Special Orthogonal Group over Real Numbers

The Special Orthogonal Group over Real Numbers, denoted as $ SO(n, \mathbb{R}) $, includes all $ n \times n $ square matrices with real elements that are both orthogonal and have a determinant of one. Orthogonality in a matrix $ A $ means it satisfies $ A^TA = I $, which means the matrix multiplied by its transpose results in the identity matrix $ I $.

The determinant condition, $ \det(A) = 1 $, is crucial because it ensures the transformation maintains the orientation of space.

Essentially, this group represents the collection of rigid motions within $ \mathbb{R}^n $ that preserve the original orientation of objects and keep the origin static.

Orientation Preservation
This implies that such movements do not flip the object; its initial orientation remains unchanged.
For instance, a shape is not mirrored or turned over during the transformation.
Origin Stability
This means these movements do not displace the origin point of the $ R^n $ space; hence, if the movement is applied to the entire space, the origin remains unmoved.
For example, the Special Orthogonal Group consists of all rotations around the origin in n-dimensional space that do not flip the object.

In other words, after a rigid motion, the shape and size of an object remain unchanged, although its position in space may be altered.

Such movements are vital in numerous applications because they allow the manipulation of objects in space without changing their shape or volume.

Example

Consider the following matrix:

$$ A = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} $$

This is a specific 2x2 example in $ SO(2, \mathbb{R}) $ because it fulfills all the criteria of the Special Orthogonal Group over Real Numbers.

It's a square matrix. The number of rows is equal to the number of columns.
The determinant of the matrix is one $$ \det(A) = 0 \cdot 0 - (-1) \cdot 1 = 1 $$
It's orthogonal because the product of matrix A with its inverse $ A^{-1} $ equals the identity matrix. $$ AA^{-1} = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I $$

This matrix represents a 90-degree (or $ \frac{\pi}{2} $ radians) counterclockwise rotation around the origin of the Cartesian plane.

When we apply this matrix to a point, say $ (x, y) $, it yields the new coordinates of the point after rotation.

For instance, applying the matrix to point $ P(1, 0) $:

\[
\begin{pmatrix}
0 & -1 \\
1 & 0
\end{pmatrix}
\begin{pmatrix}
1 \\
0
\end{pmatrix}
=
\begin{pmatrix}
0 \cdot 1 + (-1) \cdot 0 \\
1 \cdot 1 + 0 \cdot 0
\end{pmatrix}
=
\begin{pmatrix}
0 \\
1
\end{pmatrix}
\]

Thus, the point $ P(1, 0) $, after undergoing a 90-degree counterclockwise rotation, moves to $ P' (0, 1) $, landing exactly 90 degrees from its starting position in the standard coordinate system of the plane.

the point moves 90 degrees

The vector $ \mathbf{v} $ representing the position of point $ P $ rotates by 90 degrees, transforming into the vector $ \mathbf{u} $, which now points towards the new location $ P' $.