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I don't understand why the determinant of the 0x0 matrix equals 1.

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    $\begingroup$ What do you expect it to return, and why? $\endgroup$ – Ben Grossmann Apr 28 '16 at 11:06
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The only linear map $\{0\}\to\{0\}$ is the identity map and the identity map has determinant $1$, because this is a required property of the determinant.

From a slightly different point of view, since the basis of $\{0\}$ is empty, the representative matrix of the unique endomorphism is the empty matrix (which is invertible and its inverse is, of course, the empty matrix). Invertible matrices should have nonzero determinant, shouldn't they?

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Since the determinant of a matrix $M$ is the product of the eigenvalues of $M$, this yields an empty product for $\det([\ ])$, which is by definition 1.

The reason the empty product is defined to 1 (this is one reason of many) is because it makes the definition of exponentiation in terms of multiplication work. Since $$a^n = \underbrace{a\cdot a\cdots a}_{\text{$n$ times}}$$ And therefore, $a^0$ is $a$ multiplied $0$ times, which we know is $1$. Hence, empty product is $1$

More conceptually, we have the integers $\mathbb Z$ form a monoid with the multiplication operation, with multiplcative identity $1$. Thus, the empty product better evaluate to $1$, since in general, the only sensible way to define the product of a set of elements by induction is to announce:

\begin{align*} &(S, *, e) \text{ is a monoid} \\ &\mathsf{prod}: \mathsf{List}(S) \rightarrow S \\ &\mathsf{prod}([]) = e \\ &\mathsf{prod}(x::\mathsf{rest}) = x * \mathsf{prod}(\mathsf{rest}) \\ \end{align*}

where $[]$ is the empty list, and $(x::\mathsf{rest})$ is syntax for "a list starting with element $x$ followed by a list called $\mathsf{rest}$.

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