A group is a set with a defined product. This product is associative. The group is closed under this product. We have an identity element and each element has an inverse. Formally,

Definition (1.10.1)

A group is a (nonempty) set $G$ with a binary operation (a product) $G \times G \rightarrow G$ satisfying the following properties:

The group is closed under the operation.
The product is associative. For all $a, b, c \in G$, we have $(ab)c = a(bc)$.
There is an identity element $e \in G$ with property that for all $a \in G$, $ea = ae = a$.
For each element $a \in G$, there is an element $a^{-1}\ \in G$ satisfying $aa^{-1} = a^{-1}a = e$ (The inverse).

What are examples of the groups?

$\mathbf{R}$ with the addition operation.
$\mathbf{R}^{x} = \mathbf{R} - \{0\}$ with the multiplication operation.
$\mathbf{Z}$ with the addition operation.
The set of invertible $n \times n$ matrices with entries in $\mathbf{R}$ with matrix multiplication as the product.
Any vector space is a group if you forget about the scalar multiplication.
For any set $T$, define the set of all bijections $g: T \rightarrow T$. The set of all bijections is the symmetric group of $T$. The operation here is the composition of these maps.

Example

Suppose we define the group $(\mathbf{R}, \ast)$ where $\ast$ is defined as

$$ \begin{align*} x \ast y = \sqrt[3]{x^3 + y^3} \end{align*} $$

We can further check that this product satisfies the three axioms of a group.

Example

Let $F$ be a field (for example $\mathbf{R}$ or $\mathbf{C}$). Let $Mat_{n \times n}(F)$ be the set of matrices with entries in $F$. Then define $GL_n(F)$ (General Linear Group) as a subset of $A \in M_{n \times n}(F)$ where $A$ is an invertible matrix.

The set $GL_n(F)$ equipped with matrix multiplication is a group. It satisfies the three axioms

The group is closed under multiplication because multiplying two invertible matrices is another invertible matrix. $(AB)^{-1} = B^{-1}A^{-1}$
The identity element is the identity matrix.
Matrix multiplication is associative.

Rotations in Plane

We can define a plane rotation by an angle counter clockwise by the left multiplication by the following rotation matrix

$$ \begin{align*} Rot(\theta) &= \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix} \end{align*} $$

The rotation matrix has some properties

$\text{Rot}(0) = I$ is the identity matrix.
$\text{Rot}(\alpha)Rot(\beta) = \text{Rot}(\alpha + \beta)$.
$\text{Rot}(\alpha)^{-1} = \text{Rot}(-\alpha)$.
$\text{Rot}(\alpha + 2\pi n) = \text{Rot}(\alpha)$ where $n \in \mathbf{Z}$.

From this we see that the collection of rotation matrices forms a group with matrix multiplication.

Rotations in Space

$$ \begin{align*} \text{Rot}_{e_3}(\theta) &= \begin{pmatrix} \cos\theta & -\sin\theta & 0 \\ \sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{pmatrix} \end{align*} $$

This defines a rotation around the $z-$axis by angle $\theta$ counter clockwise viewed from the head of the vector $e_3$ where $e_3 = (0,0,1)$. But what if we wanted to rotate around a different vector? let $u \in \mathbf{R}^3$ such that $\lVert {u} \rVert = 1$ and let $\text{Rot}_u(\theta)$ be a rotation matrix around the axis through $u$ by angle $\theta$ counterclockwise when viewed from the head of $u$. How do we compute such a matrix?

How Do you Compute The Rotation Matrix?

Recall from Linear Algebra that

Definition

An orthogonal matrix is a square matrix $P \in Mat_{n \times n}(\mathbf{R})$ such that $P^{T}P = I$ or $P^{-1} = P^{T}$.

Equivalently the columns of $P$ are an orthonormal basis of $\mathbf{R}^n$ and the rows of $P$ are also orthonormal basis of $\mathbf{R}^n$. Based on this, we have a useful formula where if we’re given $u \in \mathbf{R}^3$ such that $\lVert u \rVert = 1$, $\theta \in \mathbf{R}$ and $P$ an orthogonal matrix, then $P \text{Rot}_u(\theta) P^{-1}$ is actually a rotation matrix around the vector $Pu$. So

$$ \begin{align*} \text{Rot}_{Pu}(\theta) = P \text{Rot}_u(\theta) P^{-1} \end{align*} $$

Why is this a useful formula? Given $\lVert u \rVert = 1$ and some angle $\theta$, construct an orthonormal basis using Gram-Schmidt by setting $u_3 = u$ and finding two more perpendicular vectors (normalized) $u_1, u_2 \in \mathbf{R}^3$. Let $P = [u_1 \quad u_2 \quad u_3]$. This means that $Pe_3 = u_3 = u$. Therefore

$$ \begin{align*} \text{Rot}_{u}(\theta) = P \text{Rot}_{e_3}(\theta) P^{T} \end{align*} $$

So we’re taking the standard rotation around the $z$-axis or $e_3$ and changing the basis from the standard vectors $e_1, e_2, e_3$ to $u_1, u_2, u_3$ so now the rotation is actually around $u_3$ instead.

Example

So now suppose we want to rotate around $\frac{(e_1 + e_2)}{\sqrt{2}}$ (which is a unit vector) by $\theta = \frac{\pi}{3}$. Then, we’ll need two more orthonormal vectors in addition to $\frac{(e_1 + e_2)}{\sqrt{2}}$ to construct an orthonormal basis. We can use Gram-Schmidt to come up with the following orthonormal vectors and set them to be the column vectors of $P$ as follows

$$ \begin{align*} P &= \begin{pmatrix} 0 & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ 0 & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ 1 & 0 & 0 \end{pmatrix} \end{align*} $$

The rotation $\text{Rot}_{e_3}$ is as follows

$$ \begin{align*} \text{Rot}_{e_3}(\frac{\pi}{3}\big) &= \begin{pmatrix} \frac{1}{2} & -\frac{\sqrt{3}}{2} & 0 \\ \frac{\sqrt{3}}{2} & \frac{1}{2} & 0 \\ 0 & 0 & 1 \end{pmatrix} \end{align*} $$

Therefore,

$$ \begin{align*} \text{Rot}_{\frac{e_1 + e_2}{\sqrt{2}}}\big(\frac{\pi}{3}\big) &= P \text{Rot}_{e_3}(\frac{\pi}{3}\big) P^{T} \\ &= \begin{pmatrix} \frac{3}{4} & \frac{1}{4} & \frac{\sqrt{3}}{8} \\ \frac{3}{4} & \frac{3}{4} & -\frac{\sqrt{3}}{8} \\ -\frac{\sqrt{3}}{8} & \frac{\sqrt{3}}{8} & \frac{1}{2} \end{pmatrix} \end{align*} $$

Rotations in Space

From last lecture, we saw the symmetries of the square. These symmetries can be represented with rotation matrices. We have the identity rotation $I$. We can also define the rotation around the axis coming through the centroid of the face ($z-$axis in the lecture) as $R = \text{Rot}_{e_3}(\frac{\pi}{2})$. Therefore, $R^2 = \text{Rot}_{e_3}(\pi)$ and $R^3 = \text{Rot}_{e_3}(\frac{3\pi}{2})$.

What about the $180$ degrees rotation around the $x-axis$ labeled as $a$ from last time? We can define it as $A = \text{Rot}_{e_1}(\pi)$. Similarly, $B = \text{Rot}_{e_2}(\pi)$, $C = \text{Rot}_{\frac{e_1 - e_2}{\sqrt{2}}}(\pi)$ and $D = \text{Rot}_{\frac{e_1 + e_2}{\sqrt{2}}}(\pi)$

Rotations in space have also the following properties / identities

$\text{Rot}_u(0) = I$ is the identity matrix where $u$ is any unit vector.
$\text{Rot}_u(\theta + 2\pi n) = \text{Rot}_u(\theta)$ where $n \in \mathbf{Z}$.
$\text{Rot}_u(\theta) = \text{Rot}_{-u}(-\theta)$.
$\text{Rot}_u(\theta)^{-1} = \text{Rot}_{u}(-\theta) = \text{Rot}_{-u}(\theta)$.
$\text{Rot}_u(\alpha)Rot_v(\beta) = a$ is a rotation matrix.
$\text{Rot}_u(\alpha)Rot_u(\beta) = Rot_u(\alpha + \beta)$.

But why is the last statement true? To figure it out, we need to introduce another definition

Definition

A special orthogonal matrix is an $A \in Mat_{n \times n}(\mathbf{R})$ such that

$A^TA = I$. (The orthogonal property)
$\det(A) = \pm 1$. (This is as a result of the fact that $det(A^{-1}) = \frac{1}{\det(A)}$ and that $\det(A) = \det(A^T)$

And now we have the following proposition

Proposition

$A \in Mat_{3 \times 3}(\mathbf{R})$ is a rotation matrix if and only if it is special orthogonal.

We denote these matrices special matrices with

$$ \begin{align*} SO(n) = \{ A \in Mat_{n \times n} \text{ special orthogonal} \} \subseteq O(n) \subseteq GL_n(\mathbf{R}) \end{align*} $$

Observation: $A, B \in SO(n)$ implies that $AB \in SO(n)$. To show that this is true, we need to show that properties of special orthogonal matrices hold. To see that, observe that

$ (AB)^TAB = B^TA^TAB = B^TB = I $
$ \det(AB) = \det(A)\det(B) = 1 $

Also observe that the identity matrix is in $SO(n)$ and that for any $A \in SO(n)$, $A^{-1} = A^{T} \in SO(n)$. Therefore $SO(n)$ is a group with matrix multiplication.

So now we know that the collection of special orthogonal matrices is a group. Therefore, if the proposition we introduced earlier holds (where we said that $A \in Mat_{3 \times 3}(\mathbf{R})$ is a rotation matrix if and only if it is special orthogonal), then we can also conclude that the product of two rotation matrices is also a rotation matrix. So let’s sketch the proof of the proposition

Proof
$\Rightarrow$ ($\text{Rot}_u(\theta) \in SO(3)$):
Let $\text{Rot}_u(\theta) = P\text{Rot}_{e_3}(\theta)P^{-1}$ for some chosen $P \in O(3)$. We want to show that $P\text{Rot}_{e_3}(\theta)P^{-1} \in SO(3)$. To do so we need to show that if $A \in SO(n)$ and $P \in O(n)$, then $PAP^{-1} = PAP^{T} \in SO(3)$. We can easily show this by verifying the two properties of special orthogonal matrices. For example,

$$ \begin{align*} (PAP^{T})^{T}PAP^{T} &= PA^TP^TPAP^{T} \\ &= PA^TAP^{T} \quad \text{ (because $P \in O(n)$)} \\ &= PP^{T} = I. \end{align*} $$

The verification that the determinat is 1 is also the same. So now what’s left is to show that $\text{Rot}_{e_3}(\theta)$ is in $SO(3)$. We can verify this because we know the exact matrix of $\text{Rot}_{e_3}(\theta)$ so we can computationally verify that it is a special orthogonal matrix. Finally, since $P \in O(n)$ and we just showed that $\text{Rot}_{e_3}(\theta) \in SO(3)$, then $P\text{Rot}_{e_3}(\theta)P^{T}$ is in $SO(3)$ as we wanted to show.

$\Leftarrow$ ($A \in SO(3) \implies A = \text{Rot}_u(\theta)$):
For this we’ll use the fact that if $A \in SO(3)$, then $1$ is an eigenvalue of $A$. So let $u$ be the eigenvector of $A$ corresponding to $\lambda = 1$ so $Au = u$. Choose $\lVert u \rVert = 1$. Now let $u_3 = u$ and form an orthonormal basis $\{u_1, u_2, u_3\}$. Now Let $P = [u_1 \quad u_2 \quad u_3]$ and let $B = P^{-1}AP = P^{T}AP$. We know that $Pe_3 = u_3$ but $u_3$ is an eigenvector of $A$.

$$ \begin{align*} Be_3 = P^{-1}APe_3 = P^{-1}Au_3 = P^{-1}u_3 = e_3 \end{align*} $$

So $e_3$ is an eigenvector of $B$ … then by the fact we used in the left direction, we can conclude that $B \in SO(3)$ (Why?). So this means that $B$ has orthonormal columns since it’s in $SO(3)$ and the third column specifically is $e_3$. So

$$ \begin{align*} B &= \begin{pmatrix} a & b & 0 \\ c & d & 0 \\ 0 & 0 & 1 \end{pmatrix} \end{align*} $$

By an algebraic argument we can show that $a = d = \cos\theta$ and $c = -b = \sin\theta$. So $B$ must be $\text{Rot}_{e_3}$. Therefore, $A = P\text{Rot}_{e_3}P^{T}$….

References

MATH417 by Charles Rezk
Algebra: Abstract and Concrete by Frederick M. Goodman