Find Rotational Eigenvalues
$begingroup$
What are the complex eigenvalues of the matrix $A$ that represents a rotation of $mathbb{R}^3$ through the angle $theta$ about a vector $u$?
I know that an eigenvalue should be $1$, and the other $2$ should be complex, where one is the conjugate of the other. I can't seem to get it however.
eigenvalues-eigenvectors rotations
edited Nov 23 '18 at 20:17
Bernard
119k639112
asked Nov 23 '18 at 19:05
IUissoprettyIUissopretty
536
$endgroup$
add a comment |
$begingroup$
What are the complex eigenvalues of the matrix $A$ that represents a rotation of $mathbb{R}^3$ through the angle $theta$ about a vector $u$?
I know that an eigenvalue should be $1$, and the other $2$ should be complex, where one is the conjugate of the other. I can't seem to get it however.
eigenvalues-eigenvectors rotations
edited Nov 23 '18 at 20:17
Bernard
119k639112
asked Nov 23 '18 at 19:05
IUissoprettyIUissopretty
536
$endgroup$
$begingroup$
Can you do this in two dimensions? What are the complex eigenvalues of a typical rotation matrix?
$endgroup$
– Lord Shark the Unknown
Nov 23 '18 at 19:08
add a comment |
$begingroup$
What are the complex eigenvalues of the matrix $A$ that represents a rotation of $mathbb{R}^3$ through the angle $theta$ about a vector $u$?
I know that an eigenvalue should be $1$, and the other $2$ should be complex, where one is the conjugate of the other. I can't seem to get it however.
eigenvalues-eigenvectors rotations
edited Nov 23 '18 at 20:17
Bernard
119k639112
asked Nov 23 '18 at 19:05
IUissoprettyIUissopretty
536
$endgroup$
What are the complex eigenvalues of the matrix $A$ that represents a rotation of $mathbb{R}^3$ through the angle $theta$ about a vector $u$?
I know that an eigenvalue should be $1$, and the other $2$ should be complex, where one is the conjugate of the other. I can't seem to get it however.
eigenvalues-eigenvectors rotations
eigenvalues-eigenvectors rotations
edited Nov 23 '18 at 20:17
Bernard
119k639112
asked Nov 23 '18 at 19:05
IUissoprettyIUissopretty
536
edited Nov 23 '18 at 20:17
Bernard
119k639112
asked Nov 23 '18 at 19:05
IUissoprettyIUissopretty
536
edited Nov 23 '18 at 20:17
Bernard
119k639112
edited Nov 23 '18 at 20:17
Bernard
119k639112
edited Nov 23 '18 at 20:17
Bernard
119k639112
119k639112
asked Nov 23 '18 at 19:05
IUissoprettyIUissopretty
536
asked Nov 23 '18 at 19:05
IUissoprettyIUissopretty
536
asked Nov 23 '18 at 19:05
IUissoprettyIUissopretty
536
536
$begingroup$
Can you do this in two dimensions? What are the complex eigenvalues of a typical rotation matrix?
$endgroup$
– Lord Shark the Unknown
Nov 23 '18 at 19:08
add a comment |
$begingroup$
Can you do this in two dimensions? What are the complex eigenvalues of a typical rotation matrix?
$endgroup$
– Lord Shark the Unknown
Nov 23 '18 at 19:08
$begingroup$
Can you do this in two dimensions? What are the complex eigenvalues of a typical rotation matrix?
$endgroup$
– Lord Shark the Unknown
Nov 23 '18 at 19:08
$begingroup$
Can you do this in two dimensions? What are the complex eigenvalues of a typical rotation matrix?
$endgroup$
– Lord Shark the Unknown
Nov 23 '18 at 19:08
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The eigenvalues of a plane rotation of angle $theta$ are the eigenvalues of $left(begin{smallmatrix}costheta&-sintheta\sintheta&costhetaend{smallmatrix}right)$: $e^{pm itheta}$. So, since a rotation around $u$ with angle $theta$ maps $u$ into itself (and therefore $u$ is an engenvector with eigenvalue $1$), its eigenvalues are $1$ and $e^{pm itheta}$.
answered Nov 23 '18 at 19:10
José Carlos SantosJosé Carlos Santos
153k22123225
$endgroup$
$begingroup$
But since the rotation is with respect to some arbitrary axis, wouldn't the complex eigenvalue be different? Or does it not matter. That's the part I'm confused about.
$endgroup$
– IUissopretty
Nov 23 '18 at 19:55
$begingroup$
Let $e_1=frac u{lVert urVert}$, let $e_2$ be a vector with norm $1$ orthogonal to $e_1$ and let $e_3$ be a vector with norm $1$ orthogonal to both $e_1$ and $e_2$. Then the matrix of $A$ with respect to the basis $(e_1,e_2,e_3)$ is$$begin{bmatrix}1&0&0\0&costheta&mpsintheta\0&pmsintheta&costhetaend{bmatrix},$$whose eigenvalues are $1$ and $e^{pm itheta}$.
$endgroup$
– José Carlos Santos
Nov 23 '18 at 20:43
1
$begingroup$
It doesn't matter. As far as eigenvalues are concerned, a rotation in one plane is the same as a rotation by the same angle in another plane.
$endgroup$
– Robert Israel
Nov 23 '18 at 20:43
$begingroup$
Ok, I understand now. Thanks!
$endgroup$
– IUissopretty
Nov 23 '18 at 20:52
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%2f3010714%2ffind-rotational-eigenvalues%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 eigenvalues of a plane rotation of angle $theta$ are the eigenvalues of $left(begin{smallmatrix}costheta&-sintheta\sintheta&costhetaend{smallmatrix}right)$: $e^{pm itheta}$. So, since a rotation around $u$ with angle $theta$ maps $u$ into itself (and therefore $u$ is an engenvector with eigenvalue $1$), its eigenvalues are $1$ and $e^{pm itheta}$.
answered Nov 23 '18 at 19:10
José Carlos SantosJosé Carlos Santos
153k22123225
$endgroup$
$begingroup$
But since the rotation is with respect to some arbitrary axis, wouldn't the complex eigenvalue be different? Or does it not matter. That's the part I'm confused about.
$endgroup$
– IUissopretty
Nov 23 '18 at 19:55
$begingroup$
Let $e_1=frac u{lVert urVert}$, let $e_2$ be a vector with norm $1$ orthogonal to $e_1$ and let $e_3$ be a vector with norm $1$ orthogonal to both $e_1$ and $e_2$. Then the matrix of $A$ with respect to the basis $(e_1,e_2,e_3)$ is$$begin{bmatrix}1&0&0\0&costheta&mpsintheta\0&pmsintheta&costhetaend{bmatrix},$$whose eigenvalues are $1$ and $e^{pm itheta}$.
$endgroup$
– José Carlos Santos
Nov 23 '18 at 20:43
1
$begingroup$
It doesn't matter. As far as eigenvalues are concerned, a rotation in one plane is the same as a rotation by the same angle in another plane.
$endgroup$
– Robert Israel
Nov 23 '18 at 20:43
$begingroup$
Ok, I understand now. Thanks!
$endgroup$
– IUissopretty
Nov 23 '18 at 20:52
add a comment |
$begingroup$
The eigenvalues of a plane rotation of angle $theta$ are the eigenvalues of $left(begin{smallmatrix}costheta&-sintheta\sintheta&costhetaend{smallmatrix}right)$: $e^{pm itheta}$. So, since a rotation around $u$ with angle $theta$ maps $u$ into itself (and therefore $u$ is an engenvector with eigenvalue $1$), its eigenvalues are $1$ and $e^{pm itheta}$.
answered Nov 23 '18 at 19:10
José Carlos SantosJosé Carlos Santos
153k22123225
$endgroup$
$begingroup$
But since the rotation is with respect to some arbitrary axis, wouldn't the complex eigenvalue be different? Or does it not matter. That's the part I'm confused about.
$endgroup$
– IUissopretty
Nov 23 '18 at 19:55
$begingroup$
Let $e_1=frac u{lVert urVert}$, let $e_2$ be a vector with norm $1$ orthogonal to $e_1$ and let $e_3$ be a vector with norm $1$ orthogonal to both $e_1$ and $e_2$. Then the matrix of $A$ with respect to the basis $(e_1,e_2,e_3)$ is$$begin{bmatrix}1&0&0\0&costheta&mpsintheta\0&pmsintheta&costhetaend{bmatrix},$$whose eigenvalues are $1$ and $e^{pm itheta}$.
$endgroup$
– José Carlos Santos
Nov 23 '18 at 20:43
1
$begingroup$
It doesn't matter. As far as eigenvalues are concerned, a rotation in one plane is the same as a rotation by the same angle in another plane.
$endgroup$
– Robert Israel
Nov 23 '18 at 20:43
$begingroup$
Ok, I understand now. Thanks!
$endgroup$
– IUissopretty
Nov 23 '18 at 20:52
add a comment |
$begingroup$
The eigenvalues of a plane rotation of angle $theta$ are the eigenvalues of $left(begin{smallmatrix}costheta&-sintheta\sintheta&costhetaend{smallmatrix}right)$: $e^{pm itheta}$. So, since a rotation around $u$ with angle $theta$ maps $u$ into itself (and therefore $u$ is an engenvector with eigenvalue $1$), its eigenvalues are $1$ and $e^{pm itheta}$.
answered Nov 23 '18 at 19:10
José Carlos SantosJosé Carlos Santos
153k22123225
$endgroup$
The eigenvalues of a plane rotation of angle $theta$ are the eigenvalues of $left(begin{smallmatrix}costheta&-sintheta\sintheta&costhetaend{smallmatrix}right)$: $e^{pm itheta}$. So, since a rotation around $u$ with angle $theta$ maps $u$ into itself (and therefore $u$ is an engenvector with eigenvalue $1$), its eigenvalues are $1$ and $e^{pm itheta}$.
answered Nov 23 '18 at 19:10
José Carlos SantosJosé Carlos Santos
153k22123225
answered Nov 23 '18 at 19:10
José Carlos SantosJosé Carlos Santos
153k22123225
answered Nov 23 '18 at 19:10
José Carlos SantosJosé Carlos Santos
153k22123225
answered Nov 23 '18 at 19:10
José Carlos SantosJosé Carlos Santos
153k22123225
153k22123225
$begingroup$
But since the rotation is with respect to some arbitrary axis, wouldn't the complex eigenvalue be different? Or does it not matter. That's the part I'm confused about.
$endgroup$
– IUissopretty
Nov 23 '18 at 19:55
$begingroup$
Let $e_1=frac u{lVert urVert}$, let $e_2$ be a vector with norm $1$ orthogonal to $e_1$ and let $e_3$ be a vector with norm $1$ orthogonal to both $e_1$ and $e_2$. Then the matrix of $A$ with respect to the basis $(e_1,e_2,e_3)$ is$$begin{bmatrix}1&0&0\0&costheta&mpsintheta\0&pmsintheta&costhetaend{bmatrix},$$whose eigenvalues are $1$ and $e^{pm itheta}$.
$endgroup$
– José Carlos Santos
Nov 23 '18 at 20:43
1
$begingroup$
It doesn't matter. As far as eigenvalues are concerned, a rotation in one plane is the same as a rotation by the same angle in another plane.
$endgroup$
– Robert Israel
Nov 23 '18 at 20:43
$begingroup$
Ok, I understand now. Thanks!
$endgroup$
– IUissopretty
Nov 23 '18 at 20:52
add a comment |
$begingroup$
But since the rotation is with respect to some arbitrary axis, wouldn't the complex eigenvalue be different? Or does it not matter. That's the part I'm confused about.
$endgroup$
– IUissopretty
Nov 23 '18 at 19:55
$begingroup$
Let $e_1=frac u{lVert urVert}$, let $e_2$ be a vector with norm $1$ orthogonal to $e_1$ and let $e_3$ be a vector with norm $1$ orthogonal to both $e_1$ and $e_2$. Then the matrix of $A$ with respect to the basis $(e_1,e_2,e_3)$ is$$begin{bmatrix}1&0&0\0&costheta&mpsintheta\0&pmsintheta&costhetaend{bmatrix},$$whose eigenvalues are $1$ and $e^{pm itheta}$.
$endgroup$
– José Carlos Santos
Nov 23 '18 at 20:43
1
$begingroup$
It doesn't matter. As far as eigenvalues are concerned, a rotation in one plane is the same as a rotation by the same angle in another plane.
$endgroup$
– Robert Israel
Nov 23 '18 at 20:43
$begingroup$
Ok, I understand now. Thanks!
$endgroup$
– IUissopretty
Nov 23 '18 at 20:52
$begingroup$
But since the rotation is with respect to some arbitrary axis, wouldn't the complex eigenvalue be different? Or does it not matter. That's the part I'm confused about.
$endgroup$
– IUissopretty
Nov 23 '18 at 19:55
$begingroup$
But since the rotation is with respect to some arbitrary axis, wouldn't the complex eigenvalue be different? Or does it not matter. That's the part I'm confused about.
$endgroup$
– IUissopretty
Nov 23 '18 at 19:55
$begingroup$
Let $e_1=frac u{lVert urVert}$, let $e_2$ be a vector with norm $1$ orthogonal to $e_1$ and let $e_3$ be a vector with norm $1$ orthogonal to both $e_1$ and $e_2$. Then the matrix of $A$ with respect to the basis $(e_1,e_2,e_3)$ is$$begin{bmatrix}1&0&0\0&costheta&mpsintheta\0&pmsintheta&costhetaend{bmatrix},$$whose eigenvalues are $1$ and $e^{pm itheta}$.
$endgroup$
– José Carlos Santos
Nov 23 '18 at 20:43
$begingroup$
Let $e_1=frac u{lVert urVert}$, let $e_2$ be a vector with norm $1$ orthogonal to $e_1$ and let $e_3$ be a vector with norm $1$ orthogonal to both $e_1$ and $e_2$. Then the matrix of $A$ with respect to the basis $(e_1,e_2,e_3)$ is$$begin{bmatrix}1&0&0\0&costheta&mpsintheta\0&pmsintheta&costhetaend{bmatrix},$$whose eigenvalues are $1$ and $e^{pm itheta}$.
$endgroup$
– José Carlos Santos
Nov 23 '18 at 20:43
1
1
$begingroup$
It doesn't matter. As far as eigenvalues are concerned, a rotation in one plane is the same as a rotation by the same angle in another plane.
$endgroup$
– Robert Israel
Nov 23 '18 at 20:43
$begingroup$
It doesn't matter. As far as eigenvalues are concerned, a rotation in one plane is the same as a rotation by the same angle in another plane.
$endgroup$
– Robert Israel
Nov 23 '18 at 20:43
$begingroup$
Ok, I understand now. Thanks!
$endgroup$
– IUissopretty
Nov 23 '18 at 20:52
$begingroup$
Ok, I understand now. Thanks!
$endgroup$
– IUissopretty
Nov 23 '18 at 20:52
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%2f3010714%2ffind-rotational-eigenvalues%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
$begingroup$
Can you do this in two dimensions? What are the complex eigenvalues of a typical rotation matrix?
$endgroup$
– Lord Shark the Unknown
Nov 23 '18 at 19:08