" tommy quaternion "

[tommy1729]

Hi, why you don't specify that \( C*B=A \) in your list of identities? You forgot to add it or it is possible to derive it from the others?

Also, you see it as a non-associative algebra over the reals or the complex?

Trivial question (sorry but I'm new to linear algebra): being an algebra means that \( \mathbb T \) is a \( k \)-vector space of dimension d=4 on which is defined a bilinear application \( *:{\mathbb T} \times {\mathbb T}\to {\mathbb T} \) that is not associative... but it must be bilinear because we want it distributive:

1. An element is of the form \( T=x1+yA+zB+wC \). Let  \( T,R,S,U\in{\mathbb T} \), the application being distributive means that
   
   \( (T+R)*(S+U)=T*S+R*S+T*U+R*U \)
   
   i.e. every translation by a Tommy quaternion is a endomorphism of  the addition group;
   
   
   
2. and we also want that for every scalar \( \lambda \)
   
   \( \lambda R*S= R*\lambda S=\lambda (R*S) \)
   
   i.e we want it to contain a copy of the base field, i.e multiplication by scalar  is multiplication by \( \lambda1 \) where \( 1\in \mathbb T \). So the previous means that the Tommy quaternions of the form \( T=x1+0A+0B+0C \) commute and "associate" with everything.
   

But this means that the multiplication is bilinear: every element should it should have a representation.. e.g. multiplication by \( T \) has a 4x4 matrix \( M_T \).
Obviously if it is not associative we "should not" have \( M_{T*S}=M_{T}M_{S} \) in general...

Probably my understanding of non-associative alg. is so poor that I'm missing something obvious.

yeah ... I forgot to add C*B sorry.
And thanks for pointing it out.

It is corrected now.

Well matrix representation is based on representing the multiplication.

But matrix multiplication is associative , however these numbers are not associative by multiplication.

Functional composition is also associative btw.

So easy representations might not exist.

In my opinion this makes these commutative algebras interesting.

regards

tommy1729