A Matrix proof of Euler's theorem of rotation

[Kashmir]

The question arises the way Goldstein proves Euler theorem (3rd Ed pg 150-156 ) which says:
" In three-dimensional space, any displacement of a rigid body such that
a point on the rigid body remains fixed, is equivalent to a single rotation about some axis that runs through the fixed point".

The Matrix proof essentially takes an arbitrary $3 \times 3$ orthogonal matrix with real entries and shows that there is at least one vector $n\neq 0$ with $Rn=n$ that is an eigenvector with +1 as its eigenvalue .The author states that this proves the Eulers theorem, which I am not sure why this is true, since it seems that all we have shown is that are some vectors that are invariant along some line.

 

[PeroK]

The question arises the way Goldstein proves Euler theorem (3rd Ed pg 150-156 ) which says:
" In three-dimensional space, any displacement of a rigid body such that
a point on the rigid body remains fixed, is equivalent to a single rotation about some axis that runs through the fixed point".

The Matrix proof essentially takes an arbitrary $3 \times 3$ orthogonal matrix with real entries and shows that there is at least one vector $n\neq 0$ with $Rn=n$ that is an eigenvector with +1 as its eigenvalue .The author states that this proves the Eulers theorem, which I am not sure why this is true, since it seems that all we have shown is that are some vectors that are invariant along some line.

That's your axis of rotation. Sure, you need to show that if points along a given axis are fixed, then the transformation of the rigid body is a rotation about that axis.

 

[Kashmir]

That's your axis of rotation. Sure, you need to show that if points along a given axis are fixed, then the transformation of the rigid body is a rotation about that axis.

yes,that's what i thought . These conditions are necessary but might not be sufficient ,but then why doesn't author prove sufficiency ?

 

[Kashmir]

That's your axis of rotation. Sure, you need to show that if points along a given axis are fixed, then the transformation of the rigid body is a rotation about that axis.

Also I see that the matrix A is orthogonal so it preserves norm, which is a necessary condition but still it doesn't prove that the transformation A is a rotation.

 

[PeroK]

Also I see that the matrix A is orthogonal so it preserves norm, which is a necessary condition but still it doesn't prove that the transformation A is a rotation.

Once you have a fixed axis, there's not a lot of room for manoeuvre.

 

[vanhees71]

$$\newcommand{\uvec}[1]{\underline{#1}}$$
That's more or less Cayley's theorem: A rigid body with one point fixed is rotating around this point. Taking this point as the origin of both the space-fixed (inertial) reference frame and the body-fixed reference frame, there must be a rotation matrix mapping the body-fixed Cartesian (righthanded) basis to the space-fixed one
$$\vec{e}_i'=D_{ji} \vec{e}_j.$$
The position vector of a fixed point in the body $\vec{r}=r_i' \vec{e}_i'$ has $r_i'=\text{const}$ (that's the rigidity condition). The same point is described in the space-fixed frame by
$$\vec{r}=r_i' \vec{e}_i' = r_{i}' D_{ji} \vec{e}_j \; \Rightarrow\; r_j=D_{ji} r_i'.$$
So if we write $\uvec{r}=(r_1,r_2,r_3)^{\text{T}}$ and $\uvec{r}'=(r_1',r_2',r_3')$ you have
$$\uvec{r} = \hat{D} \uvec{r}'.$$
Now $\hat{D}$ is a orthogonal $(3 \times 3$-matrix with $\mathrm{det} \hat{D}=1$. Thus it's a normal matrix, i.e., $\hat{D} \hat{D}^{\text{T}}=\hat{D}^{\text{T}} \hat{D}=\hat{1}$, and thus it's diagonalizable. The characteristic polynomial of a real matrix is a real polynomial, and thus you have at least one real eigenvalue, because a real polynomial has only pairs of roots that are conjugate complex or real ones. Since it's a polynomial of degree 3 you must have at least one real eigenvalue. All eigenvalues obey $|\lambda|=1$. So the three eigenvalues must be $\lambda_1=\exp(\mathrm{i} \varphi)$, $\lambda_2=\exp(-\mathrm{i} \varphi)$, and $\lambda_3 \in \{\pm 1 \}$ with some real phase $\varphi$. Further $\lambda_1 \lambda_2 \lambda_3=\mathrm{det} \hat{D}=+1$, and from that $\lambda_3=1$. That means an $SO(3)$ matrix has always one eigenvalue 1. This means the matrix describes a rotation around the direction given by the corresponding real eigenvector.

Of course $\hat{D}$ is in general a function of time, and in general also the rotation axis will be in general time dependent. This momentary rotation axis is given by the momentary angular velocity of the rotating body, but that's another story.

 

[Gaussian97]

Also I see that the matrix A is orthogonal so it preserves norm, which is a necessary condition but still it doesn't prove that the transformation A is a rotation.

You need to prove that $A$ also has determinant 1. If you prove that, then $A$ will be a rotation, by definition.

$$\newcommand{\uvec}[1]{\underline{#1}}$$
That's more or less Cayley's theorem: A rigid body with one point fixed is rotating around this point. Taking this point as the origin of both the space-fixed (inertial) reference frame and the body-fixed reference frame, there must be a rotation matrix mapping the body-fixed Cartesian (righthanded) basis to the space-fixed one
$$\vec{e}_i'=D_{ji} \vec{e}_j.$$
The position vector of a fixed point in the body $\vec{r}=r_i' \vec{e}_i'$ has $r_i'=\text{const}$ (that's the rigidity condition). The same point is described in the space-fixed frame by
$$\vec{r}=r_i' \vec{e}_i' = r_{i}' D_{ji} \vec{e}_j \; \Rightarrow\; r_j=D_{ji} r_i'.$$
So if we write $\uvec{r}=(r_1,r_2,r_3)^{\text{T}}$ and $\uvec{r}'=(r_1',r_2',r_3')$ you have
$$\uvec{r} = \hat{D} \uvec{r}'.$$
Now $\hat{D}$ is a orthogonal $(3 \times 3$-matrix with $\mathrm{det} \hat{D}=1$. Thus it's a normal matrix, i.e., $\hat{D} \hat{D}^{\text{T}}=\hat{D}^{\text{T}} \hat{D}=\hat{1}$, and thus it's diagonalizable. The characteristic polynomial of a real matrix is a real polynomial, and thus you have at least one real eigenvalue, because a real polynomial has only pairs of roots that are conjugate complex or real ones. Since it's a polynomial of degree 3 you must have at least one real eigenvalue. All eigenvalues obey $|\lambda|=1$. So the three eigenvalues must be $\lambda_1=\exp(\mathrm{i} \varphi)$, $\lambda_2=\exp(-\mathrm{i} \varphi)$, and $\lambda_3 \in \{\pm 1 \}$ with some real phase $\varphi$. Further $\lambda_1 \lambda_2 \lambda_3=\mathrm{det} \hat{D}=+1$, and from that $\lambda_3=1$. That means an $SO(3)$ matrix has always one eigenvalue 1. This means the matrix describes a rotation around the direction given by the corresponding real eigenvector.

Of course $\hat{D}$ is in general a function of time, and in general also the rotation axis will be in general time dependent. This momentary rotation axis is given by the momentary angular velocity of the rotating body, but that's another story.

If I understand it correctly, the OP is not asking to prove why there is a fixed axis. But a proof that the 3D rotation is equivalent to a 2D rotation about that axis.

 

[vanhees71]

That's easy. If you have a rotation around a fixed axis, you can make it to the 3-axis of a Cartesian coordinate system. Then the rotation acts on vector components with a matrix
$$\hat{D}=\begin{pmatrix} a & b & 0 \\ c & d & 0 \\0 & 0 & 1\end{pmatrix},$$
i.e., only the 1- and 2-components are changed, i.e., the rotation is in this plane. It's also easy to show that
$a=\cos \varphi$, $b=\sin \varphi$, $b=-\sin \varphi$, $d=\cos \varphi$ is the most general rotation matrix of this kind ($\varphi \in [0,2 \pi)$).