1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Can you represent $e^{At}$ in terms of $e^{Jt}$? That may help you. $\endgroup$ – Jonas Oct 1 at 15:51
2
$\begingroup$

With

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

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

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

since

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

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

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

$OE\Delta$.

$\endgroup$
  • 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.