Suppose I apply SVD (singular value decomposition) on some real-valued matrix $M$, that is, $M = USV^T$. Now, if $S$ is an identity matrix, what does it mean? Does $M$ have some special properties? Could SVD be then calculated in some peculiar (faster) way?
Linear Algebra – Implications of All Singular Values Being Ones
linear algebramatricesnumerical linear algebrasvd
Best Answer
Recall that the singular values of a real matrix $M$ are precisely the eigenvalues of the positive-semidefinite real matrix $M^T M$. Then the singular values of $M$ are all $1$ if and only if the eigenvalues of $M^T M$ are all $1$, if and only if $M^T M = 1$, if and only if $M$ is orthogonal.
EDIT: As Vedran Šego points out, there is, of course, a more straightforward way:
On the one hand, if the singular values of $M$ are all $1$, then any singular value decomposition $M = USV^T$ of $M$ will have $S = 1$, so that $M = UV^T$ is a product of orthogonal matrices, and thus orthogonal. On the other hand, if $M$ is orthogonal, then $M^T M =1$, so that the singular values of $M$ are all $1$.
EDIT 2: And as Memming points out, if $M$ is orthogonal, then $M = M 1 1^T$ is trivially a singular value decomposition of $M$.