Motivated by the problem of covariance estimation.

Let $\Sigma$ be a positive-definite matrix whose diagonal entries are identically 1. (i.e. $\Sigma$ is a correlation matrix.)

If $L$ is an lower-triangular matrix such that $L^T L = \Sigma$, can the entries of $L$ be bounded? (i.e. do there exist real numbers $l, u$ such that $l \geq (L)_{ij} \leq u$?)

  • $\begingroup$ do you mean $l \leq L_{ij} \leq u$? $\endgroup$
    – Macro
    Jul 18, 2011 at 17:09
  • $\begingroup$ This question is very, very close to stats.stackexchange.com/q/5747/919 . The answers there are in the affirmative: $[-1,1]$ are obvious bounds. That they are tight is demonstrated by considering multivariate normal distributions. $\endgroup$
    – whuber
    Jul 18, 2011 at 17:18
  • 4
    $\begingroup$ By definition you have $\sum_i L_{i,j}^2 = 1$. Since $L$ is real, you must have $-1 \le L_{i,j} \le 1$, as @whuber points out. $\endgroup$
    – shabbychef
    Jul 18, 2011 at 20:47

1 Answer 1


OK, I'll post it as an answer. Since, by definition of $L$ and $\Sigma$, $\sum_i L_{i,j}^2 = \Sigma_{j,j} = 1,\,\,\forall j,$ we must have $|L_{i,j}| \le 1$.


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