Return to Revisions

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.