TITLE: Does the unique map on the zero space have determinant 1?
QUESTION [11 upvotes]: 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?

REPLY [3 votes]: I don't recall offhand any reference which discusses this issue but I'll add my two cents in support of taking $\det{f} = 1$ as the definition for the unique map $f \colon V \rightarrow V$ on a zero-dimensional space.
This definition is consistent with various standard theorems in linear algebra so that one doesn't need to exclude the zero dimensional case as an exception. In fact, I can't think of a single theorem which will become false taking $\det{f} = 1$ as the definition while most will break if you take $\det{f} = 0$ as the definition and don't exclude the zero dimensional case. For example,

An endomorphism on a finite dimensional vector space is invertible iff $\det(f) \neq 0$.
The characteristic polynomial of an operator $f \colon V \rightarrow V$ on a finite dimensional vector space is monic of degree $\dim V$ and a scalar $\lambda \in \mathbb{F}$ is a root of the characteristic polynomial iff $\lambda$ is an eigenvalue of $f$. The characteristic polynomial of $f$ is defined as $\chi_f(x) = \det(x \cdot \operatorname{id} - f)$ so using the "1" convention it becomes $\chi_f(x) = 1$ which is indeed monic of degree zero and doesn't have any roots. Using the $0$ convention gives $\chi_f(x) = 0$ which has degree $-\infty$ and all scalars as roots.
The minimal polynomial of $f$ divides the characteristic polynomial. Again, the minimal polynomial is the unique monic polynomial $m_f$ of minimal (non-zero) degree such that $m_f(f) = 0$. In the zero dimensional case, it becomes $1$ (indeed $m_f(f) = \operatorname{id}_V = 0$) and it divides the characteristic polynomial $1$. If the characteristic polynomial would be zero, this would be false.
The characteristic polynomial of the restriction of an operator $g$ to an $g$-invariant subspace divides the characteristic polynomial of $g$. Since $\{ 0 \}$ is a legitimate $g$-invariant subspace, it makes sense to take the characteristic polynomial of $g|_{\{0\}} = f$ to be $1$ and not $0$.
An orthogonal map on an inner product space should have determinant one.
The determinant of the identity map is $1$.
It might look silly but the idea of orienting a zero-dimensional vector space is important (for example, to state a general version of Stokes' theorem which includes the fundamental theorem of calculus as a special case). An orientation on a zero-dimensional real vector space is just a choice of $\pm 1$ which states whether the point is "positive or negative". The unique map on the zero-dimensional vector space does nothing (it is the identity map) so it should be orientation preserving and a map is orientation preserving iff $\det(f) > 0$.

Finally, let me say that in my mind, if one wants to define the determinant of the unique map $f \colon V \rightarrow V$ on the zero dimensional space, then the definition shouldn't depend on the choice of field $\mathbb{F}$ over which we are working on (this is more of a meta-mathematical statement). It would be ridiculous to define say $\det(f) = 2$ if $\mathbb{F} = \mathbb{R}$ while $\det(f) = 3$ if $\mathbb{F} = \mathbb{Z}_5$. Thus it leaves one with two sensible choices: $\det(f) = 0$ or $\det(f) = 1$. Since $\det(f) = 0$ breaks up so many theorems, it doesn't make sense to take it as the definition, better just leave it undefined and that's it.

Added: Almost any textbook which embraces the definition of the determinant via the exterior algebra will have as a result a definition for the zero dimensional case which will be $\det(f) = 1$. For such textbooks, it won't be a convention but a (trivial) result. For a specific example, see Algebra 1, Chapters 1-3 by Bourbaki. On page 525 they state that $\det([]) = 1$ but for them, this is not a convention or an ad hoc definition but a result proved from their definition of the determinant and the notion of a matrix.