Given a matrix $A$, we can write the Jordan decomposition as $$A=SJS^{-1}$$ My question is whether the followings now holds: $$\text{eig}(e^{At})=\text{eig}(e^{Jt})$$ I've tried relating the determinant to the eigenvalues, but I'm not able to determine whether all of the eigenvalues are equivalent. How could I go about proving or disproving this?

  • $\begingroup$ Can you represent $e^{At}$ in terms of $e^{Jt}$? That may help you. $\endgroup$ – Jonas Oct 1 at 15:51


$A = SJS^{-1}, \tag 1$

we have

$Se^{At}S^{-1} = S \left (\displaystyle \sum_0^\infty \dfrac{(At)^n}{n!} \right ) S^{-1}$ $= \displaystyle \sum_0^\infty \dfrac{(SAS^{-1}t)^n}{n!} = \sum_0^\infty \dfrac{(Jt)^n}{n!} = e^{Jt}; \tag 2$

that is, $e^{At}$ is similar to $e^{Jt}$; thus their characteristic polynomials are the same, since

$\det ( e^{Jt} - \lambda I) = \det (Se^{At}S^{-1} - \lambda SS^{-1}) = \det (S(e^{At} - \lambda I)S^{-1})$ $= \det (S) \det(e^{At} - \lambda I) \det (S^{-1}) = \det(e^{At} - \lambda I), \tag 3$


$\det(S) \det(S^{-1}) = \det (SS^{-1}) = \det(I) = 1; \tag 4$

since the characteristic polynomials are the same, the eigenvalues, being the roots of said polynomials, are also the same, that is

$\text{eig}(e^{At}) = \text{eig}(e^{Jt}), \tag 5$


  • 1
    $\begingroup$ I don't understand how you go from (1) to (2). $\endgroup$ – Rodrigo de Azevedo Oct 2 at 6:41
  • $\begingroup$ Use the fact that $S(At)^nS^{-1} = (S(At)S^{-1})^n$ applied termwise. Does this help? $\endgroup$ – Robert Lewis Oct 2 at 6:45
  • $\begingroup$ Should't $S$ be on the right and $S^{-1}$ on the left? $\endgroup$ – Rodrigo de Azevedo Oct 2 at 6:47
  • $\begingroup$ @RodrigodeAzevedo: I don't think so; it follows the pattern of (1). $\endgroup$ – Robert Lewis Oct 2 at 6:55
  • $\begingroup$ @RodrigodeAzevedo: Oh wait I may have got 'em switched up . . . let e see . . . $\endgroup$ – Robert Lewis Oct 2 at 6:56

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