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Let $M\in M_n(F)$ and define $\phi:M_n(F)\to M_n(F)$ by $\phi(X)=AX$ for all $X\in M_n(F)$. Prove that $\det(\phi)=\det(A)^n$.

I can prove it by considering the matrix representation with respect to the usual basis, which turns out to be a block diagonal form consisting of $n$ copies of $A$. Nevertheless, I'm looking for a clean (basis-free) approach to this problem.

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    How do you define $\det$ without a basis?2012-06-07
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    @copper.hat: using the universal property of the exterior power (see http://en.wikipedia.org/wiki/Exterior_algebra).2012-06-07
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    I'm guessing that "Let $M\in M_n(F)$..." is supposed to be "Let $A\in M_n(F)$"?2012-06-07
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    @QiaochuYuan: Thanks!2012-06-07

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