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 $m=\binom{n-1}{i-1}$. 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.
answered Jul 21, 2011 at 11:05
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5$\begingroup$ This result is called the Sylvester--Franke theorem. $\endgroup$– KConradCommented Jul 21, 2011 at 13:02
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4$\begingroup$ 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. $\endgroup$ Commented Jul 21, 2011 at 13:32
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4$\begingroup$ 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. $\endgroup$– KConradCommented Jul 21, 2011 at 22:28