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If you check the quaternion product derivation at wikipedia:

http://en.wikipedia.org/wiki/Quaternion#Hamilton_product

You can see that it is derived from a multiplication table between the quaternions 1,i,j,k. All books I have on the topic do the same. But the derivation assumes that quaternion multiplication distributes over quaternion addition (or am I wrong?). Is this a "valid" assumption, that does not need to be proven?

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The multiplication is defined, in fact, as the unique product which distributes along sums, behaves nicely with scalars and takes the correct values at the basic elements.

That this is actually such a multiplication is a simple result in linear algebra, similar to thaht which says that a linear map need only be given on a basis of its domain.

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No, that is not what I wrote, really. What I am saying is that the multiplication of the quaternions is defined as the unique product which distributes with sums, plays nicely with scalars and which extends the multiplcation table one finds in the Wikipedia page. There exists other multiplications which interact as badly as you may want with addition! –  Mariano Suárez-Alvarez Sep 27 '12 at 7:00
    
Well, as I wrote in my answer, one has to check that such a multiplication exists... –  Mariano Suárez-Alvarez Sep 27 '12 at 7:18
1  
It is a very simple matter: if $V$ is a vector space and $\{v_1,\dots,v_n\}$ is a basis of $V$, given any array $(w_{i,j})_{1\leq i,j\leq n}$ of elements in $V$, there is exactly one function $m:V\times V\to V$ which is bilinear and such that $m(v_i,v_j)=w_{i,j}$ for all choices of $i,j\in\{1,\dots,n\}$. This should appear, in one form on another, is pretty much any linear algebra book (near where inner products are treated, or bilinear forms, I guess) –  Mariano Suárez-Alvarez Sep 27 '12 at 7:53
    
Can you provide a link or reference to a full proof? –  user1095108 Sep 28 '12 at 2:44

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