# Change in eigenvalues due to perturbation to a correlation matrix

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: $$\hat{\lambda}_1 \geq \lambda_1 \quad \text{and} \quad \hat{\lambda}_i \leq \lambda_i \quad i\in[2,\dots,n].$$

Example:

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?

• 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. – user617446 Dec 9 '18 at 12:53
• @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. – hari Dec 9 '18 at 15:32