Why is the eigenvector of the smallest eigenvalue of this matrix gives the intersection of a set of lines in...












2












$begingroup$


Consider $L = {l_1 cdots l_n}$ be a set of lines in the plane in homogeneous coordinates. In this definition, a line $ax+by+c = 0$ is given by its direction $l^T = (a/c,b/c,1)^T$ such that it consists of the points where $l^T(x,y,1)^T=0$. Now suppose that I want to find the intersection of lines in $L$, assuming that it exists. Otherwise, I want to find the point given by the least square solution had I written all this as a system of linear equations.



I have found a PDF that claims the eigenvector of the smallest eigenvalue of this matrix is the answer:



$$M = sum_i l_i l_i^T$$



I wonder why this is true. In case anyone wonders where I have found this, here's the link: click here.










share|cite|improve this question











$endgroup$












  • $begingroup$
    I don’t see where those slides claim that this gives you the least-squares solution to the intersection problem. Also, there are better estimates of vanishing points than the point that minimizes the sum of the squared distances to the lines. See the end of section 8.6 in Hartley & Zisserman for a brief discussion.
    $endgroup$
    – amd
    Dec 14 '18 at 21:04
















2












$begingroup$


Consider $L = {l_1 cdots l_n}$ be a set of lines in the plane in homogeneous coordinates. In this definition, a line $ax+by+c = 0$ is given by its direction $l^T = (a/c,b/c,1)^T$ such that it consists of the points where $l^T(x,y,1)^T=0$. Now suppose that I want to find the intersection of lines in $L$, assuming that it exists. Otherwise, I want to find the point given by the least square solution had I written all this as a system of linear equations.



I have found a PDF that claims the eigenvector of the smallest eigenvalue of this matrix is the answer:



$$M = sum_i l_i l_i^T$$



I wonder why this is true. In case anyone wonders where I have found this, here's the link: click here.










share|cite|improve this question











$endgroup$












  • $begingroup$
    I don’t see where those slides claim that this gives you the least-squares solution to the intersection problem. Also, there are better estimates of vanishing points than the point that minimizes the sum of the squared distances to the lines. See the end of section 8.6 in Hartley & Zisserman for a brief discussion.
    $endgroup$
    – amd
    Dec 14 '18 at 21:04














2












2








2


0



$begingroup$


Consider $L = {l_1 cdots l_n}$ be a set of lines in the plane in homogeneous coordinates. In this definition, a line $ax+by+c = 0$ is given by its direction $l^T = (a/c,b/c,1)^T$ such that it consists of the points where $l^T(x,y,1)^T=0$. Now suppose that I want to find the intersection of lines in $L$, assuming that it exists. Otherwise, I want to find the point given by the least square solution had I written all this as a system of linear equations.



I have found a PDF that claims the eigenvector of the smallest eigenvalue of this matrix is the answer:



$$M = sum_i l_i l_i^T$$



I wonder why this is true. In case anyone wonders where I have found this, here's the link: click here.










share|cite|improve this question











$endgroup$




Consider $L = {l_1 cdots l_n}$ be a set of lines in the plane in homogeneous coordinates. In this definition, a line $ax+by+c = 0$ is given by its direction $l^T = (a/c,b/c,1)^T$ such that it consists of the points where $l^T(x,y,1)^T=0$. Now suppose that I want to find the intersection of lines in $L$, assuming that it exists. Otherwise, I want to find the point given by the least square solution had I written all this as a system of linear equations.



I have found a PDF that claims the eigenvector of the smallest eigenvalue of this matrix is the answer:



$$M = sum_i l_i l_i^T$$



I wonder why this is true. In case anyone wonders where I have found this, here's the link: click here.







linear-algebra geometry analytic-geometry projective-geometry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 14 '18 at 9:32







stressed out

















asked Dec 14 '18 at 9:25









stressed outstressed out

6,3601839




6,3601839












  • $begingroup$
    I don’t see where those slides claim that this gives you the least-squares solution to the intersection problem. Also, there are better estimates of vanishing points than the point that minimizes the sum of the squared distances to the lines. See the end of section 8.6 in Hartley & Zisserman for a brief discussion.
    $endgroup$
    – amd
    Dec 14 '18 at 21:04


















  • $begingroup$
    I don’t see where those slides claim that this gives you the least-squares solution to the intersection problem. Also, there are better estimates of vanishing points than the point that minimizes the sum of the squared distances to the lines. See the end of section 8.6 in Hartley & Zisserman for a brief discussion.
    $endgroup$
    – amd
    Dec 14 '18 at 21:04
















$begingroup$
I don’t see where those slides claim that this gives you the least-squares solution to the intersection problem. Also, there are better estimates of vanishing points than the point that minimizes the sum of the squared distances to the lines. See the end of section 8.6 in Hartley & Zisserman for a brief discussion.
$endgroup$
– amd
Dec 14 '18 at 21:04




$begingroup$
I don’t see where those slides claim that this gives you the least-squares solution to the intersection problem. Also, there are better estimates of vanishing points than the point that minimizes the sum of the squared distances to the lines. See the end of section 8.6 in Hartley & Zisserman for a brief discussion.
$endgroup$
– amd
Dec 14 '18 at 21:04










1 Answer
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active

oldest

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1












$begingroup$

If $v=(x,y,1)^T$ is the intersection of all lines, then $l_i^Tv=0$ for every $i$, i.e. $l_il_i^Tv=vec 0$ for every $i$. It follows that $Mv=vec 0$, i.e. $v$ is an eigenvector of $M$ with zero eigenvalue.



If the known coefficient of the lines are only approximations of their real values, then the eigenvalue is not zero, but very near to it (of course we are supposing the approximations to be very good). Hence the claim is true, provided we specify "the smallest eigenvalue in modulus".



Very likely, this is also the point given by the least square solution, in case the lines don't coverge: it shouldn't be difficult to carry out the computation and verify the claim.



EDIT.



I must retract my last sentence above: in general, the eigenvector with smallest eigenvalue does not yield the least square solution.



Here's a counterexample: lets take three non-coverging lines $l_1=(1,-1,1)$, $l_2=(-1,-1,1)$, $l_3=(0,-2,1)$. If $v=(x,y,1)^T$ and $l_iv=a_i$, then to find the least square $v$ we have to minimize $S=sum_i a_i^2$. But:
$$
{partial Soverpartial x}=2sum_i a_i{partial a_ioverpartial x}=
2sum_i v^Tl_i^Tl_i{partial voverpartial x}=2v^TMpmatrix{1\0\0}
quadtext{and}quad
{partial Soverpartial y}=2v^TMpmatrix{0\1\0}.
$$

You can verify that imposing $partial S/partial x=0$ and
$partial S/partial y=0$ gives as solution $v=(0,2/3,1)$.



On the other hand, the smallest eigenvalue of $M$ is $(9-sqrt{73})/2$, corresponding to the eigenvector
$v=(0,(sqrt{73}-3)/8,1)$, which is different from the solution found above.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks for the answer (+1), but my problem is actually the last paragraph of your answer. No offense, but I already knew the other paragraphs.
    $endgroup$
    – stressed out
    Dec 14 '18 at 17:51












  • $begingroup$
    The slides you linked never mention the least square solution: is that a conjecture or do you have some other source?
    $endgroup$
    – Aretino
    Dec 14 '18 at 18:08










  • $begingroup$
    Yes, you're correct. It never mentions it. No, I don't have another source unfortunately. So, yes. It's a conjecture, but it's the most reasonable interpretation of what it claims. In case it means something else, I still would like to see a proof.
    $endgroup$
    – stressed out
    Dec 14 '18 at 18:35








  • 1




    $begingroup$
    See my edited answer: the conjecture is false.
    $endgroup$
    – Aretino
    Dec 14 '18 at 18:39










  • $begingroup$
    Beautiful counter-example. I wish I could upvote you twice. Anyway, I still feel kind of uncomfortable with the method above. I don't know what kind of solution it gives when the lines don't have an intersection. I have no intuition about it. Do you have any intuition about what it does? Besides just being a mere approximation.
    $endgroup$
    – stressed out
    Dec 14 '18 at 18:45













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1












$begingroup$

If $v=(x,y,1)^T$ is the intersection of all lines, then $l_i^Tv=0$ for every $i$, i.e. $l_il_i^Tv=vec 0$ for every $i$. It follows that $Mv=vec 0$, i.e. $v$ is an eigenvector of $M$ with zero eigenvalue.



If the known coefficient of the lines are only approximations of their real values, then the eigenvalue is not zero, but very near to it (of course we are supposing the approximations to be very good). Hence the claim is true, provided we specify "the smallest eigenvalue in modulus".



Very likely, this is also the point given by the least square solution, in case the lines don't coverge: it shouldn't be difficult to carry out the computation and verify the claim.



EDIT.



I must retract my last sentence above: in general, the eigenvector with smallest eigenvalue does not yield the least square solution.



Here's a counterexample: lets take three non-coverging lines $l_1=(1,-1,1)$, $l_2=(-1,-1,1)$, $l_3=(0,-2,1)$. If $v=(x,y,1)^T$ and $l_iv=a_i$, then to find the least square $v$ we have to minimize $S=sum_i a_i^2$. But:
$$
{partial Soverpartial x}=2sum_i a_i{partial a_ioverpartial x}=
2sum_i v^Tl_i^Tl_i{partial voverpartial x}=2v^TMpmatrix{1\0\0}
quadtext{and}quad
{partial Soverpartial y}=2v^TMpmatrix{0\1\0}.
$$

You can verify that imposing $partial S/partial x=0$ and
$partial S/partial y=0$ gives as solution $v=(0,2/3,1)$.



On the other hand, the smallest eigenvalue of $M$ is $(9-sqrt{73})/2$, corresponding to the eigenvector
$v=(0,(sqrt{73}-3)/8,1)$, which is different from the solution found above.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks for the answer (+1), but my problem is actually the last paragraph of your answer. No offense, but I already knew the other paragraphs.
    $endgroup$
    – stressed out
    Dec 14 '18 at 17:51












  • $begingroup$
    The slides you linked never mention the least square solution: is that a conjecture or do you have some other source?
    $endgroup$
    – Aretino
    Dec 14 '18 at 18:08










  • $begingroup$
    Yes, you're correct. It never mentions it. No, I don't have another source unfortunately. So, yes. It's a conjecture, but it's the most reasonable interpretation of what it claims. In case it means something else, I still would like to see a proof.
    $endgroup$
    – stressed out
    Dec 14 '18 at 18:35








  • 1




    $begingroup$
    See my edited answer: the conjecture is false.
    $endgroup$
    – Aretino
    Dec 14 '18 at 18:39










  • $begingroup$
    Beautiful counter-example. I wish I could upvote you twice. Anyway, I still feel kind of uncomfortable with the method above. I don't know what kind of solution it gives when the lines don't have an intersection. I have no intuition about it. Do you have any intuition about what it does? Besides just being a mere approximation.
    $endgroup$
    – stressed out
    Dec 14 '18 at 18:45


















1












$begingroup$

If $v=(x,y,1)^T$ is the intersection of all lines, then $l_i^Tv=0$ for every $i$, i.e. $l_il_i^Tv=vec 0$ for every $i$. It follows that $Mv=vec 0$, i.e. $v$ is an eigenvector of $M$ with zero eigenvalue.



If the known coefficient of the lines are only approximations of their real values, then the eigenvalue is not zero, but very near to it (of course we are supposing the approximations to be very good). Hence the claim is true, provided we specify "the smallest eigenvalue in modulus".



Very likely, this is also the point given by the least square solution, in case the lines don't coverge: it shouldn't be difficult to carry out the computation and verify the claim.



EDIT.



I must retract my last sentence above: in general, the eigenvector with smallest eigenvalue does not yield the least square solution.



Here's a counterexample: lets take three non-coverging lines $l_1=(1,-1,1)$, $l_2=(-1,-1,1)$, $l_3=(0,-2,1)$. If $v=(x,y,1)^T$ and $l_iv=a_i$, then to find the least square $v$ we have to minimize $S=sum_i a_i^2$. But:
$$
{partial Soverpartial x}=2sum_i a_i{partial a_ioverpartial x}=
2sum_i v^Tl_i^Tl_i{partial voverpartial x}=2v^TMpmatrix{1\0\0}
quadtext{and}quad
{partial Soverpartial y}=2v^TMpmatrix{0\1\0}.
$$

You can verify that imposing $partial S/partial x=0$ and
$partial S/partial y=0$ gives as solution $v=(0,2/3,1)$.



On the other hand, the smallest eigenvalue of $M$ is $(9-sqrt{73})/2$, corresponding to the eigenvector
$v=(0,(sqrt{73}-3)/8,1)$, which is different from the solution found above.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks for the answer (+1), but my problem is actually the last paragraph of your answer. No offense, but I already knew the other paragraphs.
    $endgroup$
    – stressed out
    Dec 14 '18 at 17:51












  • $begingroup$
    The slides you linked never mention the least square solution: is that a conjecture or do you have some other source?
    $endgroup$
    – Aretino
    Dec 14 '18 at 18:08










  • $begingroup$
    Yes, you're correct. It never mentions it. No, I don't have another source unfortunately. So, yes. It's a conjecture, but it's the most reasonable interpretation of what it claims. In case it means something else, I still would like to see a proof.
    $endgroup$
    – stressed out
    Dec 14 '18 at 18:35








  • 1




    $begingroup$
    See my edited answer: the conjecture is false.
    $endgroup$
    – Aretino
    Dec 14 '18 at 18:39










  • $begingroup$
    Beautiful counter-example. I wish I could upvote you twice. Anyway, I still feel kind of uncomfortable with the method above. I don't know what kind of solution it gives when the lines don't have an intersection. I have no intuition about it. Do you have any intuition about what it does? Besides just being a mere approximation.
    $endgroup$
    – stressed out
    Dec 14 '18 at 18:45
















1












1








1





$begingroup$

If $v=(x,y,1)^T$ is the intersection of all lines, then $l_i^Tv=0$ for every $i$, i.e. $l_il_i^Tv=vec 0$ for every $i$. It follows that $Mv=vec 0$, i.e. $v$ is an eigenvector of $M$ with zero eigenvalue.



If the known coefficient of the lines are only approximations of their real values, then the eigenvalue is not zero, but very near to it (of course we are supposing the approximations to be very good). Hence the claim is true, provided we specify "the smallest eigenvalue in modulus".



Very likely, this is also the point given by the least square solution, in case the lines don't coverge: it shouldn't be difficult to carry out the computation and verify the claim.



EDIT.



I must retract my last sentence above: in general, the eigenvector with smallest eigenvalue does not yield the least square solution.



Here's a counterexample: lets take three non-coverging lines $l_1=(1,-1,1)$, $l_2=(-1,-1,1)$, $l_3=(0,-2,1)$. If $v=(x,y,1)^T$ and $l_iv=a_i$, then to find the least square $v$ we have to minimize $S=sum_i a_i^2$. But:
$$
{partial Soverpartial x}=2sum_i a_i{partial a_ioverpartial x}=
2sum_i v^Tl_i^Tl_i{partial voverpartial x}=2v^TMpmatrix{1\0\0}
quadtext{and}quad
{partial Soverpartial y}=2v^TMpmatrix{0\1\0}.
$$

You can verify that imposing $partial S/partial x=0$ and
$partial S/partial y=0$ gives as solution $v=(0,2/3,1)$.



On the other hand, the smallest eigenvalue of $M$ is $(9-sqrt{73})/2$, corresponding to the eigenvector
$v=(0,(sqrt{73}-3)/8,1)$, which is different from the solution found above.






share|cite|improve this answer











$endgroup$



If $v=(x,y,1)^T$ is the intersection of all lines, then $l_i^Tv=0$ for every $i$, i.e. $l_il_i^Tv=vec 0$ for every $i$. It follows that $Mv=vec 0$, i.e. $v$ is an eigenvector of $M$ with zero eigenvalue.



If the known coefficient of the lines are only approximations of their real values, then the eigenvalue is not zero, but very near to it (of course we are supposing the approximations to be very good). Hence the claim is true, provided we specify "the smallest eigenvalue in modulus".



Very likely, this is also the point given by the least square solution, in case the lines don't coverge: it shouldn't be difficult to carry out the computation and verify the claim.



EDIT.



I must retract my last sentence above: in general, the eigenvector with smallest eigenvalue does not yield the least square solution.



Here's a counterexample: lets take three non-coverging lines $l_1=(1,-1,1)$, $l_2=(-1,-1,1)$, $l_3=(0,-2,1)$. If $v=(x,y,1)^T$ and $l_iv=a_i$, then to find the least square $v$ we have to minimize $S=sum_i a_i^2$. But:
$$
{partial Soverpartial x}=2sum_i a_i{partial a_ioverpartial x}=
2sum_i v^Tl_i^Tl_i{partial voverpartial x}=2v^TMpmatrix{1\0\0}
quadtext{and}quad
{partial Soverpartial y}=2v^TMpmatrix{0\1\0}.
$$

You can verify that imposing $partial S/partial x=0$ and
$partial S/partial y=0$ gives as solution $v=(0,2/3,1)$.



On the other hand, the smallest eigenvalue of $M$ is $(9-sqrt{73})/2$, corresponding to the eigenvector
$v=(0,(sqrt{73}-3)/8,1)$, which is different from the solution found above.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 14 '18 at 18:39

























answered Dec 14 '18 at 17:21









AretinoAretino

24.7k21444




24.7k21444












  • $begingroup$
    Thanks for the answer (+1), but my problem is actually the last paragraph of your answer. No offense, but I already knew the other paragraphs.
    $endgroup$
    – stressed out
    Dec 14 '18 at 17:51












  • $begingroup$
    The slides you linked never mention the least square solution: is that a conjecture or do you have some other source?
    $endgroup$
    – Aretino
    Dec 14 '18 at 18:08










  • $begingroup$
    Yes, you're correct. It never mentions it. No, I don't have another source unfortunately. So, yes. It's a conjecture, but it's the most reasonable interpretation of what it claims. In case it means something else, I still would like to see a proof.
    $endgroup$
    – stressed out
    Dec 14 '18 at 18:35








  • 1




    $begingroup$
    See my edited answer: the conjecture is false.
    $endgroup$
    – Aretino
    Dec 14 '18 at 18:39










  • $begingroup$
    Beautiful counter-example. I wish I could upvote you twice. Anyway, I still feel kind of uncomfortable with the method above. I don't know what kind of solution it gives when the lines don't have an intersection. I have no intuition about it. Do you have any intuition about what it does? Besides just being a mere approximation.
    $endgroup$
    – stressed out
    Dec 14 '18 at 18:45




















  • $begingroup$
    Thanks for the answer (+1), but my problem is actually the last paragraph of your answer. No offense, but I already knew the other paragraphs.
    $endgroup$
    – stressed out
    Dec 14 '18 at 17:51












  • $begingroup$
    The slides you linked never mention the least square solution: is that a conjecture or do you have some other source?
    $endgroup$
    – Aretino
    Dec 14 '18 at 18:08










  • $begingroup$
    Yes, you're correct. It never mentions it. No, I don't have another source unfortunately. So, yes. It's a conjecture, but it's the most reasonable interpretation of what it claims. In case it means something else, I still would like to see a proof.
    $endgroup$
    – stressed out
    Dec 14 '18 at 18:35








  • 1




    $begingroup$
    See my edited answer: the conjecture is false.
    $endgroup$
    – Aretino
    Dec 14 '18 at 18:39










  • $begingroup$
    Beautiful counter-example. I wish I could upvote you twice. Anyway, I still feel kind of uncomfortable with the method above. I don't know what kind of solution it gives when the lines don't have an intersection. I have no intuition about it. Do you have any intuition about what it does? Besides just being a mere approximation.
    $endgroup$
    – stressed out
    Dec 14 '18 at 18:45


















$begingroup$
Thanks for the answer (+1), but my problem is actually the last paragraph of your answer. No offense, but I already knew the other paragraphs.
$endgroup$
– stressed out
Dec 14 '18 at 17:51






$begingroup$
Thanks for the answer (+1), but my problem is actually the last paragraph of your answer. No offense, but I already knew the other paragraphs.
$endgroup$
– stressed out
Dec 14 '18 at 17:51














$begingroup$
The slides you linked never mention the least square solution: is that a conjecture or do you have some other source?
$endgroup$
– Aretino
Dec 14 '18 at 18:08




$begingroup$
The slides you linked never mention the least square solution: is that a conjecture or do you have some other source?
$endgroup$
– Aretino
Dec 14 '18 at 18:08












$begingroup$
Yes, you're correct. It never mentions it. No, I don't have another source unfortunately. So, yes. It's a conjecture, but it's the most reasonable interpretation of what it claims. In case it means something else, I still would like to see a proof.
$endgroup$
– stressed out
Dec 14 '18 at 18:35






$begingroup$
Yes, you're correct. It never mentions it. No, I don't have another source unfortunately. So, yes. It's a conjecture, but it's the most reasonable interpretation of what it claims. In case it means something else, I still would like to see a proof.
$endgroup$
– stressed out
Dec 14 '18 at 18:35






1




1




$begingroup$
See my edited answer: the conjecture is false.
$endgroup$
– Aretino
Dec 14 '18 at 18:39




$begingroup$
See my edited answer: the conjecture is false.
$endgroup$
– Aretino
Dec 14 '18 at 18:39












$begingroup$
Beautiful counter-example. I wish I could upvote you twice. Anyway, I still feel kind of uncomfortable with the method above. I don't know what kind of solution it gives when the lines don't have an intersection. I have no intuition about it. Do you have any intuition about what it does? Besides just being a mere approximation.
$endgroup$
– stressed out
Dec 14 '18 at 18:45






$begingroup$
Beautiful counter-example. I wish I could upvote you twice. Anyway, I still feel kind of uncomfortable with the method above. I don't know what kind of solution it gives when the lines don't have an intersection. I have no intuition about it. Do you have any intuition about what it does? Besides just being a mere approximation.
$endgroup$
– stressed out
Dec 14 '18 at 18:45




















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