1
$\begingroup$

This problem is motivated from one of my pattern mining research projects. Any helpful suggestions will be highly appreciated.

Consider an $n \times n$ correlation matrix A such that all the off-diagonal entries are between [-1,0]. (Note: A correlation matrix is a positive semi-definite symmetric matrix, with diagonal entries 1 and all off-diagonal entries between [-1,1]).

Let $\alpha_i = \frac{\sum_{j=1,j \neq i}^{n}|A_{ij}|}{n-1}$ denote the mean of magnitudes of off-diagonal entries in $i^{th}$ column.

Let $v_{min} = [v_1,v_2,...,v_n]^T$ be the unit eigenvector corresponding to the least eigenvalue $\lambda_{min}$ of A. Let $v_k$ be the weight with minimum magnitude in $v_{min}$.

Then empirically, I am observing that $\alpha_k$ is also minimum among all $\alpha_i$'s.

I am wondering if this is indeed true and can be proved, or otherwise, if there is any counterexample where this will break?

$\endgroup$
3
$\begingroup$

This is false.

I randomly generated a few correlation matrices having non-positive off-diagonal elements and discovered a counterexample. I then rounded the elements of this counterexample correlation matrix to 2 decimal places, and still had a counterexample.

Here is the MATLAB output.

>> disp(A), [eigenvec,eigenval] = eig(A), alpha = (sum(abs(A),1)-1)/4
    1.0000   -0.2500   -0.2100   -0.5300   -0.4500
   -0.2500    1.0000   -0.4100   -0.1900   -0.1700
   -0.2100   -0.4100    1.0000   -0.0800   -0.0100
   -0.5300   -0.1900   -0.0800    1.0000   -0.0100
   -0.4500   -0.1700   -0.0100   -0.0100    1.0000

eigenvec =
    0.5932   -0.3316   -0.0595   -0.2524    0.6862
    0.4474    0.4764    0.1461    0.7320    0.1253
    0.3542    0.6621    0.0594   -0.6233   -0.2103
    0.4405   -0.2344   -0.6893    0.1089   -0.5138
    0.3584   -0.4120    0.7046   -0.0140   -0.4530

eigenval =
    0.0207         0         0         0         0
        0     0.8447         0         0         0
        0          0    1.0117         0         0
        0          0         0    1.4103         0
        0          0         0         0    1.7126

alpha =
   0.3600     0.2550    0.1775    0.2025    0.1600

As can be seen, the first eigenvalue, 0.0207, is the smallest. The corresponding eigenvector is the first column of eigenvec, whose smallest magnitude element is the third, 0.3542. But the fifth element of alpha, 0.1600, is the smallest (smaller than the third element, 0.1775).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.