# Eigenvalues of a matrix $A$ and $e^{A}$

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

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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

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 $\pm 2\pi i$.

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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 $2\pi i$.

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