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?

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.
