Consider all $n$ by $n$ matrices whose elements are real values in $[0,1]$. Now for a given $n$, consider all those matrices with maximum determinant.
Are all the elements of these matrices always either exactly $0$ or $1$?
From numerical experiments this seems to be the case but I don't see how to prove it.
A very nice counterexample was given to my question as I didn't phrase it correctly. It should have been:
Can the maximum determinant always be reached by a matrix with only
$1$ and $0$ entries?
Best Answer
Since every term in the determinant formula is a product of distinct elements of the matrix $A$, its Laplacian, (when the determinant is viewed as a function of $\mathbb{R}^{n \times n}$) is $0$. i.e. It is harmonic. By the maximum principle, its maximum occurs on the boundary, or it is a constant function (whereas the maximum would still be on the boundary).
The boundary of the $[0,1]^{n^2}$ hypercube are those matrices that have at least one $0$ or $1$. So you can apply this argument inductively. The determinant function restricted to each subdimensional face is still harmonic. So, the max/min occurs on the boundary of each subdimensional face's boundary, etc. All the way down to the corner points.
Alternatively, this can be viewed as a standard linear optimization problem. The optimization function is linear and the constraints are given by standard inequalities. For such a problem, the max/min always occur at vertices of the constrained region. This is relied on by the simplex method.