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Given a linear system $\frac{dx}{dt}=Mx$, what's the relationship between the dynamic's property and the singular value decomposition/rank of $M$ ?

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I am afraid this is not a good setting to use the SVD. The point is that here the only natural transformation to make is changing basis for $x$, i.e., $x\to Ux$, and this leads to a transformation of the associated matrix of the form $M \mapsto UMU^{-1}$. So the basis-independent characteristics of the system will depend on the canonical form of $M$ under that kind of transformations, i.e., the Jordan form. Even if you want to restrict to orthogonal changes of basis, you'd get the canonical form under the action of $O(n)$ as $M\to UMU^T$, which is the Schur form.

The SVD, instead, is the canonical form under the action of $O(n)\times O(n)$ as $M\mapsto UMV^T$, which would correspond in your problem to choosing different bases for $x$ and for $\dot{x}$, and I am afraid that it makes little sense to do it in this setting.

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    $\begingroup$ Thanks! I get your point. But I think that if $M$ has a bigger $\sigma_1$, we can get a estimation of the increasing/declining about $x$ ,i.e., $||\dot{x}||_2 \le \sigma_1 ||x||_2$. For the rank of $M$, the lower rank brings the more eigenvalue $ \lambda =0 $ corresponding the constant component of $x$. So the singular value and rank of $M$ indeed detainment some property of the dynamic. $\endgroup$
    – Bo Yang
    Commented Jan 11, 2014 at 15:36

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