Map same thing as a Matrix.

Question

A rigid motion of $\mathbb {R^{n}}$ is a map

$M = T ◦ R : \mathbb {R^{n}} \to \mathbb {R^{n}}$

where $ \mathbb {R^{n}} \to \mathbb {R^{n}} $ is linear and satisfies $|R(v)| = |v| \space \space \space \space \space \space ∀v \in \mathbb {R^{n}}$

And T : $ \mathbb {R^{n}} \to \mathbb {R^{n}} $ is translation, i.e. there exists $ w \in \mathbb {R^{n}}$ such that $T (v) = v + w$ $\space \space \space \space \space∀v \in \mathbb {R^{n}} $

Argue that the function R is represented by a matrix $ R(v) = A(v)$ where A is n x n and satisfies $ A^T A = I $

Im not really familiar with rotational mappings but in Dynamical Systems we used rotational and translation matrices to move solutions around.

How do i argue that a rotational map is the same thing as a rotation matrix?

Answer 1

Let $R$ be the map and $e_1,e_2,\dots,e_n$ be the standard basis. We have that $\langle R(e_j),R(e_j)\rangle=1$ for all $j$.

We also have $\sqrt{2}=||e_i+e_j||=||R(e_i+e_j)||=\sqrt{R(e_i+e_j),R(e_i+e_j)\rangle}=\sqrt{\langle R(e_i),R(e_i)\rangle +\langle R(e_j),R(e_j)\rangle +2\langle R(e_i),R(e_j)\rangle}$.

This implies that $\langle e_i, e_j\rangle=0$ if $i\neq j$ and $1$ if $i=j$.

Now notice that the element $i,j$ of the matrix $R^tR$ is equal to $\langle e_i,e_j\rangle$, so we are good to go.

Answer 2

This doesn't really have to do with dynamical systems per se. There are two well known problems here, and I suggest you tackle them separately.

$(1):$ Show that every isometry of $\mathbb{R}^n$ that fixes the origin is linear.

$(2):$ Show that for linear maps $T:\mathbb{R}^n\longrightarrow\mathbb{R}^n$ the following are equivalent

• $T$ is orthogonal, that is, $TT^t=T^tT=I$
• $T$ preserves angles, that is, $\langle u,v\rangle=\langle Tu,Tv\rangle$
• $T$ is an isometry, that is, $\lVert v\rVert = \lVert Tv\rVert$