# Is there any relation between weights in the eigenvector (corresponding to least eigenvalue) and the columns of a correlation matrix?

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

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

So far, I am observing empirically that $v_k \leq \frac{1}{\sqrt(n)}$.

I am wondering if this is indeed true in general, or otherwise, if there is any counterexample where this will break?

Note: For the cases with algebraic multiplicities (e.g. identity matrix), since the set of eigenvectors are not unique for such matrices and hence technically one might choose an appropriate eigenvector that would satisfy the above observation.

• $\alpha_i$ corresponds to the i-th row, not column. Since the matrix is symmetric this is not an issue. Still, it's misleading saying things like this. – Integral Dec 17 '17 at 3:16
• I don't see any reason for introducing the alpha terms. Anyway, consider the identity matrix. – Integral Dec 17 '17 at 3:24
• Thanks for pointing that $\alpha_i$ issue. I have fixed it now. Identity matrix is a weird case because there is no unique set of eigenvectors as every unit vector is an eigenvector. So technically, it is possible to choose one that would satisfy the observation. – Saurabh Agrawal Dec 17 '17 at 19:08