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$A$ is a matrix with $1$ along diagonal and arbitrary unequal numbers less than $1$ on its non-diagonal. Let $B$ is a matrix with same $1$ along diagonal but having maximum non-diagonal element of $A$ as its non-diagonal element. What would be the relation between $\det (A)$ and $\det(B)$? Suppose $A$ and $B$ are correlation matrices. Does increasing the off-diagonal entries i.e. correlation coefficient decreases determinant?

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What type of relation are you looking for? –  Davide Giraudo Aug 26 '11 at 13:49
    
Like $det(A)$ greater or smaller than $det (B)$ –  shakera Aug 26 '11 at 14:08
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It could be greater, smaller or the same. –  Robert Israel Aug 26 '11 at 14:30
    
At what conditions it will greater or smaller. It is noted that all elements are positive. –  shakera Aug 26 '11 at 14:32
    
Do I understand correctly that if $A = \frac{1}{10}\left(\begin{array}{ccc} 10 \ 2 \ 3 \\ 4 \ 10 \ 5 \\ 6 \ 7 \ 10 \end{array}\right)$ then $B = \frac{1}{10}\left(\begin{array}{ccc} 10 \ 7 \ 7 \\ 7 \ 10 \ 7 \\ 7 \ 7 \ 10 \end{array}\right)$? –  TMM Aug 26 '11 at 15:16

1 Answer 1

up vote 2 down vote accepted

Enumerate the matrices $n \times n$ matrices $A_m$ with diagonal elements 1 and off-diagonal elements either $b$ or $0$, with at least one $0$ (there are $2^{n^2-n} - 1$ such matrices, but we can use symmetry to make the calculations somewhat more tractable). For each $m$, $\det(A_m) = 1 - k b^2 + O(b^3)$ where $k$ is the number of pairs $(i,j)$ such that $(A_m)_{ij} = (A_m)_{ji} = b$, while $\det(B) = 1 - \frac{n^2-n}{2} b^2 + O(b^3)$. Thus $\det(A_m) \ge \det(B)$ for $0 \le b \le b_m$ where $ b_m > 0$. Then any $A$ with diagonal elements 1 and off-diagonal elements in $[0,b]$ will have $\det(A) \ge \det(B)$ as long as $0 \le b \le \min_m b_m$. I don't know if there's a closed-form formula for $\epsilon(n) = \min_m b_m$. The first few values, obtained by exhaustive enumeration, are $\epsilon(2) = \infty$, $\epsilon(3) = 1/2$, $\epsilon(4) = (4 - \sqrt{10})/3$, $\epsilon(5) = -3/4\,\cos \left( 1/3\,\arctan \left( 4/7\,\sqrt {2} \right) \right) +5/4-3/4\,\sqrt {3}\sin \left( 1/3\,\arctan \left( 4/7\,\sqrt {2} \right) \right)$. This last is the least positive root of $-2+12 b-15 b^2+4 b^3$, corresponding to $A_m = \left[ \begin {array}{ccccc} 1&b&b&b&b\\ b&1&b&b&0\\ b&b&1&0&b\\ b&b&0&1&b \\ b&0&b&b&1\end {array} \right]$

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Thanks for the detailed answer. –  shakera Aug 28 '11 at 19:52
    
Excellent answer!!! –  shakera Aug 28 '11 at 19:59

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