1

Is there any matrix decomposition that is invariant to rotation in domain space? SVD is invariant to rotation in range space (I believe), but not in domain space.

In other words, I am looking for a matrix decomposition for which applying some operation on the components is equivalent to rotating the elements of the matrix.

PS: My matrix is actually an image kind of array. Thus, when I rotate the matrix elements, I interpolate the values to get the new matrix.

Nagabhushan S N
  • 796
  • I don't understand your question. What do you mean by that “SVD is invariant to rotation in range space”? E.g., for $Q\in SO(3)\setminus{I}$ and $A\in M_3(\mathbb R)$, the SVDs of $A$ and $QA$ are in general different. In particular, they are always different when $A$ is invertible. – user1551 Dec 17 '22 at 14:20
  • This answer says that rotation in range space is defined as $QAQ^T$. If SVD of $A=U \Sigma V^T$, then $Q A Q^T = Q U Q^T Q \Sigma Q^T Q V^T Q^T$ i.e. we can apply the rotation operation to each of the components of $A$. – Nagabhushan S N Dec 17 '22 at 20:03

0 Answers0