Extract determinant and trace from structure constants of basis of matrix algebra Let $K$ be a field, $n$ a natural number. Suppose that $(e_i)_{i=1}^{n^2}$ is a $K$-basis of the matrix algebra $\mathbb{M}_n(K)$ (on which multiplication is defined in the usual manner). One does not know what these elements $e_i$ look like, but we do know the $n^6$ elements $(a_{i, j, k})_{i, j, k = 1}^{n^2}$ of $K$ such that
$$
\forall i, j=1, \ldots, n^2 : e_i \cdot e_j = \sum_{k=1}^{n^2} a_{i, j, k}e_k .
$$
One calls the $(a_{i, j, k})_{i, j, k=1}^{n^2}$ the structure constants of the basis. If $n > 1$ they do not uniquely determine the $e_i$: if $S \in \mathbb{M}_n(K)$ is invertible, then the basis of $\mathbb{M}_n(K)$ given by $(S^{-1}e_iS)_{i=1}^{n^2}$ has the same structure constants. By the Skolem-Noether Theorem, the converse also holds: if $(e_i')_{i=1}^{n^2}$ is another basis with the same structure constants, then we have for all $i$ that $e_i' = S^{-1}e_iS$ for some invertible $S \in \mathbb{M}_n(K)$.
It follows that norm and trace (even characteristic polynomial) of the $e_i$ are uniquely determined by these structure constants. But how to calculate them?
 A: Yes, it's possible.
Let $M^{i,j}\in \mathbb{M}_n(K)$, where $1\leq i,j\leq n$, be the matrix which has $1$ at the intersection of $i$-th row and $j$-th column. Then if $B\in \mathbb{M}_n(K)$ is a matrix with entries $b_{k,l}$, then $B\cdot M^{i,j} = \sum_{k=1}^n b_{k,i}M^{k,j}$. It means that if we consider the linear operator $\text{ad}_B: \mathbb{M}_n(K)\to \mathbb{M}_n(K)$ given by $\text{ad}_B (X) = B\cdot X$, then in the basis $(M^{1,1}, M^{2,1}, \dots, M^{n,1}, M^{1,2}, \dots \dots, M^{n,n})$, the matrix of $\text{ad}_B$ is $n^2\times n^2$ matrix with $n$ copies of matrix $B$ along the main diagonal and zeros everywhere else. The trace of this matrix is $n~\text{tr }B$, the determinant is $(\det B)^n$, and the characteristic polynomial is the $n$-th power of the characteristic polynomial of $B$.
This can be used to find traces and determinants of $e_i$. In the basis $(e_k)_{k=1}^{n^2}$, the matrix of $\text{ad}_{e_i}$ has entries $(\text{ad}_{e_i})_{j,k} = a_{i,k,j}$. Since trace, determinant and characteristic polynomial of an operator are basis-independent, the trace etc. of this matrix are related to the trace etc. of $e_i$ as described above. If the characteristic of $K$ does not divide $n$, then you can compute $\text{tr }e_i$ as $\frac1n\text{tr ad}_{e_i}$. If $K$ has no $n$-th roots of unity other than $1$, then you can compute $\text{det }e_i$ as $\sqrt[n]{\det\text{ad}_{e_i}}$. Otherwise, you can compute the characteristic polynomial of $e_i$ as the $n$-th root of the characteristic polynomial of $\text{ad}_{e_i}$ which is a monic polynomial, and then you can find the trace and determinant of $e_i$ from its characteristic polynomial.
The same approach works for any linear combination of $e_i$.
A: The determinant of an element  $A\in M_n(K)$ can be computed by looking at $A$ in isolation. To compute this we don't even use the fact that $M_n(K)$ is a $K$-vector space, much less the algebra structure. So a different algebra structure will have no influence on any determinant values: however the property $\det(AB) =\det(A)\det(B)$ will not be valid for other algebra structures.
