# quaternion product distributivity

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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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