Undergrad Can I visualize O(3) \ SO(3) in some way?

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A 3×3 orthogonal matrix with a determinant of -1 can be visualized as a combination of a rotation and a reflection. The group SO(3) represents rotations in 3D, while O(3) encompasses both rotations and reflections. Any matrix in O(3) can be expressed as the product of a matrix in SO(3) and a reflection matrix. The structure of O(n) can be understood as a semidirect product of SO(n) and O(1), indicating that O(n) consists of two components: one with determinant 1 and another with determinant -1. This highlights the geometric distinction between the two types of transformations represented in O(3).
BerriesAndCream
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Hello.

I know that a 3×3 orthogonal matrix with determinant = 1 (so a 3×3 special orthogonal matrix) is a rotation in 3D.
I was wondering if there is a 3×3 orthogonal matrix with determinant = –1 could be visualised in some way.

Thank you!
 
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BerriesAndCream said:
Hello.

I know that a 3×3 orthogonal matrix with determinant = 1 (so a 3×3 special orthogonal matrix) is a rotation in 3D.
I was wondering if there is a 3×3 orthogonal matrix with determinant = –1 could be visualised in some way.

Thank you!
It can be visualized as a rotation of reflection.
 
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BerriesAndCream said:
Hello.

I know that a 3×3 orthogonal matrix with determinant = 1 (so a 3×3 special orthogonal matrix) is a rotation in 3D.
I was wondering if there is a 3×3 orthogonal matrix with determinant = –1 could be visualised in some way.

Thank you!
SO(3) is the group of rotations in 3D (no complex numbers). O(3) is the group of rotations and reflections, you can decompose any matrix O(3) as the product of a matrix in SO(3) and a reflection matrix.
 
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BerriesAndCream said:
Hello.

I know that a 3×3 orthogonal matrix with determinant = 1 (so a 3×3 special orthogonal matrix) is a rotation in 3D.
I was wondering if there is a 3×3 orthogonal matrix with determinant = –1 could be visualised in some way.

Thank you!
$$O(n) \cong SO(n) \rtimes O(1)$$
##O(1) \in \{1, -1\}## and ##\rtimes## is the semidirect product. So in some sense ##O(n)## can just be thought of like a direct product of ##SO(n)## and ##O(1)##. Geometrically as a manifold, Lie Group, ##O(n)## breaks into two components. ##SO(n)## is the connected component containing the Identity with ##det(A) = 1## and there is an isomorphic component that doesn't contain the identity with ##det(A) = -1##. So ##O(n)## can also be thought of as two copies of ##SO(n)##.
 
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