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The trivial vector space over any field $K$, consisting of only the zero vector, admits exactly one endomorphism, let's call it $z$, sending $0$ to itself.

  • It is the identity map, so it should have determinant $1$.
  • On the face of it, the zero map should have determinant $0$. But this is usually argued via $\lambda z = z$ for all $\lambda \in K$, so $\det z = \det (\lambda z) = \lambda^n \det z$, i.e. $(\lambda^n - 1)\det z = 0$. Normally that's enough to conclude that $\det z = 0$, but of course $n = 0$ in this case, so $\lambda^n = 1$ for all $\lambda$, and we learn nothing.
  • Despite being the zero map, it's full rank and has trivial kernel.
  • There are no nonzero vectors, so it has no eigenvectors, so it has no eigenvalues, so their product is $1$.
  • On the other hand, the determinant is meant to be multilinear, and so should map the zero matrix to zero. But should we say that $z$ is represented by a zero matrix, given that its matrix representation is $0\times 0$ and doesn't have any entries at all?

I can't help but feel like this is all very silly, but clearly the answer can't be anything other than $1$. Is there anything wrong with giving this answer? Does it cause any problems with any other typical properties of the determinant? Does it simplify any definitions or theorems?

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  • $\begingroup$ Using the Leibniz formula with $n=0$, you have $$\sum_{\sigma \in S_0} \mathrm{sign} (\sigma ) \prod_{i=1}^0 x_{i \sigma (i)} = \sum_{\sigma \in \{ 0 \}} \mathrm{sign} (\sigma ) \cdot 1 = ?$$ $\endgroup$ – Crostul Feb 5 '17 at 13:36
  • $\begingroup$ The unique permutation on the empty set must have sign 1 in order for sign to be a homomorphism! $\endgroup$ – Ben Millwood Feb 7 '17 at 12:45

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