Locus of the images of a point given by certain affine transformationsfixed points of an affine transformation is unique iff $1 notin SP(vecf )$Prove that the ratio of lengths of parallel segments is invariant under affine trasnformationsThe projective space of all lines through the originAbout the stable/invariant point sets in a plane with respect to shift/linear transformationThree points of an affine space are collinear $iff det(A)=0$, with $A$ the matrix of the barycentric coordinates.Affinity which maps one plane to anotherprojection of a vector onto a plane in $Bbb R^5$About Affine planesFind coordinates of a regular polygon in a plane.Affine transformations satisfying conditions
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Locus of the images of a point given by certain affine transformations
fixed points of an affine transformation is unique iff $1 notin SP(vecf )$Prove that the ratio of lengths of parallel segments is invariant under affine trasnformationsThe projective space of all lines through the originAbout the stable/invariant point sets in a plane with respect to shift/linear transformationThree points of an affine space are collinear $iff det(A)=0$, with $A$ the matrix of the barycentric coordinates.Affinity which maps one plane to anotherprojection of a vector onto a plane in $Bbb R^5$About Affine planesFind coordinates of a regular polygon in a plane.Affine transformations satisfying conditions
$begingroup$
In the affine plane, let P, Q, R be three aligned points. Now let's consider all the affine transformations that keep R as a fixed point, transform P into Q, and have a unique invariant line. Which is the locus of the images of a point given by these affine transformations?
I have thought about considering a reference Ref $=R;vecRP,vec??$, with $vecRP$ eigenvector with eigenvalue $fracQ$, but I'm not sure this is the right way to do the exercise, as I don't know how to continue... Could you help me? Thanks in advance!
geometry affine-geometry locus
$endgroup$
add a comment |
$begingroup$
In the affine plane, let P, Q, R be three aligned points. Now let's consider all the affine transformations that keep R as a fixed point, transform P into Q, and have a unique invariant line. Which is the locus of the images of a point given by these affine transformations?
I have thought about considering a reference Ref $=R;vecRP,vec??$, with $vecRP$ eigenvector with eigenvalue $fracQ$, but I'm not sure this is the right way to do the exercise, as I don't know how to continue... Could you help me? Thanks in advance!
geometry affine-geometry locus
$endgroup$
1
$begingroup$
Yes, that is a good approach. You can translate $R$ to the origin and assume that the transformations are linear. The invariant line must be the line $PQ$, since the vector $RQ$ is an eigenvalue because it got sent to $RP$. Now pick a point $x$. We need to distinguish cases. If $x$ is in the line $PQ$, then the image is just another point $x'$ such that $xRx'$ are in the same proportion as $PRQ$. If $x$ is outside the line $PQ$, then $x'$ can be anywhere except on the lines $PQ$ and $Rx$.
$endgroup$
– user647486
Mar 16 at 21:52
$begingroup$
By “aligned” do you mean colinear?
$endgroup$
– amd
Mar 16 at 21:56
$begingroup$
@amd That is what aligned means.
$endgroup$
– user647486
Mar 16 at 21:58
$begingroup$
@user647486 Your explanation is quite good! When you say "$x'$ can be anywhere excepte on the lines $PQ$ and $Rx$", how can we determine where does the image go? I suppose this will clarify which is the locus of the images...
$endgroup$
– Gibbs
Mar 16 at 22:26
$begingroup$
That sentence that I said is that 'location'. Pick any of those points and there is a single affine (linear after translating $R$ to the origin) that sends $P$ to $Q$ and $x$ to $x'$, and which only has the line $PQ$ and invariant line.
$endgroup$
– user647486
Mar 16 at 22:27
add a comment |
$begingroup$
In the affine plane, let P, Q, R be three aligned points. Now let's consider all the affine transformations that keep R as a fixed point, transform P into Q, and have a unique invariant line. Which is the locus of the images of a point given by these affine transformations?
I have thought about considering a reference Ref $=R;vecRP,vec??$, with $vecRP$ eigenvector with eigenvalue $fracQ$, but I'm not sure this is the right way to do the exercise, as I don't know how to continue... Could you help me? Thanks in advance!
geometry affine-geometry locus
$endgroup$
In the affine plane, let P, Q, R be three aligned points. Now let's consider all the affine transformations that keep R as a fixed point, transform P into Q, and have a unique invariant line. Which is the locus of the images of a point given by these affine transformations?
I have thought about considering a reference Ref $=R;vecRP,vec??$, with $vecRP$ eigenvector with eigenvalue $fracQ$, but I'm not sure this is the right way to do the exercise, as I don't know how to continue... Could you help me? Thanks in advance!
geometry affine-geometry locus
geometry affine-geometry locus
asked Mar 16 at 21:40
GibbsGibbs
137111
137111
1
$begingroup$
Yes, that is a good approach. You can translate $R$ to the origin and assume that the transformations are linear. The invariant line must be the line $PQ$, since the vector $RQ$ is an eigenvalue because it got sent to $RP$. Now pick a point $x$. We need to distinguish cases. If $x$ is in the line $PQ$, then the image is just another point $x'$ such that $xRx'$ are in the same proportion as $PRQ$. If $x$ is outside the line $PQ$, then $x'$ can be anywhere except on the lines $PQ$ and $Rx$.
$endgroup$
– user647486
Mar 16 at 21:52
$begingroup$
By “aligned” do you mean colinear?
$endgroup$
– amd
Mar 16 at 21:56
$begingroup$
@amd That is what aligned means.
$endgroup$
– user647486
Mar 16 at 21:58
$begingroup$
@user647486 Your explanation is quite good! When you say "$x'$ can be anywhere excepte on the lines $PQ$ and $Rx$", how can we determine where does the image go? I suppose this will clarify which is the locus of the images...
$endgroup$
– Gibbs
Mar 16 at 22:26
$begingroup$
That sentence that I said is that 'location'. Pick any of those points and there is a single affine (linear after translating $R$ to the origin) that sends $P$ to $Q$ and $x$ to $x'$, and which only has the line $PQ$ and invariant line.
$endgroup$
– user647486
Mar 16 at 22:27
add a comment |
1
$begingroup$
Yes, that is a good approach. You can translate $R$ to the origin and assume that the transformations are linear. The invariant line must be the line $PQ$, since the vector $RQ$ is an eigenvalue because it got sent to $RP$. Now pick a point $x$. We need to distinguish cases. If $x$ is in the line $PQ$, then the image is just another point $x'$ such that $xRx'$ are in the same proportion as $PRQ$. If $x$ is outside the line $PQ$, then $x'$ can be anywhere except on the lines $PQ$ and $Rx$.
$endgroup$
– user647486
Mar 16 at 21:52
$begingroup$
By “aligned” do you mean colinear?
$endgroup$
– amd
Mar 16 at 21:56
$begingroup$
@amd That is what aligned means.
$endgroup$
– user647486
Mar 16 at 21:58
$begingroup$
@user647486 Your explanation is quite good! When you say "$x'$ can be anywhere excepte on the lines $PQ$ and $Rx$", how can we determine where does the image go? I suppose this will clarify which is the locus of the images...
$endgroup$
– Gibbs
Mar 16 at 22:26
$begingroup$
That sentence that I said is that 'location'. Pick any of those points and there is a single affine (linear after translating $R$ to the origin) that sends $P$ to $Q$ and $x$ to $x'$, and which only has the line $PQ$ and invariant line.
$endgroup$
– user647486
Mar 16 at 22:27
1
1
$begingroup$
Yes, that is a good approach. You can translate $R$ to the origin and assume that the transformations are linear. The invariant line must be the line $PQ$, since the vector $RQ$ is an eigenvalue because it got sent to $RP$. Now pick a point $x$. We need to distinguish cases. If $x$ is in the line $PQ$, then the image is just another point $x'$ such that $xRx'$ are in the same proportion as $PRQ$. If $x$ is outside the line $PQ$, then $x'$ can be anywhere except on the lines $PQ$ and $Rx$.
$endgroup$
– user647486
Mar 16 at 21:52
$begingroup$
Yes, that is a good approach. You can translate $R$ to the origin and assume that the transformations are linear. The invariant line must be the line $PQ$, since the vector $RQ$ is an eigenvalue because it got sent to $RP$. Now pick a point $x$. We need to distinguish cases. If $x$ is in the line $PQ$, then the image is just another point $x'$ such that $xRx'$ are in the same proportion as $PRQ$. If $x$ is outside the line $PQ$, then $x'$ can be anywhere except on the lines $PQ$ and $Rx$.
$endgroup$
– user647486
Mar 16 at 21:52
$begingroup$
By “aligned” do you mean colinear?
$endgroup$
– amd
Mar 16 at 21:56
$begingroup$
By “aligned” do you mean colinear?
$endgroup$
– amd
Mar 16 at 21:56
$begingroup$
@amd That is what aligned means.
$endgroup$
– user647486
Mar 16 at 21:58
$begingroup$
@amd That is what aligned means.
$endgroup$
– user647486
Mar 16 at 21:58
$begingroup$
@user647486 Your explanation is quite good! When you say "$x'$ can be anywhere excepte on the lines $PQ$ and $Rx$", how can we determine where does the image go? I suppose this will clarify which is the locus of the images...
$endgroup$
– Gibbs
Mar 16 at 22:26
$begingroup$
@user647486 Your explanation is quite good! When you say "$x'$ can be anywhere excepte on the lines $PQ$ and $Rx$", how can we determine where does the image go? I suppose this will clarify which is the locus of the images...
$endgroup$
– Gibbs
Mar 16 at 22:26
$begingroup$
That sentence that I said is that 'location'. Pick any of those points and there is a single affine (linear after translating $R$ to the origin) that sends $P$ to $Q$ and $x$ to $x'$, and which only has the line $PQ$ and invariant line.
$endgroup$
– user647486
Mar 16 at 22:27
$begingroup$
That sentence that I said is that 'location'. Pick any of those points and there is a single affine (linear after translating $R$ to the origin) that sends $P$ to $Q$ and $x$ to $x'$, and which only has the line $PQ$ and invariant line.
$endgroup$
– user647486
Mar 16 at 22:27
add a comment |
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$begingroup$
Yes, that is a good approach. You can translate $R$ to the origin and assume that the transformations are linear. The invariant line must be the line $PQ$, since the vector $RQ$ is an eigenvalue because it got sent to $RP$. Now pick a point $x$. We need to distinguish cases. If $x$ is in the line $PQ$, then the image is just another point $x'$ such that $xRx'$ are in the same proportion as $PRQ$. If $x$ is outside the line $PQ$, then $x'$ can be anywhere except on the lines $PQ$ and $Rx$.
$endgroup$
– user647486
Mar 16 at 21:52
$begingroup$
By “aligned” do you mean colinear?
$endgroup$
– amd
Mar 16 at 21:56
$begingroup$
@amd That is what aligned means.
$endgroup$
– user647486
Mar 16 at 21:58
$begingroup$
@user647486 Your explanation is quite good! When you say "$x'$ can be anywhere excepte on the lines $PQ$ and $Rx$", how can we determine where does the image go? I suppose this will clarify which is the locus of the images...
$endgroup$
– Gibbs
Mar 16 at 22:26
$begingroup$
That sentence that I said is that 'location'. Pick any of those points and there is a single affine (linear after translating $R$ to the origin) that sends $P$ to $Q$ and $x$ to $x'$, and which only has the line $PQ$ and invariant line.
$endgroup$
– user647486
Mar 16 at 22:27