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Both matrix multiplication and quaternion multiplication are non-commutative; hence the use of terms like "premultiplication" and "postmultiplication". After encountering the concept of "quaternion matrices", I am a bit puzzled as to how one may multiply two of these things, since there are at least four ways to do this.

Some searching has netted this paper, but not having any access to it, I have no way towards enlightenment except to ask this question here.

If there are indeed these four ways to multiply quaternion matrices, how does one figure out which one to use in a situation, and what shorthand might be used to talk about a particular version of a multiplication?

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One should, I think, talk about (left or right) H-module morphisms of free (left or right) H-modules, where H is the quaternions. There is an obvious notion of composition here. –  Qiaochu Yuan Aug 9 '10 at 1:55
    
Apologies, my grasp of the theory of modules is not up to snuff; care to expound to a non-specialist? –  J. M. Aug 9 '10 at 2:01
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I'm not confident that the above agrees with the standard definition. The standard definition, as far as I can tell, is to take the product in the "obvious" order, e.g. in the terms of the product AB the component of A always appears before the component of B. –  Qiaochu Yuan Aug 9 '10 at 2:14
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I believe he is referring to matrices whose entries are quaternions. (Using the standard representation of quaternions as matrices, one can turn these into complex matrices.) –  Qiaochu Yuan Aug 9 '10 at 2:37
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My motivation is a bit more theoretical than my usual fare: I am figuring out how to consistently extend the standard methods of numerical linear algebra (LU, QR, SVD, etc.) to quaternion matrices. Since all of these have matrix multiplication as a necessary building block (let me worry about how to consistently divide quaternions later :P ), I want to figure out how pre- and post-multiplication by matrices should work out if the matrices have quaternion elements. I have seen some literature on quaternion matrix methods, but wish for a more organic starting point. –  J. M. Aug 9 '10 at 6:03

3 Answers 3

I guess I should expand my comment into an answer. Given two matrices $a_{ij}$ and $b_{ij}$ with entries in any (associative) ring $R$, the natural definition of the product has entries

$\displaystyle c_{ij} = \sum_k a_{ik} b_{kj}.$

This multiplication is associative, and it also agrees with the multiplication one obtains from any finite-dimensional matrix representation of $R$ by replacing each entry by the corresponding matrix.

I do not see any particular reason to consider a different notion of multiplication. Changing the order of some of the multiplications seems nonsensical to me, and multiplying in the opposite order gives you essentially the same multiplication.

This definition does not agree with the definition in my first comment; multiplication by one of the above matrices does not define an $R$-module homomorphism when $R$ is noncommutative.

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Sound reasoning, but could you maybe expand on "multiplying in the opposite order gives you essentially the same multiplication" please? –  J. M. Aug 12 '10 at 4:18
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Dear Qiaochu, If you want homomorphisms of left $R$-modules, than you should act your matrices on the right. So probably if you take the opposite ring $(M_n(H))^{op} = M_n(H^{op})$ (the isomorphism being given by transpose), you will get the ring in your first comment. Also, since $H^{op}$ is isomorphic to $H$, I think that the ring in your first comment and the matrix ring you describe above, in your answer, are actually isomorphic, although not by the identity map! –  Matt E Aug 12 '10 at 4:54
    
Matt, please pardon me for asking, but could you explain to this non-expert how the isomorphism (as I understand it, you're saying that the same set of "rules" apply to the system in Qiaochu's first comment and his answer here) is shown/applied? –  J. M. Aug 12 '10 at 5:08
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Dear Qiaochu Yuan: You could perhaps add explicitly to your answer the fact that, with the matrix multiplication you give, the $R$-linear maps from $R^n$ to $R^m$, viewed as right $R$-modules, correspond functorially to $m$ by $n$ matrices with entries in $R$. If you replace "right" by "left", you get the other multiplication. It suffices to understand the case $m=n=1$. Indeed, denoting by ${}_RR$ (resp. $R_R$) the ring $R$ viewed as a right (resp. left) module over itself, we get $End_R(R_R)=R$, $End_R({}_RR)=R^{op}$, the opposite ring. [This is essentially Matt E's comment.] –  Pierre-Yves Gaillard Aug 12 '10 at 5:20
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In reply to J. Mangaldan: mapping a quaternion to its conjugate is an antiinvolution of $H$, i.e. $\bar{z w} = \bar{w}\bar{z}$. So the map which sends an $n\times n$ quaternionic matrix to its conjugate transpose is an antiinvolution of the matrix ring: if we write it $A \mapsto A^*$, then $(AB)^* = B^* A^*$. In other words, it gives an isomorphism between matrices acting on the left, as in Qiaochu's answer above, and matrices acting on the right, as in his first comment. –  Matt E Aug 12 '10 at 5:30

This gives quite an intuitive idea about what is going on:
http://plus.maths.org/content/curious-quaternions

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I am already aware of the things discussed in both articles; unfortunately there is nothing in there on how one might multiply a matrix of quaternions. –  J. M. Aug 9 '10 at 13:33

...unfortunately I am only allowed to post one link per answer - so here is the sequel http://plus.maths.org/content/ubiquitous-octonions

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You could use `...` to un-link it. –  KennyTM Aug 9 '10 at 13:39

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