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












0












$begingroup$


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 $a_{ij}=sum_{k=1}^n l_{ik}l_{jk}=sum_{kle min {i,j}} l_{ik}l_{jk}$. I don't know what to do next.



Please help.






linear-algebra matrices positive-definite







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  • $begingroup$
    Connected : math.stackexchange.com/q/1804322
    $endgroup$
    – Jean Marie
    Jan 14 at 7:07


















0












$begingroup$


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 $a_{ij}=sum_{k=1}^n l_{ik}l_{jk}=sum_{kle min {i,j}} l_{ik}l_{jk}$. I don't know what to do next.



Please help.






linear-algebra matrices positive-definite







share|cite|improve this question











$endgroup$












  • $begingroup$
    Connected : math.stackexchange.com/q/1804322
    $endgroup$
    – Jean Marie
    Jan 14 at 7:07
















0












0








0





$begingroup$


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 $a_{ij}=sum_{k=1}^n l_{ik}l_{jk}=sum_{kle min {i,j}} l_{ik}l_{jk}$. I don't know what to do next.



Please help.






linear-algebra matrices positive-definite







share|cite|improve this question











$endgroup$




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 $a_{ij}=sum_{k=1}^n l_{ik}l_{jk}=sum_{kle min {i,j}} l_{ik}l_{jk}$. I don't know what to do next.



Please help.






linear-algebra matrices positive-definite




linear-algebra matrices positive-definite






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share|cite|improve this question













share|cite|improve this question




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edited Jan 12 at 5:50







user521337

















asked Jan 12 at 5:36









user521337user521337

1,2201417




1,2201417












  • $begingroup$
    Connected : math.stackexchange.com/q/1804322
    $endgroup$
    – Jean Marie
    Jan 14 at 7:07




















  • $begingroup$
    Connected : math.stackexchange.com/q/1804322
    $endgroup$
    – Jean Marie
    Jan 14 at 7:07


















$begingroup$
Connected : math.stackexchange.com/q/1804322
$endgroup$
– Jean Marie
Jan 14 at 7:07






$begingroup$
Connected : math.stackexchange.com/q/1804322
$endgroup$
– Jean Marie
Jan 14 at 7:07












2 Answers
2






active

oldest

votes


















2












$begingroup$

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 $kneq i, kneq j$, 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, $a_{ij} leq 0 Rightarrow l_{ij} leq 0$.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    @JeanMarie $a_{31} not leq 0$
    $endgroup$
    – jgb
    Jan 20 at 15:39










  • $begingroup$
    [+1] Probably, there is a way to express it differently but your reasoning is perfectly exact.
    $endgroup$
    – Jean Marie
    Jan 20 at 17:36










  • $begingroup$
    I noticed I had misinterpreted your implication as separated from the others.
    $endgroup$
    – Jean Marie
    Jan 20 at 17:37










  • $begingroup$
    @JeanMarie Thank you. Yes, it would have to be formulated stricter than that. But my intent was to give a direction, rather.
    $endgroup$
    – jgb
    Jan 20 at 18:27



















0












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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 $3times 3$ ... 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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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

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 $kneq i, kneq j$, 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, $a_{ij} leq 0 Rightarrow l_{ij} leq 0$.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    @JeanMarie $a_{31} not leq 0$
    $endgroup$
    – jgb
    Jan 20 at 15:39










  • $begingroup$
    [+1] Probably, there is a way to express it differently but your reasoning is perfectly exact.
    $endgroup$
    – Jean Marie
    Jan 20 at 17:36










  • $begingroup$
    I noticed I had misinterpreted your implication as separated from the others.
    $endgroup$
    – Jean Marie
    Jan 20 at 17:37










  • $begingroup$
    @JeanMarie Thank you. Yes, it would have to be formulated stricter than that. But my intent was to give a direction, rather.
    $endgroup$
    – jgb
    Jan 20 at 18:27
















2












$begingroup$

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 $kneq i, kneq j$, 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, $a_{ij} leq 0 Rightarrow l_{ij} leq 0$.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    @JeanMarie $a_{31} not leq 0$
    $endgroup$
    – jgb
    Jan 20 at 15:39










  • $begingroup$
    [+1] Probably, there is a way to express it differently but your reasoning is perfectly exact.
    $endgroup$
    – Jean Marie
    Jan 20 at 17:36










  • $begingroup$
    I noticed I had misinterpreted your implication as separated from the others.
    $endgroup$
    – Jean Marie
    Jan 20 at 17:37










  • $begingroup$
    @JeanMarie Thank you. Yes, it would have to be formulated stricter than that. But my intent was to give a direction, rather.
    $endgroup$
    – jgb
    Jan 20 at 18:27














2












2








2





$begingroup$

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 $kneq i, kneq j$, 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, $a_{ij} leq 0 Rightarrow l_{ij} leq 0$.






share|cite|improve this answer









$endgroup$



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 $kneq i, kneq j$, 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, $a_{ij} leq 0 Rightarrow l_{ij} leq 0$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 12 at 19:38









jgbjgb

14813




14813








  • 1




    $begingroup$
    @JeanMarie $a_{31} not leq 0$
    $endgroup$
    – jgb
    Jan 20 at 15:39










  • $begingroup$
    [+1] Probably, there is a way to express it differently but your reasoning is perfectly exact.
    $endgroup$
    – Jean Marie
    Jan 20 at 17:36










  • $begingroup$
    I noticed I had misinterpreted your implication as separated from the others.
    $endgroup$
    – Jean Marie
    Jan 20 at 17:37










  • $begingroup$
    @JeanMarie Thank you. Yes, it would have to be formulated stricter than that. But my intent was to give a direction, rather.
    $endgroup$
    – jgb
    Jan 20 at 18:27














  • 1




    $begingroup$
    @JeanMarie $a_{31} not leq 0$
    $endgroup$
    – jgb
    Jan 20 at 15:39










  • $begingroup$
    [+1] Probably, there is a way to express it differently but your reasoning is perfectly exact.
    $endgroup$
    – Jean Marie
    Jan 20 at 17:36










  • $begingroup$
    I noticed I had misinterpreted your implication as separated from the others.
    $endgroup$
    – Jean Marie
    Jan 20 at 17:37










  • $begingroup$
    @JeanMarie Thank you. Yes, it would have to be formulated stricter than that. But my intent was to give a direction, rather.
    $endgroup$
    – jgb
    Jan 20 at 18:27








1




1




$begingroup$
@JeanMarie $a_{31} not leq 0$
$endgroup$
– jgb
Jan 20 at 15:39




$begingroup$
@JeanMarie $a_{31} not leq 0$
$endgroup$
– jgb
Jan 20 at 15:39












$begingroup$
[+1] Probably, there is a way to express it differently but your reasoning is perfectly exact.
$endgroup$
– Jean Marie
Jan 20 at 17:36




$begingroup$
[+1] Probably, there is a way to express it differently but your reasoning is perfectly exact.
$endgroup$
– Jean Marie
Jan 20 at 17:36












$begingroup$
I noticed I had misinterpreted your implication as separated from the others.
$endgroup$
– Jean Marie
Jan 20 at 17:37




$begingroup$
I noticed I had misinterpreted your implication as separated from the others.
$endgroup$
– Jean Marie
Jan 20 at 17:37












$begingroup$
@JeanMarie Thank you. Yes, it would have to be formulated stricter than that. But my intent was to give a direction, rather.
$endgroup$
– jgb
Jan 20 at 18:27




$begingroup$
@JeanMarie Thank you. Yes, it would have to be formulated stricter than that. But my intent was to give a direction, rather.
$endgroup$
– jgb
Jan 20 at 18:27











0












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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 $3times 3$ ... 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
















0












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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 $3times 3$ ... 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














0












0








0





$begingroup$

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.






share|cite|improve this answer











$endgroup$



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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 12 at 11:27

























answered Jan 12 at 10:33









Jean MarieJean Marie

31.5k42355




31.5k42355












  • $begingroup$
    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 $3times 3$ ... 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


















  • $begingroup$
    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 $3times 3$ ... 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
















$begingroup$
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




$begingroup$
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




$begingroup$
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 $3times 3$ ... then $a_{32}=l_{31}l_{21}+l_{32}l_{22}$ ... how does your argument work now ...?
$endgroup$
– user521337
Jan 12 at 11:00




$begingroup$
suppose we're dealing with $3times 3$ ... 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 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




$begingroup$
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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