Determinants (and traces) of linear maps of matrices

Let $k$ be a field or a commutative ring with unit and let $F:M_n(k)\to M_n(k)$ be a $k$-linear map. Suppose that $F$ is given in the form $F(X) = A_1XB_1 + \cdots + A_m X B_m$ for some $A_i,B_i\in M_n(k)$ (note that any $k$-linear $F$ can be written in this form, though not uniquely).

Is there a formula that allows one to determine $\mathrm{det}(F)$ (no pun intended) or $\mathrm{tr}(F)$ directly from the matrices $A_i, B_i$ ($1\leq i\leq m$), i.e. without having to compute a representation matrix of $F$?

We have $F(X)=\sum_i A_i X B_i = \sum_i (B_i^T \otimes A_i) vec(X)$ (see here), i.e. $F \sim \sum_i (B_i^T \otimes A_i)$. Because of some formulas here we have $tr(F)=\sum_i tr(A_i)tr(B_i)$. For the determinant I don't think there is a nice formula.
• In a sense, it's impossible for there to be such a formula for the determinant, since it isn't linear. For example, suppose for a second that $A_1 = \cdots = A_m = I$, and $B_1, \ldots, B_m$ are any matrices that add up to a fixed matrix $B$. Then you're basically asking for a formula for $\det(B) = \det(B_1 + \cdots + B_m)$ in terms of $B_1, \ldots, B_m$, without having to compute $B_1 + \cdots + B_m$. You can do something if you have some control over the choice of $A_i,B_i$ matrices (e.g., you can choose them to be mutually orthogonal), but not in this general setting. – Nathaniel Johnston May 22 '17 at 17:11