2 Blast. Forgot the product must be associative too.

Consider a vector space $V$ (with dimension $n+1$ and elements $v$) on which a (commutative) 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.)

1

# Vector "product" diagonalization

Consider a vector space $V$ (with dimension $n+1$ and elements $v$) on which a (commutative) "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.)