Determining the matrix of a projection
$begingroup$
I have a question about the correctness of my ideas regarding the following exercise.
Define
$A_0=[(1,0,0,0)]$
$B_0=[(0,1,0,0)]$
$A_1=[(0,0,1,0)]$
$B_1=[(0,0,0,1)] in mathbb RP^3$ where $mathbb RP^3$ denotes real projective space.
Let $l_0$ be the line through $A_0$ and $B_0$ and $l_1$ be the line through $A_1$ and $B_1$. Let $l_2$ be an arbitrary line skew to both $l_0$ and $l_1$ and let $A_2=[a]$ and $B_2=[b]$ be two points of $l_2$ so that $P=[s cdot a+ t cdot b] $ is a general point of $l_2$. Let $pi$ be the plane spanned by $l_0$ and P and define Q to be the intersection of $pi$ with $l_1$. Let f be the map $f: l_2 to l_1$ defined by $f(P)=Q.$ Find a matrix in $PGL(3,mathbb R)$ corresponding to f.
My idea to find the matrix was this:
We know that the representation vector $x$ of point $X$ in $pi$ can be written as
$$x=s_0 cdot (1,0,0,0)+t_0 cdot (0,1,0,0)+k_0(sa+tb)tag1$$
for $s_0,t_0,k_0 in mathbb R$.
The respresentation vector $y$ of a point $Y$ on $l_1$ can be written as
$$y=s_1 cdot (0,0,1,0) + t_1 cdot (0,0,0,1)tag2$$
for $s_1, t_1 in mathbb R$.
Since the line $l_2$ is determined by the points $A_2$ and $B_2$ it is sufficient to find a matrix $A=(a_{ij})$ that maps the representation vectors $a$ and $b$ onto representation vectors of the points $(pi A_2)cap l_1$ and $(pi B_2)cap l_1$. The representation vectors of the image points must satisfy the equations $(1)$ and $(2)$. So we get the following equations
$Aa=s_0' cdot (1,0,0,0)+t_0' cdot (0,1,0,0)+a$
$Aa=s_1' cdot (0,0,1,0) + t_1' cdot (0,0,0,1)$
$Ab=s_0'' cdot (1,0,0,0)+t_0'' cdot (0,1,0,0)+b$
$Ab=s_1'' cdot (0,0,1,0) + t_1'' cdot (0,0,0,1)$
Since the matrix $A$ has 16 entries we get 16 equations in total. Solving this system with respect to the entries of $A$ should give me the matrix I need.
My question is:
Is there an easier way to solve this than solving a system of 16 equations?
geometry projective-geometry
edited Jan 13 at 21:30
MvG
31.2k450107
asked Jan 11 at 19:16
PolymorphPolymorph
1206
$endgroup$
|
show 1 more comment
$begingroup$
I have a question about the correctness of my ideas regarding the following exercise.
Define
$A_0=[(1,0,0,0)]$
$B_0=[(0,1,0,0)]$
$A_1=[(0,0,1,0)]$
$B_1=[(0,0,0,1)] in mathbb RP^3$ where $mathbb RP^3$ denotes real projective space.
Let $l_0$ be the line through $A_0$ and $B_0$ and $l_1$ be the line through $A_1$ and $B_1$. Let $l_2$ be an arbitrary line skew to both $l_0$ and $l_1$ and let $A_2=[a]$ and $B_2=[b]$ be two points of $l_2$ so that $P=[s cdot a+ t cdot b] $ is a general point of $l_2$. Let $pi$ be the plane spanned by $l_0$ and P and define Q to be the intersection of $pi$ with $l_1$. Let f be the map $f: l_2 to l_1$ defined by $f(P)=Q.$ Find a matrix in $PGL(3,mathbb R)$ corresponding to f.
My idea to find the matrix was this:
We know that the representation vector $x$ of point $X$ in $pi$ can be written as
$$x=s_0 cdot (1,0,0,0)+t_0 cdot (0,1,0,0)+k_0(sa+tb)tag1$$
for $s_0,t_0,k_0 in mathbb R$.
The respresentation vector $y$ of a point $Y$ on $l_1$ can be written as
$$y=s_1 cdot (0,0,1,0) + t_1 cdot (0,0,0,1)tag2$$
for $s_1, t_1 in mathbb R$.
Since the line $l_2$ is determined by the points $A_2$ and $B_2$ it is sufficient to find a matrix $A=(a_{ij})$ that maps the representation vectors $a$ and $b$ onto representation vectors of the points $(pi A_2)cap l_1$ and $(pi B_2)cap l_1$. The representation vectors of the image points must satisfy the equations $(1)$ and $(2)$. So we get the following equations
$Aa=s_0' cdot (1,0,0,0)+t_0' cdot (0,1,0,0)+a$
$Aa=s_1' cdot (0,0,1,0) + t_1' cdot (0,0,0,1)$
$Ab=s_0'' cdot (1,0,0,0)+t_0'' cdot (0,1,0,0)+b$
$Ab=s_1'' cdot (0,0,1,0) + t_1'' cdot (0,0,0,1)$
Since the matrix $A$ has 16 entries we get 16 equations in total. Solving this system with respect to the entries of $A$ should give me the matrix I need.
My question is:
Is there an easier way to solve this than solving a system of 16 equations?
geometry projective-geometry
edited Jan 13 at 21:30
MvG
31.2k450107
asked Jan 11 at 19:16
PolymorphPolymorph
1206
$endgroup$
$begingroup$
$A_0$ and $B_0$ are identical. Did you mean to write $B_0=[(0,1,0,0)]$?
$endgroup$
– amd
Jan 11 at 21:07
$begingroup$
The wording of this problem is troublesome. “Let $f$ be the map $f:l_2to l_1$” defines $f$ as a homography of two lines (an element of $PGL(2,mathbb R)$), but asks for a matrix in $PGL(3,mathbb R)$. Moreover, $PGL(3)$ operates on the projective plane, but the problem is set in a higher-dimensional space.
$endgroup$
– amd
Jan 11 at 21:55
$begingroup$
I also have a quibble with the phrasing “the map.” Even if we restrict ourselves to a homography between two lines, a single point correspondence doesn’t determine a unique map.
$endgroup$
– amd
Jan 11 at 21:57
$begingroup$
I’m guessing that when you write $IR$, you actually mean $mathbb R$. Usemathbb Rto get this symbol.
$endgroup$
– amd
Jan 11 at 21:58
$begingroup$
I can imagine two ways to read the question. Either you want $Q=[s'cdot a'+t'cdot b']$, probably with $A_1=[a'], B_1=[b']$, and then express $s',t'$ in terms of $s,t$, which would lead to a $2times2$ matrix. Or you want $Q$ as an arbitrary point in space, projecting the whole space to $l_1$ as described. In that case $l_2$ is irrelevant, and you get a $4times4$ matrix of rank $2$. It would be non-invertible. Like amd I can't see $textrm{PGL}(3,mathbb R)$ playing a role, since it has equivalence classes of invertible $3times3$ matrices. Please double-check the question statement.
$endgroup$
– MvG
Jan 14 at 23:01
|
show 1 more comment
$begingroup$
I have a question about the correctness of my ideas regarding the following exercise.
Define
$A_0=[(1,0,0,0)]$
$B_0=[(0,1,0,0)]$
$A_1=[(0,0,1,0)]$
$B_1=[(0,0,0,1)] in mathbb RP^3$ where $mathbb RP^3$ denotes real projective space.
Let $l_0$ be the line through $A_0$ and $B_0$ and $l_1$ be the line through $A_1$ and $B_1$. Let $l_2$ be an arbitrary line skew to both $l_0$ and $l_1$ and let $A_2=[a]$ and $B_2=[b]$ be two points of $l_2$ so that $P=[s cdot a+ t cdot b] $ is a general point of $l_2$. Let $pi$ be the plane spanned by $l_0$ and P and define Q to be the intersection of $pi$ with $l_1$. Let f be the map $f: l_2 to l_1$ defined by $f(P)=Q.$ Find a matrix in $PGL(3,mathbb R)$ corresponding to f.
My idea to find the matrix was this:
We know that the representation vector $x$ of point $X$ in $pi$ can be written as
$$x=s_0 cdot (1,0,0,0)+t_0 cdot (0,1,0,0)+k_0(sa+tb)tag1$$
for $s_0,t_0,k_0 in mathbb R$.
The respresentation vector $y$ of a point $Y$ on $l_1$ can be written as
$$y=s_1 cdot (0,0,1,0) + t_1 cdot (0,0,0,1)tag2$$
for $s_1, t_1 in mathbb R$.
Since the line $l_2$ is determined by the points $A_2$ and $B_2$ it is sufficient to find a matrix $A=(a_{ij})$ that maps the representation vectors $a$ and $b$ onto representation vectors of the points $(pi A_2)cap l_1$ and $(pi B_2)cap l_1$. The representation vectors of the image points must satisfy the equations $(1)$ and $(2)$. So we get the following equations
$Aa=s_0' cdot (1,0,0,0)+t_0' cdot (0,1,0,0)+a$
$Aa=s_1' cdot (0,0,1,0) + t_1' cdot (0,0,0,1)$
$Ab=s_0'' cdot (1,0,0,0)+t_0'' cdot (0,1,0,0)+b$
$Ab=s_1'' cdot (0,0,1,0) + t_1'' cdot (0,0,0,1)$
Since the matrix $A$ has 16 entries we get 16 equations in total. Solving this system with respect to the entries of $A$ should give me the matrix I need.
My question is:
Is there an easier way to solve this than solving a system of 16 equations?
geometry projective-geometry
edited Jan 13 at 21:30
MvG
31.2k450107
asked Jan 11 at 19:16
PolymorphPolymorph
1206
$endgroup$
I have a question about the correctness of my ideas regarding the following exercise.
Define
$A_0=[(1,0,0,0)]$
$B_0=[(0,1,0,0)]$
$A_1=[(0,0,1,0)]$
$B_1=[(0,0,0,1)] in mathbb RP^3$ where $mathbb RP^3$ denotes real projective space.
Let $l_0$ be the line through $A_0$ and $B_0$ and $l_1$ be the line through $A_1$ and $B_1$. Let $l_2$ be an arbitrary line skew to both $l_0$ and $l_1$ and let $A_2=[a]$ and $B_2=[b]$ be two points of $l_2$ so that $P=[s cdot a+ t cdot b] $ is a general point of $l_2$. Let $pi$ be the plane spanned by $l_0$ and P and define Q to be the intersection of $pi$ with $l_1$. Let f be the map $f: l_2 to l_1$ defined by $f(P)=Q.$ Find a matrix in $PGL(3,mathbb R)$ corresponding to f.
My idea to find the matrix was this:
We know that the representation vector $x$ of point $X$ in $pi$ can be written as
$$x=s_0 cdot (1,0,0,0)+t_0 cdot (0,1,0,0)+k_0(sa+tb)tag1$$
for $s_0,t_0,k_0 in mathbb R$.
The respresentation vector $y$ of a point $Y$ on $l_1$ can be written as
$$y=s_1 cdot (0,0,1,0) + t_1 cdot (0,0,0,1)tag2$$
for $s_1, t_1 in mathbb R$.
Since the line $l_2$ is determined by the points $A_2$ and $B_2$ it is sufficient to find a matrix $A=(a_{ij})$ that maps the representation vectors $a$ and $b$ onto representation vectors of the points $(pi A_2)cap l_1$ and $(pi B_2)cap l_1$. The representation vectors of the image points must satisfy the equations $(1)$ and $(2)$. So we get the following equations
$Aa=s_0' cdot (1,0,0,0)+t_0' cdot (0,1,0,0)+a$
$Aa=s_1' cdot (0,0,1,0) + t_1' cdot (0,0,0,1)$
$Ab=s_0'' cdot (1,0,0,0)+t_0'' cdot (0,1,0,0)+b$
$Ab=s_1'' cdot (0,0,1,0) + t_1'' cdot (0,0,0,1)$
Since the matrix $A$ has 16 entries we get 16 equations in total. Solving this system with respect to the entries of $A$ should give me the matrix I need.
My question is:
Is there an easier way to solve this than solving a system of 16 equations?
geometry projective-geometry
geometry projective-geometry
edited Jan 13 at 21:30
MvG
31.2k450107
asked Jan 11 at 19:16
PolymorphPolymorph
1206
edited Jan 13 at 21:30
MvG
31.2k450107
asked Jan 11 at 19:16
PolymorphPolymorph
1206
edited Jan 13 at 21:30
MvG
31.2k450107
edited Jan 13 at 21:30
MvG
31.2k450107
edited Jan 13 at 21:30
MvG
31.2k450107
31.2k450107
asked Jan 11 at 19:16
PolymorphPolymorph
1206
asked Jan 11 at 19:16
PolymorphPolymorph
1206
asked Jan 11 at 19:16
PolymorphPolymorph
1206
1206
$begingroup$
$A_0$ and $B_0$ are identical. Did you mean to write $B_0=[(0,1,0,0)]$?
$endgroup$
– amd
Jan 11 at 21:07
$begingroup$
The wording of this problem is troublesome. “Let $f$ be the map $f:l_2to l_1$” defines $f$ as a homography of two lines (an element of $PGL(2,mathbb R)$), but asks for a matrix in $PGL(3,mathbb R)$. Moreover, $PGL(3)$ operates on the projective plane, but the problem is set in a higher-dimensional space.
$endgroup$
– amd
Jan 11 at 21:55
$begingroup$
I also have a quibble with the phrasing “the map.” Even if we restrict ourselves to a homography between two lines, a single point correspondence doesn’t determine a unique map.
$endgroup$
– amd
Jan 11 at 21:57
$begingroup$
I’m guessing that when you write $IR$, you actually mean $mathbb R$. Usemathbb Rto get this symbol.
$endgroup$
– amd
Jan 11 at 21:58
$begingroup$
I can imagine two ways to read the question. Either you want $Q=[s'cdot a'+t'cdot b']$, probably with $A_1=[a'], B_1=[b']$, and then express $s',t'$ in terms of $s,t$, which would lead to a $2times2$ matrix. Or you want $Q$ as an arbitrary point in space, projecting the whole space to $l_1$ as described. In that case $l_2$ is irrelevant, and you get a $4times4$ matrix of rank $2$. It would be non-invertible. Like amd I can't see $textrm{PGL}(3,mathbb R)$ playing a role, since it has equivalence classes of invertible $3times3$ matrices. Please double-check the question statement.
$endgroup$
– MvG
Jan 14 at 23:01
|
show 1 more comment
$begingroup$
$A_0$ and $B_0$ are identical. Did you mean to write $B_0=[(0,1,0,0)]$?
$endgroup$
– amd
Jan 11 at 21:07
$begingroup$
The wording of this problem is troublesome. “Let $f$ be the map $f:l_2to l_1$” defines $f$ as a homography of two lines (an element of $PGL(2,mathbb R)$), but asks for a matrix in $PGL(3,mathbb R)$. Moreover, $PGL(3)$ operates on the projective plane, but the problem is set in a higher-dimensional space.
$endgroup$
– amd
Jan 11 at 21:55
$begingroup$
I also have a quibble with the phrasing “the map.” Even if we restrict ourselves to a homography between two lines, a single point correspondence doesn’t determine a unique map.
$endgroup$
– amd
Jan 11 at 21:57
$begingroup$
I’m guessing that when you write $IR$, you actually mean $mathbb R$. Usemathbb Rto get this symbol.
$endgroup$
– amd
Jan 11 at 21:58
$begingroup$
I can imagine two ways to read the question. Either you want $Q=[s'cdot a'+t'cdot b']$, probably with $A_1=[a'], B_1=[b']$, and then express $s',t'$ in terms of $s,t$, which would lead to a $2times2$ matrix. Or you want $Q$ as an arbitrary point in space, projecting the whole space to $l_1$ as described. In that case $l_2$ is irrelevant, and you get a $4times4$ matrix of rank $2$. It would be non-invertible. Like amd I can't see $textrm{PGL}(3,mathbb R)$ playing a role, since it has equivalence classes of invertible $3times3$ matrices. Please double-check the question statement.
$endgroup$
– MvG
Jan 14 at 23:01
$begingroup$
$A_0$ and $B_0$ are identical. Did you mean to write $B_0=[(0,1,0,0)]$?
$endgroup$
– amd
Jan 11 at 21:07
$begingroup$
$A_0$ and $B_0$ are identical. Did you mean to write $B_0=[(0,1,0,0)]$?
$endgroup$
– amd
Jan 11 at 21:07
$begingroup$
The wording of this problem is troublesome. “Let $f$ be the map $f:l_2to l_1$” defines $f$ as a homography of two lines (an element of $PGL(2,mathbb R)$), but asks for a matrix in $PGL(3,mathbb R)$. Moreover, $PGL(3)$ operates on the projective plane, but the problem is set in a higher-dimensional space.
$endgroup$
– amd
Jan 11 at 21:55
$begingroup$
The wording of this problem is troublesome. “Let $f$ be the map $f:l_2to l_1$” defines $f$ as a homography of two lines (an element of $PGL(2,mathbb R)$), but asks for a matrix in $PGL(3,mathbb R)$. Moreover, $PGL(3)$ operates on the projective plane, but the problem is set in a higher-dimensional space.
$endgroup$
– amd
Jan 11 at 21:55
$begingroup$
I also have a quibble with the phrasing “the map.” Even if we restrict ourselves to a homography between two lines, a single point correspondence doesn’t determine a unique map.
$endgroup$
– amd
Jan 11 at 21:57
$begingroup$
I also have a quibble with the phrasing “the map.” Even if we restrict ourselves to a homography between two lines, a single point correspondence doesn’t determine a unique map.
$endgroup$
– amd
Jan 11 at 21:57
$begingroup$
I’m guessing that when you write $IR$, you actually mean $mathbb R$. Use
mathbb R to get this symbol.$endgroup$
– amd
Jan 11 at 21:58
$begingroup$
I’m guessing that when you write $IR$, you actually mean $mathbb R$. Use
mathbb R to get this symbol.$endgroup$
– amd
Jan 11 at 21:58
$begingroup$
I can imagine two ways to read the question. Either you want $Q=[s'cdot a'+t'cdot b']$, probably with $A_1=[a'], B_1=[b']$, and then express $s',t'$ in terms of $s,t$, which would lead to a $2times2$ matrix. Or you want $Q$ as an arbitrary point in space, projecting the whole space to $l_1$ as described. In that case $l_2$ is irrelevant, and you get a $4times4$ matrix of rank $2$. It would be non-invertible. Like amd I can't see $textrm{PGL}(3,mathbb R)$ playing a role, since it has equivalence classes of invertible $3times3$ matrices. Please double-check the question statement.
$endgroup$
– MvG
Jan 14 at 23:01
$begingroup$
I can imagine two ways to read the question. Either you want $Q=[s'cdot a'+t'cdot b']$, probably with $A_1=[a'], B_1=[b']$, and then express $s',t'$ in terms of $s,t$, which would lead to a $2times2$ matrix. Or you want $Q$ as an arbitrary point in space, projecting the whole space to $l_1$ as described. In that case $l_2$ is irrelevant, and you get a $4times4$ matrix of rank $2$. It would be non-invertible. Like amd I can't see $textrm{PGL}(3,mathbb R)$ playing a role, since it has equivalence classes of invertible $3times3$ matrices. Please double-check the question statement.
$endgroup$
– MvG
Jan 14 at 23:01
|
show 1 more comment
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$begingroup$
$A_0$ and $B_0$ are identical. Did you mean to write $B_0=[(0,1,0,0)]$?
$endgroup$
– amd
Jan 11 at 21:07
$begingroup$
The wording of this problem is troublesome. “Let $f$ be the map $f:l_2to l_1$” defines $f$ as a homography of two lines (an element of $PGL(2,mathbb R)$), but asks for a matrix in $PGL(3,mathbb R)$. Moreover, $PGL(3)$ operates on the projective plane, but the problem is set in a higher-dimensional space.
$endgroup$
– amd
Jan 11 at 21:55
$begingroup$
I also have a quibble with the phrasing “the map.” Even if we restrict ourselves to a homography between two lines, a single point correspondence doesn’t determine a unique map.
$endgroup$
– amd
Jan 11 at 21:57
$begingroup$
I’m guessing that when you write $IR$, you actually mean $mathbb R$. Use
mathbb Rto get this symbol.$endgroup$
– amd
Jan 11 at 21:58
$begingroup$
I can imagine two ways to read the question. Either you want $Q=[s'cdot a'+t'cdot b']$, probably with $A_1=[a'], B_1=[b']$, and then express $s',t'$ in terms of $s,t$, which would lead to a $2times2$ matrix. Or you want $Q$ as an arbitrary point in space, projecting the whole space to $l_1$ as described. In that case $l_2$ is irrelevant, and you get a $4times4$ matrix of rank $2$. It would be non-invertible. Like amd I can't see $textrm{PGL}(3,mathbb R)$ playing a role, since it has equivalence classes of invertible $3times3$ matrices. Please double-check the question statement.
$endgroup$
– MvG
Jan 14 at 23:01