Find Rotational Eigenvalues












0












$begingroup$


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







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  • $begingroup$
    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
















0












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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














0












0








0





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 23 '18 at 20:17









Bernard

119k639112




119k639112










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


















  • $begingroup$
    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










1 Answer
1






active

oldest

votes


















1












$begingroup$

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 $e^{pm itheta}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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











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1 Answer
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1 Answer
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active

oldest

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active

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1












$begingroup$

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 $e^{pm itheta}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















1












$begingroup$

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 $e^{pm itheta}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














1












1








1





$begingroup$

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 $e^{pm itheta}$.






share|cite|improve this answer









$endgroup$



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 $e^{pm itheta}$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • $begingroup$
    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


















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