If a determinant of order three has only 1 or -1 as its elements, what would be its maximum value? I thought about this problem, but I am not getting an idea to start this problem. Help is appreciated. :D
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1See : hadamard-matrix – Peter Jul 03 '16 at 10:02
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1An obvious upper bound is $6$, as there are six products in the development. – Jul 03 '16 at 10:09
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I don't think that is correct. @YvesDaoust – Ryan Bendtner Jul 03 '16 at 10:11
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https://en.wikipedia.org/wiki/Hadamard%27s_maximal_determinant_problem – Martin R Jul 03 '16 at 10:13
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@RyanBendtner: can you substantiate ? Six times one is six, isn't it ? – Jul 03 '16 at 10:18
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@RyanBendtner: don't confuse maximum and upper bound. $6$ is an upper bound, that's undisputable. By the way, the same as that given by mvw. – Jul 03 '16 at 10:30
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Yeah, thanks a lot! @YvesDaoust – Ryan Bendtner Jul 03 '16 at 10:34
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Related. – Jyrki Lahtonen Jul 04 '16 at 04:24
2 Answers
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There is essentially one way to get three linearly independent rows: change one sign, then another.
$$\left|\begin{matrix}1&1&1\\\bar1&1&1\\\bar1&\bar1&1\end{matrix}\right|=4.$$
Other argument:
Perform Gauss elimination on the matrix. After one round of elimination, the new coefficients are either $0$ or $\pm2$, and if the matrix is in triangular form, the maximum determinant is $1\cdot2\cdot2$. If a second round is needed, the new coefficient is $0$ or $\pm4$, which could yield $1\cdot2\cdot4$. But this is impossible as the obvious upper bound is $6$.
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An upper bound can be obtained from the definition of the determinant by Leibniz $$ \DeclareMathOperator{sgn}{sgn} \DeclareMathOperator{det}{det} \det A = \underbrace{\sum_{\pi \in S_n}}_{\tiny\lvert S_n\rvert = n!}\underbrace{\sgn(\pi)}_{\tiny \in \{ -1,1\}} \underbrace{ \prod_{i=1}^n a_{i\pi(i)}}_{\tiny \in \{ -1,1\}} \le n! $$ where $S_n$ is the symmetric group of order $n$, containing all permutations of size $n$.
mvw
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