I want to show that for Endomorphisms in finitely dimensional Vector spaces $f \in \operatorname{End}(U)$ and $g \in \operatorname{End}(V)$ we have $$ \operatorname{tr}(f \otimes g)=\operatorname{tr}(f) \cdot \operatorname{tr}(g), \\ \operatorname{det}(f \otimes g)=\operatorname{det}(f)^{n} \cdot \operatorname{det}(g)^{m}, $$ whereas $m=\operatorname{dim} U, n=\operatorname{dim} V$.

I tried to use $f \otimes g=(f \otimes$ id) $\circ$ (id $\otimes g)$. but I got nowhere and what I've got is not really worth mentioning as I got stuck after one line of equations. If anyone has a hint or advice then I'd help me very much.

  • $\begingroup$ The identity you tried to use is helpful to prove the determinant equation, since $\det(f\otimes \text{id}) = \det(f)^n$ and $\det(\text{id}\otimes g) = \det(g)^m$. $\endgroup$
    – kobe
    Jan 14 at 19:29

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