A correlation matrix is a positive semidefinite matrix $B\in M_{n}(\mathbb{C})$ such that the diagonal of $B$ is the identity matrix $I$. What is the supremum of the set of positive numbers $t$ such that there is a correlation matrix $B$ having $t$ as its greatest eigenvalue?


Let $\lambda_1, \ldots, \lambda_n$ denote the eigenvalues of $B$ (with multiplicities). Since $\mathrm{tr}(B) = \lambda_1 + \ldots + \lambda_n = n$ and $\lambda_i \geq 0$, we must have $\lambda_{\max}(B) \leq n$. However, the matrix $A$ consisting of $1$-s in all entries is positive semi-definite with $\lambda_{\max}(A) = n$ (the corresponding eigenvector is $(1, \ldots, 1)^t$). Thus, the supremum is a maximum and equals to $n$.

  • $\begingroup$ Great! Thank you very much! $\endgroup$ – Jon Bannon Nov 8 '15 at 14:01
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    $\begingroup$ You're welcome. Probabilistically, the matrix $A$ corresponds to the correlation matrix of a sequence $(X_1, \ldots, X_n)$ of identical random variables. $\endgroup$ – levap Nov 8 '15 at 14:31

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