Start by constructing the 3x3 correlation matrix with all off-diagonal entries equal to ρ. Then compute its eigenvalues explicitly: one eigenvalue is 1+2ρ and the other two are 1-ρ (with multiplicity 2). For positive semidefiniteness, require all eigenvalues ≥ 0, yielding ρ ≥ -1/2. Generalize to n variables by recognizing the matrix has eigenvalues 1+(n-1)ρ and 1-ρ (multiplicity n-1), so the bound becomes ρ ≥ -1/(n-1).
Pro tip: Emphasize that the bound is tight because at ρ = -1/(n-1) the matrix becomes singular (one eigenvalue zero), and mention that this is a classic result in factor analysis and portfolio theory, showing you understand its practical implications.
Write the 3x3 matrix with 1s on the diagonal and ρ on all off-diagonals. Note that it is symmetric and has a special structure (equicorrelation).
Observe that the all-ones vector is an eigenvector with eigenvalue 1+2ρ. The orthogonal subspace (vectors summing to zero) has dimension 2, and any such vector is an eigenvector with eigenvalue 1-ρ. Thus eigenvalues are 1+2ρ, 1-ρ, 1-ρ.
Require all eigenvalues ≥ 0: 1+2ρ ≥ 0 and 1-ρ ≥ 0. The second gives ρ ≤ 1, which is always true for correlations. The first gives ρ ≥ -1/2. So the tightest lower bound is -1/2.
For an n×n equicorrelation matrix, the eigenvalues are 1+(n-1)ρ (multiplicity 1) and 1-ρ (multiplicity n-1). The condition 1+(n-1)ρ ≥ 0 yields ρ ≥ -1/(n-1). The other condition gives ρ ≤ 1, which is trivial.
At ρ = -1/(n-1), the matrix is singular (one eigenvalue zero), so the bound is tight. Mention that this is the minimum possible correlation among n variables to maintain a valid covariance structure.
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