Consider a vector space $V$ (with dimension $n+1$ and elements $v$) on which a (commutative and associative) "product" $\odot$ taking $V\odot V\rightarrow V$ is defined, and an $1$ element $v\_0$ exists: $v\odot v\_0=v$ for all $v\in V$. $\odot$ is now completely defined by choosing a basis $v\_m$ $(0\le m\le n)$ ($v\_0$ is predefined - hands off - you can't change it) and giving the "structure constants" $A^{ij}\_k$ $(0\le i,j,k\le n)$ which are free parameters (except those with $i\*j=0$ who are predefined by $v\_0$ being the $1$ element): $v\_i\odot v\_j=\sum\_k A^{ij}\_k\*v\_k$.  

After you defined the $A^{ij}\_k$, you are free to chose a convenient other basis.
First question: I have $n(n+1)$ free parameters to do so. Just from *counting*, I can
use them to have $v\_k\odot v\_k=v\_0$ for all $k$ and use up exactly all "gauge" freedom this way. My linear algebra is rusty - can I really? Or are there $A^{ij}\_k$ sets where this fails?  

Second question. Define an "outer product" $\otimes$ (with is distributive over
vector sums, and follows $(v\_i\otimes v\_j)\odot v\_k:=v\_i\otimes (v\_j\odot v\_k)$ and $v\_i\odot (v\_j\otimes v\_k):=(v\_i\odot v\_j)\otimes v\_k$) and a quantity $S=\sum\_i\sum\_j a\_{ij}\*v\_i \otimes v\_j$. $S$ should be an "eigenvector": $S\odot v\_k=v\_k\odot S$ for all $k$. Express the allowed set of $a\_{ij}$ in terms of the $A^{ij}\_k$. (This is trivial for small n, where I do it with diagrams and by hand - in fact this is knot/graph theory in disguise as always when I ask :-) But a closed formula would be nice.)



Vector "product" diagonalization
Assuming you want $(A+B) ⊙ V = A⊙V + B⊙V$ (if not things get a lot more complicated , you might as well ignore the vector structure having no way to link the addition and product) ,
the product , being bi-linear , can be described as a tensor (essentially n square matrices $M\_p$, n\*n each , such that if $V\_1⊙V\_2 =C $ , then $< V\_1 | M\_k | V2 > = C\_k$ , the k'th component of $C$ . ) . I like to imagine them as a cube . 


A⊙V = V⊙A means that each of these matrices will be symmetric . 
The second constraint is $v\odot v\_0=v$ . Linear algebra ensures we can chose a convenient form for $v\_0$ , so let $v\_0= (1,0,0 ,0...) $ . We have $< v\_0 | M\_k | B > =B\_k $ , the k'th component of $B$ . This constrains the first column of $M\_k$ to be (0,0..,0,1,0..0) where the k'th component is 1. By symmetry , the first row is also constrained . So the only freedom we have is choosing the parameters of $n \hspace{5 mm} (n-1)\*(n-1)$ symmetric matrices , in total $n\*n\*(n-1)$ parameters for a fixed $v\_0$ . Since $v\_0$ can be chosen arbitrarily from a space of n dimentions , we have $(n + n\*n\*(n+1))$ degrees of freedom for the product operator . Hope that helps .


