Eigenvalues of an Equicorrelated Matrix
Consider an $n \times n$ correlation matrix where every diagonal entry is
- Find all eigenvalues of this matrix and their multiplicities.
- For what range of $\rho$ is this matrix positive semi-definite?
Hints
- The matrix is symmetric under any permutation of the coordinates. What does that tell you about the eigenvectors?
- Write the matrix as $(1-\rho)I + \rho \mathbf{1}\mathbf{1}^T$. The rank-1 part $\mathbf{1}\mathbf{1}^T$ has a known eigendecomposition.
- The eigenvectors are $\mathbf{1}$ (with eigenvalue + (n-1)\rho$) and any vector orthogonal to $\mathbf{1}$ (with eigenvalue- \rho$). Check by direct multiplication.
Worked Solution
How to Think About It: Before doing any algebra, think about what this matrix does geometrically. It treats all variables symmetrically -- every pair has the same correlation. That symmetry tells you the eigenvectors must respect the symmetry of the matrix. The "all ones" direction is special (it's the only direction invariant under permutations of coordinates), and every direction orthogonal to it must have the same eigenvalue by symmetry. So you should expect two distinct eigenvalues: one for the $\mathbf{1}$ direction and one with multiplicity $n-1$ for everything orthogonal to it.
Quick Estimate: For $n = 3$, $\rho = 0.5$, the matrix is $\begin{pmatrix} 1 & 0.5 & 0.5 \\ 0.5 & 1 & 0.5 \\ 0.5 & 0.5 & 1 \end{pmatrix}$. The trace is 3 (sum of eigenvalues = 3). The $\mathbf{1}$ eigenvector gives eigenvalue
+ 2(0.5) = 2$. The other two eigenvalues must each be $(3-2)/2 = 0.5 = 1 - 0.5$. This confirms the pattern.Approach: Decompose the matrix as a sum of the identity and a rank-1 matrix, then read off the eigenvalues.
Formal Solution:
Write the equicorrelated matrix as:
$A = (1 - \rho)I_n + \rho \mathbf{1}\mathbf{1}^T$
where $\mathbf{1}$ is the $n \times 1$ all-ones vector. This works because the diagonal entries are $(1-\rho) + \rho \cdot 1 = 1$ and the off-diagonal entries are $0 + \rho \cdot 1 = \rho$.
The rank-1 matrix $\mathbf{1}\mathbf{1}^T$ has eigenvalue $n$ for the eigenvector $\mathbf{1}$, and eigenvalue $0$ for any vector orthogonal to $\mathbf{1}$ (multiplicity $n-1$).
Since $A = (1-\rho)I + \rho \mathbf{1}\mathbf{1}^T$, and $I$ and $\mathbf{1}\mathbf{1}^T$ share the same eigenvectors:
- Eigenvector $\mathbf{1}$: $A\mathbf{1} = (1-\rho)\mathbf{1} + \rho \cdot n \cdot \mathbf{1} = [1 + (n-1)\rho]\mathbf{1}$
$\lambda_1 = 1 + (n-1)\rho \quad \text{(multiplicity 1)}$
- Eigenvectors $v \perp \mathbf{1}$: $Av = (1-\rho)v + \rho(\mathbf{1}^T v)\mathbf{1} = (1-\rho)v$
$\lambda_2 = 1 - \rho \quad \text{(multiplicity } n-1\text{)}$
Positive semi-definiteness: $A \succeq 0$ requires both eigenvalues to be non-negative: -
- \rho \geq 0 \implies \rho \leq 1$ -+ (n-1)\rho \geq 0 \implies \rho \geq -1/(n-1)$So the valid range is $\rho \in [-1/(n-1),\, 1]$.
Answer: The eigenvalues are $\lambda_1 = 1 + (n-1)\rho$ with multiplicity 1, and $\lambda_2 = 1 - \rho$ with multiplicity $n-1$. The matrix is PSD for $\rho \in [-1/(n-1), 1]$.
Intuition
This decomposition trick -- writing a structured matrix as Identity + low-rank -- shows up constantly in quant work. Any time you see a covariance or correlation matrix with a simple pattern, your first move should be to decompose it into pieces whose eigenstructure you already know. The identity has all eigenvalues equal to 1; a rank-1 matrix has one nonzero eigenvalue. Adding them gives you the answer with almost no computation.
In finance, the equicorrelated matrix is the simplest model for a portfolio of assets driven by a single common factor. The eigenvalue
+ (n-1)\rho$ corresponds to the "market mode" (all assets moving together), and- \rho$ corresponds to the $n-1$ idiosyncratic modes. The PSD constraint $\rho \geq -1/(n-1)$ is a real restriction: you cannot have all pairwise correlations be $-0.5$ in a portfolio of 10 assets, because that would imply a negative-variance direction. This comes up when fitting correlation matrices or stress-testing portfolios.