Equicorrelation matrix

The equicorrelation matrix is the \mathbb{R}^{n \times n} matrix where the entries on the diagonal are all equal to 1 and all off-diagonal entries are equal to some parameter \rho which lies in [-1, 1]. If we were to write out the matrix, it would look something like this:

{\bf \Sigma} = \begin{pmatrix} 1 & \rho & \rho & \dots & \rho \\ \rho & 1 & \rho & \dots & \rho \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ \rho & \rho & \rho &\dots & 1 \end{pmatrix}

Alternatively, we can write it as {\bf \Sigma} = \rho {\bf 11}^T + (1-\rho){\bf I}, where {\bf 1} \in \mathbb{R}^n is the column vector with all entries being 1 and \bf I is the identity matrix.

Here are some useful properties of the equicorrelation matrix:

  • It is a Toeplitz matrix, and hence has the properties that all Toeplitz matrices has (see e.g. this link).
  • It has two eigenvalues. The first eigenvalue is 1 + (n-1)\rho, with associated eigenvector v_1 = {\bf 1}. The second eigenvalue is 1 - \rho, with n-1 associated eigenvectors v_2, \dots, v_n, where the entries of v_k are\begin{aligned} (v_k)_j = \begin{cases} 1 &\text{if } j = 1, \\ -1 &\text{if } j = k, \\ 0 &\text{otherwise.} \end{cases} \end{aligned}This can be verified directly by doing the matrix multiplication.
  • \text{det} ({\bf \Sigma}) = (1-\rho)^{n-1}[1 + (n-1)\rho]. This is because the determinant of a square matrix is equal to the product of its eigenvalues.
  • \bf \Sigma is positive definite if and only if -\frac{1}{n-1} < \rho < 1. A sketch of the proof can be found in the answer here. It boils down to proving some properties of the determinant expression in the previous point.
  • {\bf \Sigma}^{-1} = \dfrac{1}{1-\rho} \left( {\bf I} - \dfrac{\rho}{1 + (n-1)\rho} {\bf 11}^T \right). This can be verified directly by matrix multiplication. It can also be derived using the Sherman-Morrison formula.

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