Let $A$ be a $m \times n$ matrix defined as $ A = \Big[\frac{a_1}{\|a_1\|} \cdots \frac{a_n}{\|a_n\|}\Big]$ and $a_k \in \mathbb{R}^{m\times 1}$ where $k \in [1,\dots,n]$.

Now, we define a correlation matrix $R = A^TA$ where each diagonal element is $1$ and it is a symmetric matrix. The trace of $R$, i.e., $\mathbb{Tr}(R) = n$.

All non-diagonal elements of $R$ represents the correlation among the columns of $A$. We define them by correlation-coefficients $\rho_{jk} = \Big(\frac{a_j}{\|a_j\|}\Big)^T\Big(\frac{a_k}{\|a_k\|}\Big)$ which satisfy $-1 \leq \rho_{jk} \leq 1$.

In my present work, I modify each column of $A$ such that the correlation among the columns of $A$ increases. Consequently, the correlation-coefficients also increases proportionally in $R$. I am interested to comment on the change in eigenvalues of $R$ with increase in correlation-coefficients.

Numerically, I observed that only largest eigenvalue of $R$ increases whereas rest of the eigenvalues decreases. But, I am unable to verify this phenomenon theoretically. Therefore, I ask you here for a hint to proceed my investigation further.

More precisely, the claim is:

Let $\lambda$ be the set of eigenvalues of $R$ where $\lambda_1 \geq \cdots \geq \lambda_n \geq 0$ and $\hat{\lambda}$ be the set of eigenvalues of $\hat{R}$ where $\hat{\lambda}_1 \geq \cdots \geq \hat{\lambda}_n \geq 0$. Assume that the correlation-coefficients in $\hat{R}$ satisfy $$ \hat{\rho}_{jk} \geq \rho_{jk} \quad {j,k} \in [1,\cdots,n] \quad \text{and} \quad j\neq k. $$ Moreover the trace of correlation matrices remains same, i.e., $$ \mathbb{Tr}(\hat{R}) = \mathbb{Tr}({R}) = n. $$ Consequently, we claim that the eigenvalues of $\hat{R}$ and $R$ satisfy the following inequalities: $$ \hat{\lambda}_1 \geq \lambda_1 \quad \text{and} \quad \hat{\lambda}_i \leq \lambda_i \quad i\in[2,\dots,n]. $$


Suppose, all columns of $A$ are orthonormal. This implies that the resulting correlation matrix would be an identity matrix and in this case, all eigenvalues are equal to $1$.

Now, suppose all columns are linearly dependent by a positive factor. This implies that the all correlation-coefficients is equal to 1 and the resulting correlation matrix is a rank-1 matrix, i.e., $\mathbb{1}\mathbb{1}^T$ where the largest eigenvalue is $n$ and rest of the eigenvalues are zero.

In this example, the largest eigenvalue increases from $1$ to $n$ when correlation matrix changes from the identity matrix to the rank-1 matrix. On the other hand, rest of the eigenvalues decreases from $1$ to $0$.

In order to prove the above mentioned claim, will it be sufficient:

if we can show that the largest eigenvalue path from the identity matrix to a matrix of ones, i.e., $\mathbb{1}\mathbb{1}^T$ is monotonically non-decreasing. Here, we will change only off-diagonal elements which always lies between -1 to 1. We also establish similar behaviour for rest of the eigenvalues ?

Can you comment on this approach? If you think, it could be a right direction. Do you have any suggestion how should I start to prove/disprove the claim?

  • $\begingroup$ It is not clear to me if your allowed perturbation preserves the $a_i/|a_i|$ condition. If it does, that you know that the trace is still $n$ under the perturbation and hence the sum of eigenvalues is unchanged. In the 2d example that you presented, How do you know that $\delta>0$? A negative $\delta$ would invalidate your claim. $\endgroup$ – user617446 Dec 9 '18 at 12:53
  • $\begingroup$ @user617446 Yes, the allowed perturbation preserves the $a_i/\|a_i\|$ condition. This is always ensured. You are right, in this condition, sum of eigenvalues is unchanged and somehow I should utilize this information to prove my arguments. In the 2nd example, $\delta$ is positive by construction. This signify that the correlation among all columns of $A$ are increasing. $\endgroup$ – hari Dec 9 '18 at 15:32

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