Consider SVD of $M$: $$ M = U \Sigma V^\top $$ And SVD of $N= \ln M$: $$ N = U^\prime \Sigma^\prime V^{^\prime\top} $$ Anyone knows/has seen/can think of any interesting connection/relation between $U$ and $U^\prime$? (Or $\Sigma$ and $\Sigma^\prime$? )
Update: Let's suppose that both decompositions exist.
Update2: By relation, I mean not the trivial relation of $\ln( U \Sigma V^\top) \approx U^\prime \Sigma^\prime V^{^\prime\top}$.
[1] http://en.wikipedia.org/wiki/Singular_value_decomposition
Assume that $\log$ denotes the principal logarithm and $M$ has no $\leq 0$ eigenvalues. Then $\log(M)$ is defined and the columns of $U,V$ are the eigenvectors of $MM^T,M^TM$ and the columns of $U',V'$ are the eigenvectors of $\log(M)\log(M)^T,\log(M^T)\log(M)$.
If $M$ is symmetric with eigenvalues $>1$, then $U=U',V=V',\Sigma'=\log(\Sigma)$. If $M=diag(1/2,3,5)$ then the condition $U=U',V=V'$ is not satisfied.
EDIT: Vedran is right. More precisely, let $f:U\subset M_n(\mathbb{R})\rightarrow M_n(\mathbb{R})$ be a matrix function (cf. the Higham's book) s.t. $f$ is defined on $spectrum(M)$. Then $f(M)$ is a polynomial in $M$. If $M$ is diagonalizable ($M=PDP^{-1}$) , then $f(M)=Pf(D)P^{-1}$. If $M$ is symmetric $\geq 0$, then $PDP^T$ ($P$ orthogonal) is a SVD of $M$ and $f(M)=Pf(D)P^T$ is a SVD of $f(M)$ only if $f(D)\geq 0$.