# Do SVD and Jordan imply that any operator can be represented through diagonal/almost diagonal matrix?

Good morning. Assume that we are given a complex vector space. I know that SVD guarantees a factorization of form $M = USV^{*},$ and the Jordan form gives us a factorization of form $M = PJP^{-1}.$ My question is the following: does this mean that for any given operator $M$ on a vector space we can have a matrix from one basis to another basis that is diagonal (via the SVD), and 'almost diagonal'/jordan via Jordan decomposition? My text material isn't quite clear on this, it seems to suggest that via SVD we can indeed get a diagonal matrix for any operator from some specific basis to other (which follow from the SVD decomposition), with the singular values on the diagonal, and a similar thing for Jordan form -- representing any operator as an almost diagonal matrix if we pick the appropriate bases. Thanks for your clarifications.

-
"...Jordan form gives us a factorization of form $M=PJP^\ast$." - er, no. That particular Jordan form is possible if $M$ were Hermitian. If not, $P$ is certainly not unitary... and thus the correct Jordan decomposition goes like $M=PJP^{-1}$. – J. M. Nov 12 '11 at 23:03
@J.M.: Sorry that's what I meant. I corrected it. – Matvey Nov 13 '11 at 0:08

Let $A = U \Sigma V^*$ be the SVD decomposition of a square matrix $A$. Then $\Sigma$ is not a representation of the linear transformation represented by $A$. Such representing matrices must be similar to $A$, but $\Sigma$ need not be similar. On the other hand, the Jordan decomposition implies that every linear transformation from a vector space to itself, i.e. every endomorphism of a vector space, can be represented by a unique block diagonal matrix, and block diagonal is the simplest form that we can have in general.