Eigenvalues of exponential matrix

Question

Let's consider a matrix $X \in M\left(2, \mathbb{R}\right)$. It's a well-known result that if $\lambda$ is an eigenvalue of $X$, then $\exp\left(\lambda\right)$ is an eigenvalue of $\exp\left(X \right)$.

But does it hold the other way round? More precisely, my question is as follows: is we know eigenvalues of $\exp\left(X\right)$, then what can we say about eigenvalues of $X$? Does existence of eigenvalues of $\exp\left(X\right)$ even guarantee existence of (real) eigenvalues of $X$?

Answer 1

What's true is that if $\mu$ is an eigenvalue of $\exp(X)$, then there is some eigenvalue $\lambda$ of $X$ such that $\mu = \exp(\lambda)$; equivalently, every eigenvalue of $\exp(X)$ is of the form $\exp(\lambda)$ for some $\lambda$ which is an eigenvalue of $X$, or equivalently again, if $\mu$ is an eigenvalue of $\exp(X)$ then $\log(\mu)$ is an eigenvalue of $X$ for one of the possible values of $\log(\mu)$.

You don't get any control over which of the possible values of $\log(\mu)$ occur, except that if $X$ is real then its non-real eigenvalues must come in complex conjugate pairs. For example, suppose

$$X = \left[ \begin{array}{cc} 0 & 2 \pi \\ - 2 \pi & 0 \end{array} \right]$$

which has eigenvalues $\pm 2 \pi i$. Then $\exp(X)$ has eigenvalues $\exp(\pm 2 \pi i) = 1$, and in fact I invite you to compute that $\exp(X) = 1$ is just the identity matrix. So $\exp(X)$ having real eigenvalues does not imply that $X$ has real eigenvalues.

Everything I've said generalizes to $n \times n$ matrices (at least).

Answer 2

As Qiaochu mentioned, the fact that the exponential maps certain non-real complex numbers to real numbers precludes guaranteeing that $X$ will have real eigenvalues.

But, other than that, a very strong relation holds:

The Jordan form of $e^X$ consists precisely of the same Jordan blocks of $X$, with each eigenvalue $\lambda$ replaced with $e^\lambda$. So, not only does each eigenvalue of $e^X$ is the exponential of an eigenvalue of $X$, but both the algebraic and geometric multiplicities are preserved.

And this is not very hard to see. First, since $e^{SXS^{-1}}=Se^XS^{-1}$, we may assume that $X$ is already in its Jordan form. It is then trivial to see that the exponential $X$ will be the block-diagonal matrix with each Jordan block $J$ replaced with $e^J$. So all we need to do is to analyze what the Jordan form of $e^J$ is.

Suppose that $J$ has size $m$, and that $\lambda$ is its eigenvalue. Then $$\tag1e^J=e^{\lambda}e^{-\lambda}e^J=e^{\lambda}\,e^{J-\lambda I}.$$ As $J-\lambda I=J_m$ is nilpotent of order $m$, we have $$\tag2 e^{J-\lambda I}=\sum_{k=0}^{m-1}\frac1{k!}\,(J-\lambda I)^k=\begin{bmatrix} 1&1/2&1/6&\cdots&1/(m-1)! \\ 0&1&1/2&\cdots&1/(m-2)!\\ 0&0&1&\cdots&\vdots\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ 0&0&0&\cdots&1 \end{bmatrix} $$ It then easy to see that $e^{J-\lambda I}-I$ is nilpotent of order $m$ (since the $m^{\rm th}$ power is zero, and all smaller powers are nonzero). It follows that the Jordan form of $e^{J-\lambda I}-I$ is $J_m$, the Jordan block of order $m$. So the Jordan form of $e^{J-\lambda I}$ is $I+J_m$, the Jordan block of size $m$ and eigenvalue $1$.

Finally, using $(1)$, we get that the Jordan form $e^J$ is $e^\lambda I+J_m$.