Each eigenvalue of $e^X$ is the exponential of an eigenvalue of $X$

Question

Let $X \in M(n)$, I was able to show that if $v$ is an eigenvector of $X$ with corresponding eigenvalue $\lambda$, that $v$ is also an eigenvector of $e^X$ with eigenvalue $e^{\lambda}$, but I couldn't show that each eigenvalue of $e^X$ is the exponential of an eigenvalue of $X$.

Answer 1

• Show the result for $X$ diagonal (hint : we know what $e^X$ is when $X$ is diagonal, and we know the eigenvalues of a diagonal matrix).

• Show the result for $X$ diagonalizable (hint : if $X = P^{-1}AP$ where $A$ is diagonal, then $e^X = P^{-1}e^AP$, so again there's a pattern between the eigenvalues of $X$ and of $e^X$)

• Show that diagonalizable matrices are dense in the set of all matrices i.e. given any matrix $A$ we can find diagonalizable $A_n$ such that $A_n \to A$ entrywise (hint : every matrix with distinct eigenvalues is diagonalizable. Take the JCF of $A$, and modify it very little to make the eigenvalues distinct, the result is diagonalizable but not too far away...)

• For the $A_n$ you constructed above, show that for every (complex) eigenvalue $\lambda$ of $A$ there are (complex) eigenvalues $\lambda_n$ of $A_n$ such that $\lambda_n \to \lambda$.

• Show that $e^{A_n} \to e^A$ pointwise (this should be obvious from the fact that taking powers preserves convergence), and here too every eigenvalue of $e^A$ is the limit of some sequence of eigenvalues of each $e^{A_n}$.

• Conclude.

Answer 2

This isn't true when the underlying field is real. E.g. $X=\pmatrix{0&-2\pi\\ 2\pi&0}$ hasn't any eigenvalue over $\mathbb R$ but $e^X=I$ has two.

The statement is true over $\mathbb C$. This can be seen by triangularising $X$.