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.

  • $\begingroup$ 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. May 21 '17 at 14:48
  • 1
    $\begingroup$ 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 May 22 '17 at 17:11
  • $\begingroup$ @NathanielJohnston what do you mean by mutually orthogonal? $\endgroup$ – T.... Oct 22 '17 at 16:01
  • $\begingroup$ @777 - Orthogonal in the Frobenius (Hilbert-Schmidt) inner product. $\endgroup$ – Nathaniel Johnston Oct 23 '17 at 23:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.