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Equicorrelation Matrix

Posted on (Update: )
Tags: Equicorrelation, Toeplitz, Woodbury Formula

kjytay’s blog summarizes some properties of equicorrelation matix, which has the following form,

where the off-diagonals $\rho \in[-1,1]$.

Rewrite in matrix form, we have

Toeplitz matrix

First of all, $\Sigma$ is the so-called Toeplitz matrix, in which each descending from left to right is constant.

Formally, any $n\times n$ matrix $A$ of the form,

is a Toeplitz matrix, and the elements satisfy

Note that a Toeplitz is not necessarily symmetric, but it is persymmetric. In R, toeplitz() returns a symmetric Toeplitz matrix with the given first row.

More properties can be found in Robert M. Gray. Toeplitz and Circulant Matrices: A review


Firstly consider the eigenvalues of $\1\1^T$, since it is a rank-1 matrix, which implies that it has $n-1$ eigenvalues which are zero. Denote the eigenvectors as $\xi_i, i=1,\ldots,n$. Suppose the nonzero eigenvalue is $\lambda_1$ with its associated eigenvector $\xi_1$, then

left multiplying $\1^T$ get

which implies that $\lambda_1 = n$.

Note that

which means that the eigenvalues of $\Sigma$ is $(n-1)\rho+1$ and $n-1$ eigenvalues with value $1-\rho$, and the same eigenvectors as $\1\1^T$.

In the words of a handout, the rank-1 modification $\rho\1\1^T$ to the matrix $(1-\rho)\I$ only changes one of the eigenvalues, i.e., updating one eigenvalue $1-\rho$ to $(n-1)\rho+1$.

Based on the eigenvalues, we can easily get

  • $\det(\Sigma) = (1-\rho)^{n-1}[1+(n-1)\rho]$
  • $\Sigma$ is positive definite iff $-\frac{1}{n-1} < \rho < 1$.


By the Woodbury matrix identity, or say Sherman-Morrison formula, we can get

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