Principal axes of deformed ellipsoid The 2019 Stack Overflow Developer Survey Results Are InTransformation of ellipsoid to sphereLinear transformations for fixing the line $y = 0$What shape do we get when we shear an ellipse? And more generally, do affine transformations always map conic sections to conic sections?linear combination of phase of complex signalsFor which point is the transformation linear?Can you convert a non “normal” complex square matrix into a “normal” one?Explain the geometrical meaning of Singular Value Decomposition (SVD)How is it that matrix multiplication represents linear transforms?Find a matrix $P$ that diagonalizes the matrix $A$, and determine $P^-1AP$Ellipsoid and linear transformation
Worn-tile Scrabble
Why isn't the circumferential light around the M87 black hole's event horizon symmetric?
Apparent duplicates between Haynes service instructions and MOT
What is the closest word meaning "respect for time / mindful"
Does the shape of a die affect the probability of a number being rolled?
How are circuits which use complex ICs normally simulated?
What do the Banks children have against barley water?
"as much details as you can remember"
Is there any way to tell whether the shot is going to hit you or not?
What is the motivation for a law requiring 2 parties to consent for recording a conversation
If a Druid sees an animal’s corpse, can they wild shape into that animal?
Is this app Icon Browser Safe/Legit?
Identify boardgame from Big movie
What did it mean to "align" a radio?
Resizing object distorts it (Illustrator CC 2018)
Is there a symbol for a right arrow with a square in the middle?
Origin of "cooter" meaning "vagina"
Multiply Two Integer Polynomials
Does a dangling wire really electrocute me if I'm standing in water?
Is flight data recorder erased after every flight?
How to support a colleague who finds meetings extremely tiring?
Are there any other methods to apply to solving simultaneous equations?
Why is the maximum length of OpenWrt’s root password 8 characters?
Are there incongruent pythagorean triangles with the same perimeter and same area?
Principal axes of deformed ellipsoid
The 2019 Stack Overflow Developer Survey Results Are InTransformation of ellipsoid to sphereLinear transformations for fixing the line $y = 0$What shape do we get when we shear an ellipse? And more generally, do affine transformations always map conic sections to conic sections?linear combination of phase of complex signalsFor which point is the transformation linear?Can you convert a non “normal” complex square matrix into a “normal” one?Explain the geometrical meaning of Singular Value Decomposition (SVD)How is it that matrix multiplication represents linear transforms?Find a matrix $P$ that diagonalizes the matrix $A$, and determine $P^-1AP$Ellipsoid and linear transformation
$begingroup$
I have a transformation matrix $T$ which stretches and rotates the unit sphere into an ellipsoid. I have a second, diagonal matrix
$$S=beginbmatrixs_x&0&0\
0&s_y&0\
0&0&s_z
endbmatrix$$
which then stretches the first ellipsoid into another one. I need to find the principle axis directions and lengths of the second ellipsoid.
My first thought was to compose the two matrices
$$T'=ST$$
and find the eigenvalues and eigenvectors of $T'$, but this just gave the axes of the first ellipsoid deformed by $S$. (This seems obvious in hindsight, since the composition preserves the complete mapping, not just the resulting shape.)
My second thought is to work out the algebraic equation of the ellipsoid described by $T'$ and massage it back into a standard form.
Can anyone help?
linear-transformations
asked Mar 23 at 17:42
FragileXFragileX
33
$endgroup$
add a comment |
$begingroup$
I have a transformation matrix $T$ which stretches and rotates the unit sphere into an ellipsoid. I have a second, diagonal matrix
$$S=beginbmatrixs_x&0&0\
0&s_y&0\
0&0&s_z
endbmatrix$$
which then stretches the first ellipsoid into another one. I need to find the principle axis directions and lengths of the second ellipsoid.
My first thought was to compose the two matrices
$$T'=ST$$
and find the eigenvalues and eigenvectors of $T'$, but this just gave the axes of the first ellipsoid deformed by $S$. (This seems obvious in hindsight, since the composition preserves the complete mapping, not just the resulting shape.)
My second thought is to work out the algebraic equation of the ellipsoid described by $T'$ and massage it back into a standard form.
Can anyone help?
linear-transformations
asked Mar 23 at 17:42
FragileXFragileX
33
$endgroup$
$begingroup$
BTW, I'm expressing my vectors as columns, so the order of T'=ST is correct.
$endgroup$
– FragileX
Mar 23 at 18:39
$begingroup$
The resulting ellipsoid can be represented by the matrix $(ST)^-T(ST)^-1$.
$endgroup$
– amd
Mar 24 at 3:41
$begingroup$
@amd That produces eigenvectors pointing in the right direction, but the eigenvalues are not the right size.
$endgroup$
– FragileX
Mar 24 at 18:55
$begingroup$
The eigenvalues from $(ST)^-T(ST)^-1$ become the expected axis semi-lengths if I take the reciprocal of their square roots. I kind of figured that out by trial-and-error, though. Can someone who understands what's going on please write these comments up as a full answer?
$endgroup$
– FragileX
Mar 24 at 19:30
add a comment |
$begingroup$
I have a transformation matrix $T$ which stretches and rotates the unit sphere into an ellipsoid. I have a second, diagonal matrix
$$S=beginbmatrixs_x&0&0\
0&s_y&0\
0&0&s_z
endbmatrix$$
which then stretches the first ellipsoid into another one. I need to find the principle axis directions and lengths of the second ellipsoid.
My first thought was to compose the two matrices
$$T'=ST$$
and find the eigenvalues and eigenvectors of $T'$, but this just gave the axes of the first ellipsoid deformed by $S$. (This seems obvious in hindsight, since the composition preserves the complete mapping, not just the resulting shape.)
My second thought is to work out the algebraic equation of the ellipsoid described by $T'$ and massage it back into a standard form.
Can anyone help?
linear-transformations
asked Mar 23 at 17:42
FragileXFragileX
33
$endgroup$
I have a transformation matrix $T$ which stretches and rotates the unit sphere into an ellipsoid. I have a second, diagonal matrix
$$S=beginbmatrixs_x&0&0\
0&s_y&0\
0&0&s_z
endbmatrix$$
which then stretches the first ellipsoid into another one. I need to find the principle axis directions and lengths of the second ellipsoid.
My first thought was to compose the two matrices
$$T'=ST$$
and find the eigenvalues and eigenvectors of $T'$, but this just gave the axes of the first ellipsoid deformed by $S$. (This seems obvious in hindsight, since the composition preserves the complete mapping, not just the resulting shape.)
My second thought is to work out the algebraic equation of the ellipsoid described by $T'$ and massage it back into a standard form.
Can anyone help?
linear-transformations
linear-transformations
asked Mar 23 at 17:42
FragileXFragileX
33
asked Mar 23 at 17:42
FragileXFragileX
33
edited Mar 23 at 18:05
FragileX
asked Mar 23 at 17:42
FragileXFragileX
33
asked Mar 23 at 17:42
FragileXFragileX
33
asked Mar 23 at 17:42
FragileXFragileX
33
33
$begingroup$
BTW, I'm expressing my vectors as columns, so the order of T'=ST is correct.
$endgroup$
– FragileX
Mar 23 at 18:39
$begingroup$
The resulting ellipsoid can be represented by the matrix $(ST)^-T(ST)^-1$.
$endgroup$
– amd
Mar 24 at 3:41
$begingroup$
@amd That produces eigenvectors pointing in the right direction, but the eigenvalues are not the right size.
$endgroup$
– FragileX
Mar 24 at 18:55
$begingroup$
The eigenvalues from $(ST)^-T(ST)^-1$ become the expected axis semi-lengths if I take the reciprocal of their square roots. I kind of figured that out by trial-and-error, though. Can someone who understands what's going on please write these comments up as a full answer?
$endgroup$
– FragileX
Mar 24 at 19:30
add a comment |
$begingroup$
BTW, I'm expressing my vectors as columns, so the order of T'=ST is correct.
$endgroup$
– FragileX
Mar 23 at 18:39
$begingroup$
The resulting ellipsoid can be represented by the matrix $(ST)^-T(ST)^-1$.
$endgroup$
– amd
Mar 24 at 3:41
$begingroup$
@amd That produces eigenvectors pointing in the right direction, but the eigenvalues are not the right size.
$endgroup$
– FragileX
Mar 24 at 18:55
$begingroup$
The eigenvalues from $(ST)^-T(ST)^-1$ become the expected axis semi-lengths if I take the reciprocal of their square roots. I kind of figured that out by trial-and-error, though. Can someone who understands what's going on please write these comments up as a full answer?
$endgroup$
– FragileX
Mar 24 at 19:30
$begingroup$
BTW, I'm expressing my vectors as columns, so the order of T'=ST is correct.
$endgroup$
– FragileX
Mar 23 at 18:39
$begingroup$
BTW, I'm expressing my vectors as columns, so the order of T'=ST is correct.
$endgroup$
– FragileX
Mar 23 at 18:39
$begingroup$
The resulting ellipsoid can be represented by the matrix $(ST)^-T(ST)^-1$.
$endgroup$
– amd
Mar 24 at 3:41
$begingroup$
The resulting ellipsoid can be represented by the matrix $(ST)^-T(ST)^-1$.
$endgroup$
– amd
Mar 24 at 3:41
$begingroup$
@amd That produces eigenvectors pointing in the right direction, but the eigenvalues are not the right size.
$endgroup$
– FragileX
Mar 24 at 18:55
$begingroup$
@amd That produces eigenvectors pointing in the right direction, but the eigenvalues are not the right size.
$endgroup$
– FragileX
Mar 24 at 18:55
$begingroup$
The eigenvalues from $(ST)^-T(ST)^-1$ become the expected axis semi-lengths if I take the reciprocal of their square roots. I kind of figured that out by trial-and-error, though. Can someone who understands what's going on please write these comments up as a full answer?
$endgroup$
– FragileX
Mar 24 at 19:30
$begingroup$
The eigenvalues from $(ST)^-T(ST)^-1$ become the expected axis semi-lengths if I take the reciprocal of their square roots. I kind of figured that out by trial-and-error, though. Can someone who understands what's going on please write these comments up as a full answer?
$endgroup$
– FragileX
Mar 24 at 19:30
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The equation $Ax^2+By^2+Cz^2+Exy+Fxz+Gyz=1$ of a quadric surface can be expressed in matrix form as $$beginbmatrixx&y&zendbmatrix beginbmatrixA&frac E2&frac F2\frac E2&B&frac G2\frac F2&frac G2&Cendbmatrix beginbmatrixx\y\zendbmatrix = mathbf x^TQmathbf x = 1.$$ The symmetric matrix $Q$ can be said to represent the quadric.† In particular, the unit sphere is represented by the identity matrix.
The principal axis theorem tells us that there’s an orthonormal basis of eigenvectors of $Q$ that gives the axes of this quadric and that the half-axis lengths are given by the reciprocal square roots of the (absolute values) of the corresponding eigenvalues: after a change of basis that’s an isometry because the basis is orthonormal, the equation of the quadric takes the form $lambda_1x'^2+lambda_2y'^2+lambda_3z'^2=1$.
Given the invertible point transformation $mathbf x'=Mmathbf x$, we have $$mathbf x^TQmathbf x = (M^-1mathbf x')^T Q (M^-1mathbf x') = mathbf x'^T (M^-TQM^-1) mathbf x',$$ so the quadric transforms as $Q' = M^-TQM^-1$. In your case, $M=ST$, so when we transform the unit sphere $Q=I$, we get $$Q' = M^-TQM^-1 = (ST)^-T(ST)^-1.$$ The reciprocal square roots of the eigenvalues of this matrix then are the half-axis lengths of the resulting ellipsoid and the corresponding eigenvectors, chosen to form an orthogonal set in case of repeated eigenvalues, are its principal axes.
† The fully-general quadric equation that includes linear terms and and arbitrary constant term can also be represented by a single $4times4$ matrix by using homogeneous coordinates. All of this applies to quadrics of any dimension.
answered Mar 27 at 18:11
amdamd
31.6k21052
$endgroup$
add a comment |
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%2f3159618%2fprincipal-axes-of-deformed-ellipsoid%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The equation $Ax^2+By^2+Cz^2+Exy+Fxz+Gyz=1$ of a quadric surface can be expressed in matrix form as $$beginbmatrixx&y&zendbmatrix beginbmatrixA&frac E2&frac F2\frac E2&B&frac G2\frac F2&frac G2&Cendbmatrix beginbmatrixx\y\zendbmatrix = mathbf x^TQmathbf x = 1.$$ The symmetric matrix $Q$ can be said to represent the quadric.† In particular, the unit sphere is represented by the identity matrix.
The principal axis theorem tells us that there’s an orthonormal basis of eigenvectors of $Q$ that gives the axes of this quadric and that the half-axis lengths are given by the reciprocal square roots of the (absolute values) of the corresponding eigenvalues: after a change of basis that’s an isometry because the basis is orthonormal, the equation of the quadric takes the form $lambda_1x'^2+lambda_2y'^2+lambda_3z'^2=1$.
Given the invertible point transformation $mathbf x'=Mmathbf x$, we have $$mathbf x^TQmathbf x = (M^-1mathbf x')^T Q (M^-1mathbf x') = mathbf x'^T (M^-TQM^-1) mathbf x',$$ so the quadric transforms as $Q' = M^-TQM^-1$. In your case, $M=ST$, so when we transform the unit sphere $Q=I$, we get $$Q' = M^-TQM^-1 = (ST)^-T(ST)^-1.$$ The reciprocal square roots of the eigenvalues of this matrix then are the half-axis lengths of the resulting ellipsoid and the corresponding eigenvectors, chosen to form an orthogonal set in case of repeated eigenvalues, are its principal axes.
† The fully-general quadric equation that includes linear terms and and arbitrary constant term can also be represented by a single $4times4$ matrix by using homogeneous coordinates. All of this applies to quadrics of any dimension.
answered Mar 27 at 18:11
amdamd
31.6k21052
$endgroup$
add a comment |
$begingroup$
The equation $Ax^2+By^2+Cz^2+Exy+Fxz+Gyz=1$ of a quadric surface can be expressed in matrix form as $$beginbmatrixx&y&zendbmatrix beginbmatrixA&frac E2&frac F2\frac E2&B&frac G2\frac F2&frac G2&Cendbmatrix beginbmatrixx\y\zendbmatrix = mathbf x^TQmathbf x = 1.$$ The symmetric matrix $Q$ can be said to represent the quadric.† In particular, the unit sphere is represented by the identity matrix.
The principal axis theorem tells us that there’s an orthonormal basis of eigenvectors of $Q$ that gives the axes of this quadric and that the half-axis lengths are given by the reciprocal square roots of the (absolute values) of the corresponding eigenvalues: after a change of basis that’s an isometry because the basis is orthonormal, the equation of the quadric takes the form $lambda_1x'^2+lambda_2y'^2+lambda_3z'^2=1$.
Given the invertible point transformation $mathbf x'=Mmathbf x$, we have $$mathbf x^TQmathbf x = (M^-1mathbf x')^T Q (M^-1mathbf x') = mathbf x'^T (M^-TQM^-1) mathbf x',$$ so the quadric transforms as $Q' = M^-TQM^-1$. In your case, $M=ST$, so when we transform the unit sphere $Q=I$, we get $$Q' = M^-TQM^-1 = (ST)^-T(ST)^-1.$$ The reciprocal square roots of the eigenvalues of this matrix then are the half-axis lengths of the resulting ellipsoid and the corresponding eigenvectors, chosen to form an orthogonal set in case of repeated eigenvalues, are its principal axes.
† The fully-general quadric equation that includes linear terms and and arbitrary constant term can also be represented by a single $4times4$ matrix by using homogeneous coordinates. All of this applies to quadrics of any dimension.
answered Mar 27 at 18:11
amdamd
31.6k21052
$endgroup$
add a comment |
$begingroup$
The equation $Ax^2+By^2+Cz^2+Exy+Fxz+Gyz=1$ of a quadric surface can be expressed in matrix form as $$beginbmatrixx&y&zendbmatrix beginbmatrixA&frac E2&frac F2\frac E2&B&frac G2\frac F2&frac G2&Cendbmatrix beginbmatrixx\y\zendbmatrix = mathbf x^TQmathbf x = 1.$$ The symmetric matrix $Q$ can be said to represent the quadric.† In particular, the unit sphere is represented by the identity matrix.
The principal axis theorem tells us that there’s an orthonormal basis of eigenvectors of $Q$ that gives the axes of this quadric and that the half-axis lengths are given by the reciprocal square roots of the (absolute values) of the corresponding eigenvalues: after a change of basis that’s an isometry because the basis is orthonormal, the equation of the quadric takes the form $lambda_1x'^2+lambda_2y'^2+lambda_3z'^2=1$.
Given the invertible point transformation $mathbf x'=Mmathbf x$, we have $$mathbf x^TQmathbf x = (M^-1mathbf x')^T Q (M^-1mathbf x') = mathbf x'^T (M^-TQM^-1) mathbf x',$$ so the quadric transforms as $Q' = M^-TQM^-1$. In your case, $M=ST$, so when we transform the unit sphere $Q=I$, we get $$Q' = M^-TQM^-1 = (ST)^-T(ST)^-1.$$ The reciprocal square roots of the eigenvalues of this matrix then are the half-axis lengths of the resulting ellipsoid and the corresponding eigenvectors, chosen to form an orthogonal set in case of repeated eigenvalues, are its principal axes.
† The fully-general quadric equation that includes linear terms and and arbitrary constant term can also be represented by a single $4times4$ matrix by using homogeneous coordinates. All of this applies to quadrics of any dimension.
answered Mar 27 at 18:11
amdamd
31.6k21052
$endgroup$
The equation $Ax^2+By^2+Cz^2+Exy+Fxz+Gyz=1$ of a quadric surface can be expressed in matrix form as $$beginbmatrixx&y&zendbmatrix beginbmatrixA&frac E2&frac F2\frac E2&B&frac G2\frac F2&frac G2&Cendbmatrix beginbmatrixx\y\zendbmatrix = mathbf x^TQmathbf x = 1.$$ The symmetric matrix $Q$ can be said to represent the quadric.† In particular, the unit sphere is represented by the identity matrix.
The principal axis theorem tells us that there’s an orthonormal basis of eigenvectors of $Q$ that gives the axes of this quadric and that the half-axis lengths are given by the reciprocal square roots of the (absolute values) of the corresponding eigenvalues: after a change of basis that’s an isometry because the basis is orthonormal, the equation of the quadric takes the form $lambda_1x'^2+lambda_2y'^2+lambda_3z'^2=1$.
Given the invertible point transformation $mathbf x'=Mmathbf x$, we have $$mathbf x^TQmathbf x = (M^-1mathbf x')^T Q (M^-1mathbf x') = mathbf x'^T (M^-TQM^-1) mathbf x',$$ so the quadric transforms as $Q' = M^-TQM^-1$. In your case, $M=ST$, so when we transform the unit sphere $Q=I$, we get $$Q' = M^-TQM^-1 = (ST)^-T(ST)^-1.$$ The reciprocal square roots of the eigenvalues of this matrix then are the half-axis lengths of the resulting ellipsoid and the corresponding eigenvectors, chosen to form an orthogonal set in case of repeated eigenvalues, are its principal axes.
† The fully-general quadric equation that includes linear terms and and arbitrary constant term can also be represented by a single $4times4$ matrix by using homogeneous coordinates. All of this applies to quadrics of any dimension.
answered Mar 27 at 18:11
amdamd
31.6k21052
edited Mar 27 at 18:16
answered Mar 27 at 18:11
amdamd
31.6k21052
answered Mar 27 at 18:11
amdamd
31.6k21052
answered Mar 27 at 18:11
amdamd
31.6k21052
31.6k21052
add a comment |
add a comment |
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%2f3159618%2fprincipal-axes-of-deformed-ellipsoid%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
$begingroup$
BTW, I'm expressing my vectors as columns, so the order of T'=ST is correct.
$endgroup$
– FragileX
Mar 23 at 18:39
$begingroup$
The resulting ellipsoid can be represented by the matrix $(ST)^-T(ST)^-1$.
$endgroup$
– amd
Mar 24 at 3:41
$begingroup$
@amd That produces eigenvectors pointing in the right direction, but the eigenvalues are not the right size.
$endgroup$
– FragileX
Mar 24 at 18:55
$begingroup$
The eigenvalues from $(ST)^-T(ST)^-1$ become the expected axis semi-lengths if I take the reciprocal of their square roots. I kind of figured that out by trial-and-error, though. Can someone who understands what's going on please write these comments up as a full answer?
$endgroup$
– FragileX
Mar 24 at 19:30