with basis ${1,i,j,k}$. Then think of the natural action of the Quatenions on this vector space. The matrices you get for the actions of $1,i,j,k$ are just the ones you found in Wikipedia. $\endgroup$– user3533Commented Jan 26, 2013 at 21:30
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1$\begingroup$ Could you explain what a natural action is? I'm unfamiliar with the term. $\endgroup$– chubbycantorsetCommented Jan 26, 2013 at 21:35
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2$\begingroup$ The action of multiplication. Let's take the case of multiplication by $i$ on the left ($x\mapsto ix$). This is a linear action on this vector space, so it can be represented by a matrix relative to the basis ${1,i,j,k}$. Find this matrix. $\endgroup$– user3533Commented Jan 26, 2013 at 21:39
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1$\begingroup$ @user3533: How do you convert a linear action into a matrix? $\endgroup$– chubbycantorsetCommented Jan 26, 2013 at 21:47
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2$\begingroup$ The usual way you represent a linear operator as a matrix. In the first column put the coefficients of $i \cdot 1 = i$, in the second column, those of $i \cdot i = -1$, in the third those of $i \cdot j = k$ and in the forth those of $i \cdot k = -j$ $\endgroup$– user3533Commented Jan 26, 2013 at 21:56
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3 Answers
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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 $i^2 = -1$.
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$\begingroup$ The question asks about the real matrix representation; not the complex matrix representation. And by "why", I mean to say that surely that 4x4 matrix wasn't constructed accidentally. What is the logic behind its construction? Why are the minus signs and letters $a,b,c,d$ placed where they are? $\endgroup$ Commented Jan 26, 2013 at 21:21
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$\begingroup$ @chubbycantorset: I updated my answer. Does it help now? $\endgroup$– ThomasCommented Jan 26, 2013 at 21:26
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$\begingroup$ Not exactly. I've checked those things before, but my question asks how the map was constructed as such. To make an analogy, perhaps the formula to add the first $n$ digits was created by "experimental observation," so to say. Maybe someone just looked at the sequence of sums, and found a pattern to deduce that it should be $n(n+1)/2$. I'm looking for a similar explanation here, since I can't see how you can "experimentally observe" that this has to be the matrix for the quaternions. $\endgroup$ Commented Jan 26, 2013 at 21:33
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$\begingroup$ @chubbycantorset: So you are asking for how people historically came up with the representation? $\endgroup$– ThomasCommented Jan 26, 2013 at 21:36
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$\begingroup$ How they may have come up with it. Not necessarily how they did it exactly. I want to know if observing some property of the Quarternions can lead one to deduce that this should be the representation matrix. Although the placement of the minus signs on $b$ does help somewhat understand what's going on. $\endgroup$ Commented Jan 26, 2013 at 21:41
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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
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$\begingroup$ @chubbycantorset: does this answer to your question? $\endgroup$– G CabCommented Aug 22, 2017 at 17:45
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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.
