Need suggestion of theorems for lower bound of the largest eigenvalue of the correlation matrix (symmetric, diagonal all being one, all other values in range $[-1,1]$).

Of course the eigenvalue has to be non-negative since the matrix is positive semi-definite, but need help with a tighter lower bound.

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    $\begingroup$ If it's an $n\times n$ matrix where every entry has modulus at most $1$, then the largest eigenvalue is at most $n$. This is attained when every entry is $1$. $\endgroup$ – Michael Burr Jan 13 '18 at 4:48

(Completely modified answer)

Correlation matrix $C$ being semi-definite positive, its eigenvalues are : $0 \leq \lambda_1 \leq \cdots \leq \lambda_n.$

The trace of $C$ being $n$, we thus have: $\lambda_n \geq 1$.

This bound $1$ is attained each time $C=I_n$ (identity matrix), for example by taking any pair of independently generated $n$-dimensional vectors.

Thus $1$ is the answer to your question.

  • $\begingroup$ First, thank you for your answer. Second, maybe I misunderstood your answer, but I stated "all other values in $[-1,1]$", i.e. the matrix is not ${\bf 1}^T{\bf 1}$, which is a special case. $\endgroup$ – Tony Jan 13 '18 at 14:38
  • $\begingroup$ I have understood that all other values values are in $[-1,1]$, but I realize now that I have misunderstood that you wanted a lower bound for $\max \lambda_k$. This lower bound is 1. I modify my answer $\endgroup$ – Jean Marie Jan 13 '18 at 16:14
  • $\begingroup$ Correction done. Sorry for having misunderstood your question... $\endgroup$ – Jean Marie Jan 13 '18 at 16:34

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