Rotation matrix as product of reflection matrices

Asked May 26 '23 at 08:04

Active May 26 '23 at 09:34

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I would like to prove that any rotation matrix can be represented as finite product of reflection matrices.
Almost everything I found earlier is that product of 2 reflections is a rotation, but this is not about decomposition of rotation matrix. I tried to prove this using induction, but didn't finish this attempt.
Can anybody help me, please?

EDIT: Proof check. Looks like we can prove that any unitary matrix can be decomposed into product of reflection matrices. If $U$ - unitary, then we can choose reflection matrix $H$, such that it's first column is column of $U$, and product $HU$ has first column of identity. After this we remove 1 column and row, and do induction. We get finite number of reflection matrices multiplied by 1x1 matrix. Determinant of reflection matrix is 1. Removing first column and row doesn't affects the determinant of matrix. This is why the determinant of 1x1 matrix is $\pm det(U)$, thus $\pm 1$. And rotation matrix is unitary. Is this fully correct?

edited May 26 '23 at 09:34

asked May 26 '23 at 08:04

nezudem

1

in dimension 2 or dimension 3 ? or dimension n? – GreginGre May 26 '23 at 08:15
@GreginGre In n – nezudem May 26 '23 at 08:17
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As every special orthogonal matrix is orthogonally similar to a direct sum of 2D rotation matrices and an identity matrix, it suffices to consider the 2D case. – user1551 May 26 '23 at 08:34
Added proof, can anybody check it please? – nezudem May 26 '23 at 09:34
Complicated, but yes. Thank you. – nezudem May 26 '23 at 11:03

Rotation matrix as product of reflection matrices

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