Equicorrelation Matrix

Let $$R$$ be the $n\times n$ correlation matrix with $$1$$ on the diagonal and $\rho$ on every off-diagonal entry. Find the range of $\rho$ for which $$R$$ is a valid correlation matrix.

Answer

The matrix $$R$$ is symmetric with $$1$$ 's on the diagonal, so it suffices to check positive semidefiniteness.

Step 1: Decomposition

Let $$I$$ be the identity matrix and $$J$$ the all-ones matrix. Then

R = (1-\rho)I + \rho J.

Step 2: Eigenvalues and eigenvectors of $$I$$ and $$J$$

$$I$$ : every vector is an eigenvector with eigenvalue $$1$$ .
$$J$$ : every scalar multiple of $\mathbf{1}=(1,1,\dots,1)^\top$ is an eigenvector with eigenvalue $$n$$ .
$$J$$ : every vector orthogonal to $\mathbf{1}$ (i.e., $\sum_{i=1}^n v_i=0$ ) is an eigenvector with eigenvalue $$0$$ .

Step 3: Eigenvalues of $$R$$

Fix the vector $\mathbf{1}=(1,1,\dots,1)^\top$ . Then

R\mathbf{1}=\bigl((1-\rho)I+\rho J\bigr)\mathbf{1} =(1-\rho)\mathbf{1}+\rho(n\mathbf{1}) =\bigl(1+(n-1)\rho\bigr)\mathbf{1}.

Thus $\mathbf{1}$ is an eigenvector of $$R$$ , and the corresponding eigenvalue is

\lambda_1=1+(n-1)\rho.

Now fix any vector $$v$$ such that $v\perp \mathbf{1}$ . Then $$Jv=0$$ , so

Rv=\bigl((1-\rho)I+\rho J\bigr)v=(1-\rho)v+\rho\cdot 0=(1-\rho)v.

Thus every $v\perp \mathbf{1}$ is an eigenvector of $$R$$ , and the corresponding eigenvalue is

\lambda_2=1-\rho .

Step 4: PSD conditions

Since $R \succeq 0$ if and only if all eigenvalues are nonnegative,

1-\rho \ge 0 \quad \Rightarrow \quad \rho \le 1,

1+(n-1)\rho \ge 0 \quad \Rightarrow \quad \rho \ge -\frac{1}{n-1}.

Therefore,

-\frac{1}{n-1} \le \rho \le 1.