Well, you can calculate the volume of the parallelepiped spanned by the standard basis vectors $e_i$ with $1$ in the $i$'th coordinate and $0$ elsewhere, right?
So given a parallelepiped spanned by $n$ vectors $v_i$, what's the matrix which maps $e_i$ on $v_i$?
The solution to this will give you yet another interpretation of how the determinate measures volumes.
EDIT: I reread your question and there are some things that I want to clear up.
- The expression $vol(x)$ is meaningless for a vector.
- This means that the expression $Ax = y \Rightarrow |y| = |det(A)|vol(x)$ is meaningless as well.
- It takes $n$ vectors to specify a parallelepiped with $n$-volume. The volume will be zero only if all of the vectors are independent. A square has zero 3-volume for instance.
- That parallelepiped can be defined as $P(v_1, \ldots, v_n) = \{x: x = c_1e_1 + \ldots + c_ne_n, 0 \leq c_1, \ldots, c_n \leq 1 \}$
- The determinate measure oriented $n$-volume. It does not measure oriented $n-d$ volume.
For size 7 it was proved by Metropolis, Stein, and Wells that the set of non-negative determinant values is $[0,18]\cup\{20,24,32\}$, which means that the answer to your question in the gray box is no. See this paper:
N. Metropolis, , Spectra of determinant values in (0, 1) matrices. In A. O. L. Atkin and B. J. Birch, editors, Computers in Number Theory: Proceedings of the Science Research Atlas Symposium No. 2 held at Oxford, from 18-23 August, 1969, pages 271–276, London, 1971. Academic Press.
For general size, the data I have suggest that the set of determinant values is "dense" up to about half of Hadamard's bound and sparse thereafter, but I have no argument why that should be the case. And my data only go up to size 22. Who knows what happens in size 200?
Here are some entries in the OEIS that might be helpful.
- A003432 - maximimal determinants of n-by-n binary matrix
- A089472 - number of different values taken by the determinant of n-by-n binary matrix
- A013588 - smallest positive integer not the determinant of an n-by-n binary matrix.
I gave a talk couple of years ago, Range and distribution of determinants of binary matrices (available here), in which I discussed this topic. That discussion starts on slide 27, but other parts of the talk may also be relevant to your question.
I've also created a web page listing determinant values that have been constructed in matrix sizes up to 16. These lists are proved to be complete in sizes up to 10 and also in size 12, but I believe that the lists are probably complete in the other sizes as well. Be careful: my page discusses $\{-1,1\}$-matrices. You'll have to subtract 1 from the matrix size if you're interested in $\{0,1\}$-matrices.
See also the Mathworld and Wikipedia pages on Hadamard's maximal determinant problem. Also relevant are the papers of Tao and Vu mentioned in my slides. There is recent work on lower bounds on the maximal determinant by Brent, Osborn, and Smith. You can find these on the arXiv.
Best Answer
Consider an identity or an inequality of the following form:
$$\tag{*} \det A_n=C\cdot \det B_m, \qquad [\text{or }\le,\ \text{or }\ge], \qquad \forall A_n\in \mathcal{A},\ \forall B_m\in \mathcal{B}.$$
Here $C$ is a constant and $\mathcal{A}, \mathcal{B}$ are some families of $n\times n$ and $m\times m$ matrices.
If $n\ne m$ then the relation $(*)$ has a serious chance of being incorrect and should be regarded with suspicion if it arises in actual computations. Namely, if $\mathcal{A}$ and $\mathcal{B}$ are closed under scaling, that is
$$ (A\in \mathcal{A},\ B\in \mathcal{B},\ \lambda\ge 0)\ \Rightarrow\ \lambda A\in \mathcal{A}\ \text{and}\ \lambda B \in \mathcal{B}, $$
then $(*)$ is incorrect (except for trivial cases, such as having $C=0$ and the like). Indeed, if $(*)$ holds, then one should have
$$ \lambda^n\det A_n = C\cdot \lambda^m\det B_m, \qquad [\text{or }\le,\ \text{or }\ge],$$
which forces a contradiction in the limit $\lambda \to 0$ or $\lambda \to \infty$ (except the trivial case of an identity $0=0$, of course).
This is usually called scaling argument.