Mixing
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
30.9k450104
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
30.9k450104
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
30.9k450104
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
30.9k450104
asked Jan 11 at 19:16
PolymorphPolymorph
1206
edited Jan 13 at 21:30
MvG
30.9k450104
asked Jan 11 at 19:16
PolymorphPolymorph
1206
edited Jan 13 at 21:30
MvG
30.9k450104
edited Jan 13 at 21:30
MvG
30.9k450104
edited Jan 13 at 21:30
MvG
30.9k450104
30.9k450104
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
0
active
oldest
votes
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3070240%2fdetermining-the-matrix-of-a-projection%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3070240%2fdetermining-the-matrix-of-a-projection%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
$A_0$ 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