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Generalized variance is the determinant of correlation matrix. Does increasing the off-diagonal entries (correlation coefficients) decreases the determinant? Is a proof available? All elements are positive. Can we deduce from Hadamard inequality of determinant?

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The determinant of the covariance matrix could be considered a generalization of variance, in that it's equal to the scalar variance in the case of dimension 1. But the determinant of the correlation matrix, as opposed to the covariance matrix, is not in that sense a generalization of the variance. – Michael Hardy Aug 27 '11 at 11:46
Thanks for the proper definition. – shakera Aug 27 '11 at 12:15

It can do either. $\;$ Suppose the correlation matrix is $\begin{bmatrix} 1 & x \\ x & 1 \end{bmatrix}$.

$\operatorname{det}\left(\begin{bmatrix} 1 & x \\ x & 1 \end{bmatrix}\right) = 1\cdot 1-x\cdot x = 1-x^2$

If $x<0$ then increasing the off-diagonal entries increases the determinant.
If $0<x$ then increasing the off-diagonal entires decreases the determinant.

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Thanks... Does it generalizes for $N$. Let us supposes all elements of the correlation matrix are positive. – shakera Aug 27 '11 at 9:25

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