Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

If I know the eigenvalues of $e^{A}$, what can I say about the eigenvalues of $A$ itself?

share|improve this question
    
If A can be diagonalized, then the diagonal = eigenvalues and then the diagonal of $e^A$ is eingenvalues mapped through $x\mapsto e^x$. However, I am not sure in what conditions that reasoning can be reversed. –  dtldarek Mar 10 '12 at 15:08
    
Notice that ${\rm exp}(T^{-1}AT) = T^{-1} {\rm exp}(A)T,$ so we can suppose that $A$ is upper triangular ( or even in Jordan normal form, but triangular is enough). Then Davide's statement is clear. –  Geoff Robinson Mar 10 '12 at 15:20
    
Do you assume complex scalars? –  GEdgar Mar 10 '12 at 19:41

2 Answers 2

up vote 8 down vote accepted

If $A$ is upper triangular, then it's easy: then $e^A$ is upper triangular too, and the diagonal elements of $e^A$ are the exponentials of the diagonal elements of $A$.

But this is always the case, because we can choose a basis in which $A$ is in Jordan canonical form (which is in particular upper triangular).

So the eigenvalues of $A$ are logarithms of the eigenvalues of $e^A$.

But beware that the eigenvalues of $A$ can be complex even if $A$ and $e^A$ are both real, and they are not necessarily the principal logarithms of $e^A$'s eigenvalues. For example, if $A=\pmatrix{0&2\pi\\-2\pi&0}$ then $e^A=I$, but the eigenvalues of $A$ are $\pm 2\pi i$.

share|improve this answer

The exponential of a Jordan block $$ \begin{bmatrix} \lambda&1&0&0&\dots\\ 0&\lambda&1&0&\dots\\ 0&0&\lambda&1&\dots\\ 0&0&0&\lambda&\dots\\ &&\vdots&&\ddots \end{bmatrix}\tag{1} $$ is $$ e^\lambda\begin{bmatrix} 1&\frac{1}{1!}&\frac{1}{2!}&\frac{1}{3!}&\dots\\ 0&1&\frac{1}{1!}&\frac{1}{2!}&\dots\\ 0&0&1&\frac{1}{1!}&\dots\\ 0&0&0&1&\dots\\ \vdots&\vdots&\vdots&\vdots&\ddots \end{bmatrix}\tag{2} $$ Since the Jordan Normal Form of a matrix is composed of Jordan blocks $(1)$ along the diagonal, and similarly placed square blocks along the diagonal do not interact when added or multiplied, the exponential of the Jordan Normal Form consists of exponential blocks $(2)$ along the diagonal.

If $A=PJP^{-1}$, then $e^A=Pe^JP^{-1}$. Thus, if the eigenvalues of $A$ are $\{\lambda_j\}$, the eigenvalues of $e^A$ are $\{e^{\lambda_j}\}$. However, just as with logarithms, if the eigenvalues of $e^A$ are known, the eigenvalues of $A$ are known up to an integral multiple of $2\pi i$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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