Consider the differential equation $$ \vec x'(t) = A(t) \vec x(t) , $$ where $A(t)$ is periodic: $A(t+T) = A(t)$. Let $\Phi(t)$ be its fundamental solution, i.e. $\Phi(0) = I$ and $\Phi'(t) = A(t) \cdot \Phi(t)$.

The Floquet Theorem tells us that $$ \Phi(t) = Q(t) \cdot \mathrm e^{B t} , $$ where $Q(t+T) = Q(t)$ and $B$ is the time-independent Floquet generator. (To prove the theorem, choose $B$ such that $\mathrm e^{BT} = \Phi(T)$ and define $Q(t) = \Phi(t) \cdot \mathrm e^{-B t}$.)

My question is: What is known about the properties of $B$ depending on the properties of $A$?
For example,

  • If every $A(t)$ is skew-adjoint, I know that $\Phi(T)$ is unitary and I can choose $B$ to be skew-adjoint as well.
  • If every $A(t)$ is a right (doubly) stochastic matrix, can $B$ always be chosen to be a right (doubly) stochastic matrix?
  • If the spectrum of $A(t)$ is in the left-half plane $\{ z \in \mathbb C: \Re(z) < 0 \}$ for every $t$, does the same hold for $B$? Or similarly, if every $A(t)$ is the generator of a contractive semigroup, is $B$ as well?

Are there any general results or techniques?

  • 1
    $\begingroup$ For $\Phi(T)$ to be unitary, you want $A(t)$ to be skew-hermitian, i.e. $A(t)^* = - A(t)$, not self-adjoint. $\endgroup$ Jul 19, 2017 at 23:28
  • $\begingroup$ Oops, thanks, edited. $\endgroup$
    – Noiralef
    Jul 20, 2017 at 6:25

1 Answer 1


When you choose $A(t)$ to belong to some Lie-algebra ${\mathfrak g}$, then the matrix differential equation $$ M'=A(t) M, \; M(0)={\rm Id} $$ is pretty close to choosing a connection (although it may be time dependent) in ${\rm GL}_n({\Bbb R})$ (or ${\rm GL}_n({\Bbb C})$ if $A$ is complex valued) and look at the collection of paths that you may construct in ${\rm GL_n}$ starting at the identity. In other words you get the path-connected component in the corresponding Lie-group that contains the identity.


If ${\mathfrak g}$ is the set of skew-symmetric matrics (i.e. ${\mathfrak so}(n)$) then you get ${\rm SO}(n)$, i.e. orthogonal matrices with determinant $+1$. You do not get $O(n)$ because it has two connected components, distinguished by the determinant.

If ${\mathfrak g}$ is the skew-adjoint (complex) matrices you get all of $U(n)$ since it is path connected. For given $B$ you may also construct a corresponding path $A: [0,1] \rightarrow {\mathfrak g}$ explicitly.

Your other examples are not related to a Lie-algebra and I don't know if there are general techniques to treat those. In the specific cases mentioned the answer is no in both cases. For $A$ a stochastic matrix (in particular having positive entries), $B$ is positive and has diagonal entries at least one and can never be a probability.

In the case of $A$ having negative spectrum, the question of $B$ always having the same property doesn't really make sense (and the answer is no), but you may ask if the corresponding $B$ has a spectrum inside the unit disk as is the case when you look at $\exp(t A)$ for $t>0$. This is not the case in general (I just tried it out in matlab). You do have that for any $B$ with spectrum in the open unit disk, the matrix $A=\log B$ exists and has spectrum in the open negative half-plane, and $B=\exp(A)$ corresponding to a 'path' with constant $A$. However, this property is not stable under multiplication so need not hold for a non-constant $A$.


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