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Is SVD decomposition rotationally invariant?

If so can you provide the underlying intuition or a demonstration?

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The SVD applies to an arbitrary matrix $A$ that maps one real vector space to another, say $A\colon \mathbb R^n \to \mathbb R^m$. What do you mean by rotational invariance in this case? @dineshdileep's answer deals with rotating only the target space $\mathbb R^m$, but you could also rotate just the source space $\mathbb R^n$, or both. The proof doesn't change very much, in any case. – Rahul Oct 30 '12 at 19:24
up vote 2 down vote accepted

svd is rotationally invariant in the sense that it won't change the singular values (but it does change the column space). Consider a matrix $A$ with its svd $A=U\Sigma V^{H}$. Consider a rotation matrix $R$. By virtue of it being a rotation matrix, $R$ is unitary. Now consider applying the rotation matrix $R$ on $A$, so that $B=RA=RU\Sigma V^{H}=(RU)\Sigma V^{H}$ is the svd of $B$. This is because product of unitary matrices is again unitary. So $B$ has the same singular values as $A$. Note that $B$ and $A$ will have same row space, same rank, and if $A$ is hermitian, same eigenvalues also.

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Are you saying that matrix $V$ do not change? – Aslan986 Oct 30 '12 at 16:02
No it won't, since its a left multiplication – dineshdileep Oct 30 '12 at 16:05

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