# Absolute Determinant: Geometric Argument

Using a geometric argument, explain why the matrix below has maximal absolute determinant among all matrices with entires in {-1,1}.
$\begin{bmatrix} 1 & 1 & 1 & 1 \\ 1 & -1 & 1 & -1 \\ 1 & 1 & -1 & -1 \\ 1 & -1 & -1 & 1 \end{bmatrix}$

I believe this has something to do with the fact that each sub matrix is either a rotation or a reflection and there is a symmetric number of both. Does this thinking seem in line with others?

• Check this out math.stackexchange.com/questions/250534/… – ir7 Mar 19 '18 at 22:18
• – ir7 Mar 19 '18 at 22:20
• Volume rather. See this – SphericalTriangle Mar 19 '18 at 22:21
• @ir7 Understood, but why does this specific matrix have maximal absolute determinant? Couldn't you find one with a different ordering that has a determinant of 16 as well? – Johnny G. Mar 19 '18 at 23:32

It’s a Hadamard matrix as its rows are mutually orthogonal. If you multiply it by its transpose you get the identity matrix times 4. Hence its determinant is $\pm 16$. You can then look up Hadamard’s maximal determinant problem to see why only these matrices can achieve the upper bound (maximum volume parallelotope).
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