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Suppose A is a n times n matrix.

what is the determinant of the i-th exterior power of A, in terms of determinant of A ?


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The answer is just $\det(\Lambda^i(A))=\det(A)^m$, where $m=\left(\begin{matrix}n-1\\\\ i-1\end{matrix}\right)$. Indeed, by continuity it is enough to prove this when $A$ is diagonalisable, and by conjugation-invariance it suffices to prove it when $A$ is diagonal, and in that case it is straightforward. Alternatively, you can show that $SL_n(\mathbb{C})$ is the commutator subgroup of $GL_n(\mathbb{C})$ and thus that any polynomial homomorphism $GL_n(\mathbb{C})\to\mathbb{C}^\times$ is a power of the determinant. To find the relevant power, just take $A$ to be a multiple of the identity.

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This result is called the Sylvester--Franke theorem. – KConrad Jul 21 '11 at 13:02
Neil's first argument works over an arbitrary field: first observe that WLOG the field is alg closed, and then use the Zariski topology for the continuity argument. – Kevin Buzzard Jul 21 '11 at 13:32
In that case I'd say the result is true over any commutative ring, since the formula can be regarded as a universal polynomial identity over Z and thus it suffices to check it over the complex numbers. – KConrad Jul 21 '11 at 22:28

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