Why is the eigenvector of the smallest eigenvalue of this matrix gives the intersection of a set of lines in...
$begingroup$
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.
linear-algebra geometry analytic-geometry projective-geometry
asked Dec 14 '18 at 9:25
stressed outstressed out
5,4381638
$endgroup$
add a comment |
$begingroup$
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.
linear-algebra geometry analytic-geometry projective-geometry
asked Dec 14 '18 at 9:25
stressed outstressed out
5,4381638
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
linear-algebra geometry analytic-geometry projective-geometry
asked Dec 14 '18 at 9:25
stressed outstressed out
5,4381638
$endgroup$
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
linear-algebra geometry analytic-geometry projective-geometry
asked Dec 14 '18 at 9:25
stressed outstressed out
5,4381638
asked Dec 14 '18 at 9:25
stressed outstressed out
5,4381638
edited Dec 14 '18 at 9:32
stressed out
asked Dec 14 '18 at 9:25
stressed outstressed out
5,4381638
asked Dec 14 '18 at 9:25
stressed outstressed out
5,4381638
asked Dec 14 '18 at 9:25
stressed outstressed out
5,4381638
5,4381638
$begingroup$
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
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$begingroup$
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.
answered Dec 14 '18 at 17:21
AretinoAretino
23.7k21443
$endgroup$
$begingroup$
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
|
show 3 more comments
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$begingroup$
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.
answered Dec 14 '18 at 17:21
AretinoAretino
23.7k21443
$endgroup$
$begingroup$
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
|
show 3 more comments
$begingroup$
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.
answered Dec 14 '18 at 17:21
AretinoAretino
23.7k21443
$endgroup$
$begingroup$
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
|
show 3 more comments
$begingroup$
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.
answered Dec 14 '18 at 17:21
AretinoAretino
23.7k21443
$endgroup$
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.
answered Dec 14 '18 at 17:21
AretinoAretino
23.7k21443
edited Dec 14 '18 at 18:39
answered Dec 14 '18 at 17:21
AretinoAretino
23.7k21443
answered Dec 14 '18 at 17:21
AretinoAretino
23.7k21443
answered Dec 14 '18 at 17:21
AretinoAretino
23.7k21443
23.7k21443
$begingroup$
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
|
show 3 more comments
$begingroup$
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
|
show 3 more comments
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$begingroup$
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