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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.

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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, $a_{ij} \leq 0 \Rightarrow l_{ij} \leq 0$.

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

$$\underbrace{\begin{pmatrix}l_{11}&0\\l_{21}&l_{22}\end{pmatrix}}_{L}\underbrace{\begin{pmatrix}l_{11}&l_{21}\\0&l_{22}\end{pmatrix}}_{L^T}=\underbrace{\begin{pmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{pmatrix}}_{A},$$

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.

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  • $\begingroup$ 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

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