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I am confident that, other than the identity matrix any $3\times3$ matrix $\mathfrak{T}$ of real numbers such that $\det \mathfrak{T}=1,$ and $\mathfrak{T}^{-1}=\mathfrak{T}^{T}$ expresses a rotation about some fixed axis. That is, every non-zero vector is actively transformed about the same axis, by the same angle. Apparently this means the matrix has exactly one unit eigenvector.

As is shown below, if we have an axle and an angle, we can produce a rotation tensor which has a matrix representation. Since it is designed to rotate a figure rigidly, we can argue from geometry that it has a unit determinant. Since, as is show, the inverse equals the transpose, and the determinant is the same for a matrix and its transpose, we can conclude algebraically that the determinant is 1.

So, if we have a rotation by a known angle about an axis, we can produce a matrix satisfying the conditions of orthogonality. How do we show that every orthogonal matrix of $SO\left(3\right)$ (except the identity matrix) represents a rotation about an axis?

Deriving the Rotation Tensor for Known Axle and Angle

If we are given an axis of rotation $\hat{\varrho}$ and an angle $\theta$ by which a vector $\mathfrak{r}_o$ is to be rotated, we can construct a rotation tensor appearing in Menzel's Mathematical Physics II-28. The figure shows the setup prior to rotating $\mathfrak{r}_o.$

enter image description here

Prestidigitation Identity

First we introduce the identity tensor $\mathfrak{I},$ and what I call the prestidigitation identity. The third and fourth lines justify the second. Index summation is intended.

\begin{align*} \mathfrak{I}:&=\hat{\mathfrak{e}}_i \hat{\mathfrak{e}}^i\\ \hat{\varrho }\times \mathfrak{r}&=\mathfrak{I}\cdot \hat{\varrho }\times \mathfrak{r}=\mathfrak{I}\times \hat{\varrho }\cdot \mathfrak{r}\\ &=\hat{\mathfrak{e}}_i \hat{\mathfrak{e}}^i\cdot \hat{\mathfrak{e}}_j \varrho ^j\times \mathfrak{r}\\ &=\hat{\mathfrak{e}}_i \hat{\mathfrak{e}}^i\times \hat{\mathfrak{e}}_j \varrho ^j\cdot \mathfrak{r} \end{align*}

Ad Hoc Basis

Next we resolve $\mathfrak{r}$ into a component $\mathfrak{r}_{o\parallel }$ parallel to $\hat\varrho,$ and a component $\mathfrak{r}_{o\perp}$ perpendicular to $\hat\varrho.$ As an intermediate step we obtain a vector $\mathfrak{r}_{on}$ normal to the plane spanned by $\hat\varrho$ and $\mathfrak{r}_o.$

\begin{align*} \mathfrak{r}_{o\parallel }&=\hat{\varrho } \hat{\varrho }\cdot \mathfrak{r}_o\\ \mathfrak{r}_{on}&=\hat{\varrho }\times \mathfrak{r}_o\\ \mathfrak{r}_{o\perp}&=\left(\hat{\varrho }\times \mathfrak{r}_o\right)\times \hat{\varrho }\\ &=\left(\mathfrak{r}_o\hat{\varrho } -\hat{\varrho } \mathfrak{r}_o\right)\cdot \hat{\varrho } \end{align*}

Obtaining the Rotation Tensor

Using $\mathfrak{r}_{o\perp}$ and $\mathfrak{r}_{on}$ as ad hoc orthogonal (not normalized) basis vectors, we rotate $\mathfrak{r}_{o\perp}$ by $\theta$ in the plane normal to $\hat\varrho,$ and apply prestidigitation. By rearranging the resulting expression, we are able to factor out the arbitrarily chosen argument vector $\mathfrak{r}_o$ to isolate the rotation tensor $\mathfrak{R}.$

\begin{align*} \mathfrak{r}&=\mathfrak{r}_{o\parallel }+\mathfrak{r}_{o\perp}\cos (\theta )+ \mathfrak{r}_{on}\sin (\theta )\\ &=\hat{\varrho}\hat{\varrho}\cdot\mathfrak{r}_{o}+\left(\mathfrak{r}_{o}\hat{\varrho}-\hat{\varrho}\mathfrak{r}_{o}\right)\cdot\hat{\varrho}\cos(\theta)+\hat{\varrho}\times\mathfrak{r}_{o}\sin(\theta)\\&=\hat{\varrho}\hat{\varrho}\cdot\mathfrak{r}_{o}+\left(\mathfrak{I}-\hat{\varrho}\hat{\varrho}\right)\cdot\mathfrak{r}_{o}\cos(\theta)+\mathfrak{I}\times\hat{\varrho}\cdot\mathfrak{r}_{o}\sin(\theta)\\ &=\left(\hat{\varrho}\hat{\varrho}\left(1-\cos(\theta)\right)+\mathfrak{I}\cos(\theta)+\mathfrak{I}\times\hat{\varrho}\sin(\theta)\right)\cdot\mathfrak{r}_{o}\\ &=\mathfrak{R}\cdot\mathfrak{r}_{o} \end{align*}

Symmetric $\mathfrak{S}$ and Anti-symmetric $\mathfrak{A}$ Parts of $\mathfrak{R}$

The terms of the rotation tensor involving $c=\cos(\theta)$, which we shall collectively call $\mathfrak{S}=\hat{\varrho}\hat{\varrho}\left(1-c\right)+\mathfrak{I}c,$ pertain to the axial direction of the argument vector. The term involving $s=\sin(\theta),$ which we shall call $\mathfrak{A}=\mathfrak{I}\times\hat{\varrho}s,$ relates to the oriented plane of rotation.

These tensor components exhibit symmetry and anti-symmetry in two ways. Firstly, due to the trigonometric properties of cosine (symmetric,) and sine (anti-symmetric). Secondly by transposition when expressed in matrix form. Transposing $\mathfrak{S}$ means reversing the order of the dyadic product $\hat{\varrho}\hat{\varrho}$ which is obviously symmetric. Since $\mathfrak{A}$ involves a cross product, we expect it to be anti-symmetric, but it will be useful to examine this further. Write $\rho^i$ for the direction cosines of $\hat\varrho,$ and observe that dotting with $\mathfrak{r}_o$ on the right is equivalent to multiplying the associated matrix by the column representation of $\mathfrak{r}_o$.

\begin{align*} \mathfrak{I}\times\hat{\varrho}&=\hat{\mathfrak{e}}_{i}\hat{\mathfrak{e}}_{j}\times\hat{\mathfrak{e}}_{k}\delta^{ij}\rho^{k}\\&=\left(\begin{aligned}\hat{\mathfrak{e}}_{1}\hat{\mathfrak{e}}_{1} & \times\left(\hat{\mathfrak{e}}_{2}\rho^{2}+\hat{\mathfrak{e}}_{3}\rho^{3}\right)\\ +\hat{\mathfrak{e}}_{2}\hat{\mathfrak{e}}_{2} & \times\left(\hat{\mathfrak{e}}_{3}\rho^{3}+\hat{\mathfrak{e}}_{1}\rho^{1}\right)\\ +\hat{\mathfrak{e}}_{3}\hat{\mathfrak{e}}_{3} & \times\left(\hat{\mathfrak{e}}_{1}\rho^{1}+\hat{\mathfrak{e}}_{2}\rho^{2}\right) \end{aligned} \right)\\&=\left(\begin{aligned}\hat{\mathfrak{e}}_{1} & \left(\hat{\mathfrak{e}}_{3}\rho^{2}-\hat{\mathfrak{e}}_{2}\rho^{3}\right)\\ +\hat{\mathfrak{e}}_{2} & \left(\hat{\mathfrak{e}}_{1}\rho^{3}-\hat{\mathfrak{e}}_{3}\rho^{1}\right)\\ +\hat{\mathfrak{e}}_{3} & \left(\hat{\mathfrak{e}}_{2}\rho^{1}-\hat{\mathfrak{e}}_{1}\rho^{2}\right) \end{aligned} \right)\\&=\begin{bmatrix}0 & -\rho^{3} & +\rho^{2}\\ +\rho^{3} & 0 & -\rho^{1}\\ -\rho^{2} & +\rho^{1} & 0 \end{bmatrix} \end{align*}

The Transpose of $\mathfrak{R}$ is its Inverse

It is clear from geometric reasoning that reversing the sign of $\theta$ will produce the inverse rotation tensor $\mathfrak{R}^{-1}$. Since this amounts to changing the sign of $\mathfrak{A},$ it is clear that the transpose of $\mathfrak{R}$ is its inverse.

\begin{align*} \mathfrak{R}&=\mathfrak{S}+\mathfrak{A}\\ \mathfrak{R}^{-1}&=\mathfrak{S}-\mathfrak{A}\\ \mathfrak{S}^{T}&=+\mathfrak{S}\\ \mathfrak{A}^{T}&=-\mathfrak{A}\\ \mathfrak{R}^{-1}&=\mathfrak{R}^{T} \end{align*}

Using Legerdemain to Show $\mathfrak{R}\cdot\mathfrak{R}^{-1}=\mathfrak{I}$

All of this is good and well, but we should still show that our geometric argument actually produces the inverse. That is $$\mathfrak{R}\cdot\mathfrak{R}^{-1}=\left(\mathfrak{S}+\mathfrak{A}\right)\cdot\left(\mathfrak{S}-\mathfrak{A}\right)=\mathfrak{I}.$$

By linearity we have \begin{align*} \mathfrak{R}\cdot\mathfrak{R}^{-1}&=\left(\mathfrak{S}+\mathfrak{A}\right)\cdot\left(\mathfrak{S}-\mathfrak{A}\right)\\ &=\mathfrak{S}\cdot\mathfrak{S}-\mathfrak{A}\cdot\mathfrak{A}. \end{align*}

Examining the product $-\mathfrak{A}\cdot\mathfrak{A}$ first will give us the clue we need to complete the task. Computing this using the dyadic product is a bit intractable, but the matrix product is straight forward. We use the properties of direction cosines to obtain the form shown.

\begin{align*} \left(s\mathfrak{I}\times\hat{\varrho}\right)\cdot\left(-s\mathfrak{I}\times\hat{\varrho}\right) &=s^2\left[\begin{array}{ccc} 0 & -\rho^{z} & \rho^{y}\\ \rho^{z} & 0 & -\rho^{x}\\ -\rho^{y} & \rho^{x} & 0 \end{array}\right]\left[\begin{array}{ccc} 0 & \rho^{z} & -\rho^{y}\\ -\rho^{z} & 0 & \rho^{x}\\ \rho^{y} & -\rho^{x} & 0 \end{array}\right]\\ &=s^2\left[\begin{array}{ccc} 1-\rho^{x}\rho^{x} & -\rho^{x}\rho^{y} & -\rho^{x}\rho^{z}\\ -\rho^{y}\rho^{x} & 1-\rho^{y}\rho^{y} & -\rho^{y}\rho^{z}\\ -\rho^{z}\rho^{x} & -\rho^{z}\rho^{y} & 1-\rho^{z}\rho^{z} \end{array}\right]\\ &=s^2\left(\mathfrak{I}-\hat{\varrho}\hat{\varrho}\right) \end{align*}

We make the following observations \begin{align*} \hat{\varrho}\hat{\varrho}&=\hat{\varrho}\hat{\varrho}\cdot\hat{\varrho}\hat{\varrho} =\mathfrak{I}\cdot\hat{\varrho}\hat{\varrho}=\hat{\varrho}\hat{\varrho}\cdot\mathfrak{I}\\ 0 &=\hat{\varrho}\hat{\varrho}\cdot\left(\mathfrak{I}-\hat{\varrho}\hat{\varrho}\right) =\left(\mathfrak{I}-\hat{\varrho}\hat{\varrho}\right)\cdot\hat{\varrho}\hat{\varrho}\\ \left(\mathfrak{I}-\hat{\varrho}\hat{\varrho}\right)\cdot\left(\mathfrak{I}-\hat{\varrho}\hat{\varrho}\right) &=\mathfrak{I}\cdot\left(\mathfrak{I}-\hat{\varrho}\hat{\varrho}\right) -\hat{\varrho}\hat{\varrho}\cdot\left(\mathfrak{I}-\hat{\varrho}\hat{\varrho}\right)\\ &=\left(\mathfrak{I}-\hat{\varrho}\hat{\varrho}\right)\\ \mathfrak{S}&=\hat{\varrho}\hat{\varrho}\left(1-c\right)+\mathfrak{I}c\\ &=c \left(\mathfrak{I}-\hat{\varrho } \hat{\varrho }\right)+\hat{\varrho } \hat{\varrho }\\ \mathfrak{S}\cdot\mathfrak{S}&=c^{2}\left(\mathfrak{I}-\hat{\varrho}\hat{\varrho}\right)\cdot\left(\mathfrak{I}-\hat{\varrho}\hat{\varrho}\right)+\left(\mathfrak{I}-\hat{\varrho}\hat{\varrho}\right)\cdot\hat{\varrho}\hat{\varrho}+\hat{\varrho}\hat{\varrho}\cdot\left(\mathfrak{I}-\hat{\varrho}\hat{\varrho}\right)+\hat{\varrho}\hat{\varrho}\cdot\hat{\varrho}\hat{\varrho}\\ &=c^{2}\left(\mathfrak{I}-\hat{\varrho}\hat{\varrho}\right)+\hat{\varrho}\hat{\varrho} \end{align*}

Finally, we have the desired result.

\begin{align*} \mathfrak{R}\cdot\mathfrak{R}^{-1}&=\left(\mathfrak{S}+\mathfrak{A}\right)\cdot\left(\mathfrak{S}-\mathfrak{A}\right)\\ &=\mathfrak{S}\cdot\mathfrak{S}-\mathfrak{A}\cdot\mathfrak{A}\\ &=c^{2}\left(\mathfrak{I}-\hat{\varrho}\hat{\varrho}\right)+\hat{\varrho}\hat{\varrho}+s^2\left(\mathfrak{I}-\hat{\varrho}\hat{\varrho}\right)\\ &=\mathfrak{I} \end{align*}

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  • $\begingroup$ You will have two unit eigenvectors. $\endgroup$
    – Randall
    Feb 16, 2023 at 15:04
  • $\begingroup$ You can show your bold-faced question by showing that each such matrix has a 1-dim eigenspace, and that is your axis of rotation. That's not hard by characteristic polynomial stuff. $\endgroup$
    – Randall
    Feb 16, 2023 at 15:07
  • $\begingroup$ If you're ok with defining a rotation as a composition of two reflections, the you just need to prove the Cartan–Dieudonné theorem. $\endgroup$ Feb 16, 2023 at 16:17
  • $\begingroup$ en.wikipedia.org/wiki/Euler's_rotation_theorem $\endgroup$
    – anon
    Feb 16, 2023 at 17:18

1 Answer 1

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An orthogonal $3 \times 3$ matrix $A$ with determinant 1 has 1 as an eigenvalue1 Let $\mathbf w$ be a corresponding unit eigenvector. Let $\mathbf u$ be a unit vector orthogonal to $\mathbf w$. Let $\mathbf {v=w \times u}$. Then $$A\mathbf u=\cos \theta \mathbf u + \sin \theta \mathbf v,A\mathbf v=-\sin \theta \mathbf u + \cos \theta \mathbf v.$$ The angle $\theta$ is independent of the choice of $\mathbf u$.

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  • $\begingroup$ It's the part about the eigenvalue and eigenvector that I don't know how to do. I tried writing out the characteristic equation, but haven't an obvious way to use orthogonality to simplify it to a point where it makes sense to me. I am a bit rusty in that area, and can't say it was ever a strong point. $\endgroup$ Feb 17, 2023 at 0:20

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