# On the entries of $LL^t$ where $L \in GL_n (\mathbb R)$ is lower triangular with positive diagonal entries

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.

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

• @JeanMarie $a_{31} \not \leq 0$ – jgb Jan 20 at 15:39
• [+1] Probably, there is a way to express it differently but your reasoning is perfectly exact. – Jean Marie Jan 20 at 17:36
• I noticed I had misinterpreted your implication as separated from the others. – Jean Marie Jan 20 at 17:37
• @JeanMarie Thank you. Yes, it would have to be formulated stricter than that. But my intent was to give a direction, rather. – jgb Jan 20 at 18:27

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.

• 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 ... – user521337 Jan 12 at 10:47
• 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. – Jean Marie Jan 12 at 10:51
• 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 ...? – user521337 Jan 12 at 11:00
• 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. – Jean Marie Jan 12 at 11:22
• 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 – Jean Marie Jan 12 at 11:29