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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 $F$?

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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.

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  • $\begingroup$ Thanks, that's very enlightening. I will wait a little longer before accepting an answer as I am still hoping for some relationship for the determinant. In that line of thoughts, I am wondering if there isn't a different choice of vectorization that might be suitable to observing the determinant case. $\endgroup$
    – M.G.
    Commented May 21, 2017 at 14:48
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    $\begingroup$ 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. $\endgroup$
    – Nathaniel Johnston
    Commented May 22, 2017 at 17:11
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    $\begingroup$ @NathanielJohnston what do you mean by mutually orthogonal? $\endgroup$
    – Turbo
    Commented Oct 22, 2017 at 16:01
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    $\begingroup$ @777 - Orthogonal in the Frobenius (Hilbert-Schmidt) inner product. $\endgroup$
    – Nathaniel Johnston
    Commented Oct 23, 2017 at 23:49

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