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I would like to understand under which assumptions the following statement holds \begin{equation} (\lambda,v) \text{ is an eigenpair of the matrix } A \Leftrightarrow (\exp(\lambda\Delta),v) \text{ is an eigenpair of the matrix } \exp(A\Delta) \end{equation} where $\Delta$ is a positive real constant and $A$ is a $\mathbb{R}^{n\times n}$-valued matrix.

$\Rightarrow$ can be proved by induction, see for example Eigenvector and eigenvalue for exponential matrix. I am wondering under which assumption $\Leftarrow$ holds. Indeed, $\log(\exp(A\Delta))$ is not uniquely defined as the matrix $A$.

Example where I got stuck in the proof. To prove $\Leftarrow$, I assume $(\exp(\Delta\lambda),v)$ eigenpair of $\exp(A\Delta)$, therefore $\exp(A\Delta)v=\exp(\Delta\lambda)v$. Considering the definition of the exponential matrix and the taylor expansion of $\exp(\lambda\Delta)$ we get \begin{equation} \sum_{k=0}^\infty \frac{\Delta^kA^k}{k!}v=\sum_{k=0}^\infty \frac{\Delta^k\lambda^k}{k!}v \end{equation}
which is equivalent to \begin{equation} \sum_{k=0}^\infty \frac{\Delta^k\left(A^k-\lambda^kI_n\right)v}{k!}=0. \end{equation} If I assume by contradiction that $\lambda$ is not an eigenvalue of $A$, how can I prove that the previous series is different from $0$?

Luca Fenzi
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1 Answers1

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Suppose wlog that $A$ is triangular with complex values, up to a conjugation. Then if its diagonal (which is the list of complex eigenvalues with multiplicity) is $\lambda_1, \ldots, \lambda_n$, the diagonal of $\text{exp}(A)$ will be $\text{exp}(\lambda_1), \ldots, \text{exp}(\lambda_1)$ as a simple calculation yields. The eigenvectors are the standard vectors of the basis in both cases.

On balance, if you are interested to complex eigenvalues your question is true: there is a correspondence between the eigenvalues of $A$ and $\text{exp}(A)$ given by $\lambda \mapsto e^{\lambda}$. On the other hand, if you only look at real eigenvalues, something could disappear. Consider the matrix $(0 \ \ \pi , \pi \ \ 0)$: the eigenvalues are $\pm i\pi$, so nothing real. But the exponential diagonalize as $(-1 \ 0 , 0 \ -1)$

Maybe I misunderstood what you were talking about?

Andrea Marino
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  • I think you misunderstood. I am interested in the other side of the proof, i.e. $\Leftarrow$. More precisely, if I know that $(\mu, v)$ is an eigenpair of $B$ how can I prove that the $(\lambda, v)$ is an eigenvalue of the matrix $A$, where $A$ and $B$ are real matrices, $\log(B)=A$ and $\log(\mu)=\lambda$. The eigenvalues of $A$ and $B$ are allowed to be complex, hence we apply the real-logarithm only to the matrices, while we apply the complex logarithm to the eigenvalues.

    Note that if we do not impose the reality condition on the matrix $A$ then this matrix is not uniquely defined.

     – Luca Fenzi Oct 15 '19 at 15:41
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    Still, I don't see the point, sorry! This is also the other direction: there is a bijection between the two sets, so if $\mu$ is an eigenvalue of the exponential, then $\mu=e^{\lambda}$ for some $\lambda$ eigenvalue of $A$. – Andrea Marino Oct 15 '19 at 15:53