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.
Do SVD and Jordan imply that any operator can be represented through diagonal/almost diagonal matrix?
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linear-algebra
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0@J.M.: Sorry that's what I meant. I corrected it. – 2011-11-13
1 Answers
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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.