Which is the max value of the determinant with 4 lines and 4 colums ,where every term is +- 1?

Question

I understand that this problem can be solve with the volume of a tetrahedron.

But i don't know how.

please help me !

\begin{vmatrix} \pm1 & \pm1 & \pm1 & \pm1 \\ \pm1 & \pm1 & \pm1 & \pm1 \\ \pm1 & \pm1 & \pm1 & \pm1 \\ \pm1 & \pm1 & \pm1 & \pm1 \notag \end{vmatrix}

@EwanDelanoy $16$ quite obviously is an upper bound, I am curious is the absolute upper bound though? I mean can it be achieved? — Guy, Apr 04 '14 at 06:33
@EwanDelanoy you will notice I am not the OP, but I get your point. ;) — Guy, Apr 04 '14 at 06:38

score 4 · Accepted Answer · answered Apr 04 '14 at 06:33

4

You have $|\det[x_1 \cdots x_n ] | \le \|x_1\|\cdots \|x_n\|$ (Hadamard inequality).

This shows the determinant is upper bounded by 16.

Now choose $A = \begin{bmatrix} 1 & 1 & 1 & 1 \\ 1 & -1 & 1 & -1 \\ 1 & 1 & -1 & -1 \\ 1 & -1 & -1 & 1 \end{bmatrix}$, and compute $\det A = 16$.

answered Apr 04 '14 at 06:33

copper.hat

Are you saying $16$? In that case can it be achieved? – Guy Apr 04 '14 at 06:34
@Sabyasachi: My typing skills are limited. Yes, it can be achieved... – copper.hat Apr 04 '14 at 06:36
Ah very nice. +1. It is very satisfying when the first obvious trivial upperbound is the actual upper bound. – Guy Apr 04 '14 at 06:37

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