Let $L \in GL_n (\mathbb R)$ be a lower triangular matrix with positive diaginal entries and let $A :=LL^t$ . (note that $A$ is positive definite i.e. $A$ is symmetric and all eigenvalues of $A$ are positive).

Let $A=[a_{ij}] $ and $L=[l_{ij}]$ . If $a_{ij}\le 0, \forall i>j$, then how to prove that $l_{ij} \le 0, \forall i >j$ ?

My work: Writing down the product, we have $a_{ij}=\sum_{k=1}^n l_{ik}l_{jk}=\sum_{k\le \min \{i,j\}} l_{ik}l_{jk}$. I don't know what to do next.

Please help.


Since $A$ is symmetric, consider only the case $i>j$. You already know

$$a_{ij} = \sum_{k=1}^{j} l_{ik}l_{jk}$$

Go row by row. Start from the left, and use $a_{ij} \leq 0$ as well as $l_{ii} > 0$ as you proceed down and towards the main diagonal.

  1. $0 \geq a_{21} = l_{21}l_{11} \Rightarrow l_{21} \leq 0$, since $l_{11} > 0$
  2. $0 \geq a_{31} = l_{31}l_{11} \Rightarrow l_{31} \leq 0$, since $l_{11} > 0$
  3. $0 \geq a_{32} = l_{31}l_{21} + l_{32}l_{22} \Rightarrow l_{32} \leq 0$, since $l_{21} \leq 0$, $l_{31} \leq 0$ and $l_{22} > 0$

You should see a pattern emerge, that $a_{ij}$ is the sum of products $l_{ik}l_{jk}$ with $k\neq i, k\neq j$, which are products of negative numbers or zero (using previous results). Only the last term $l_{ij}l_{jj}$ can be negative. With $l_{jj}$ being positive, $a_{ij} \leq 0 \Rightarrow l_{ij} \leq 0$.

  • 1
    $\begingroup$ @JeanMarie $a_{31} \not \leq 0$ $\endgroup$ – jgb 2 days ago
  • $\begingroup$ [+1] Probably, there is a way to express it differently but your reasoning is perfectly exact. $\endgroup$ – Jean Marie 2 days ago
  • $\begingroup$ I noticed I had misinterpreted your implication as separated from the others. $\endgroup$ – Jean Marie 2 days ago
  • $\begingroup$ @JeanMarie Thank you. Yes, it would have to be formulated stricter than that. But my intent was to give a direction, rather. $\endgroup$ – jgb yesterday

I have a proof for the case $n=2$ ; if


we have in particular

$$l_{21}l_{11}=a_{21} \tag{1}$$

As $l_{11}>0$, (1) gives :

$$l_{21} \leq 0 \ \iff \ a_{21} \leq 0 \tag{2}$$

which allows to conclude.

This proof, after discussion with the OP, does not extend to more general cases.

  • $\begingroup$ but in the general case, the sum expression for $a_{ij}$ would contain many more elements ... I don't see how a similar argument proves that ... $\endgroup$ – user521337 Jan 12 at 10:47
  • $\begingroup$ I don't see why you don't agree with the re-indexing argument : the important fact is that the sign of $l_{ij}$ is defined by the sign of $a_{ij}$ exclusively. $\endgroup$ – Jean Marie Jan 12 at 10:51
  • $\begingroup$ suppose we're dealing with $3\times 3$ ... then $a_{32}=l_{31}l_{21}+l_{32}l_{22}$ ... how does your argument work now ...? $\endgroup$ – user521337 Jan 12 at 11:00
  • $\begingroup$ I am shaked in my argumentation, especialy by the fact that I don't use the "all-negative" property of $A$'s off-diagonal elements. $\endgroup$ – Jean Marie Jan 12 at 11:22
  • $\begingroup$ I have understood why I couldn't use the argument thinking to the associated quadratic form. Sorry for the inconvenience. I just modified my answer $\endgroup$ – Jean Marie Jan 12 at 11:29

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.