Matrix Exponential does not map open balls to open balls?

Solution 1:

It is true that matrix exponential is not an open map. For example, look at $\exp$ in a neighborhood of $\left( \begin{smallmatrix} 2 \pi i & 0 \\ 0 & - 2 \pi i \end{smallmatrix} \right)$. All matrices near $\left( \begin{smallmatrix} 2 \pi i & 0 \\ 0 & -2 \pi i \end{smallmatrix} \right)$ are diagonalizable (over $\mathbb{C}$) so their exponentials are diagonalizable. However, $\exp \left( \begin{smallmatrix} 2 \pi i & 0 \\ 0 & -2 \pi i \end{smallmatrix} \right) = \left( \begin{smallmatrix} 1 & 0 \\ 0 & 1 \end{smallmatrix} \right)$ and an arbitrarily small neighborhood of $\left( \begin{smallmatrix} 1 & 0 \\ 0 & 1 \end{smallmatrix} \right)$ contains matrices of the form $\left( \begin{smallmatrix} 1 & \epsilon \\ 0 & 1 \end{smallmatrix} \right)$ which are not diagonalizable.

Note that the matrix $\left( \begin{smallmatrix} 1 & \epsilon \\ 0 & 1 \end{smallmatrix} \right)$ does have logarithms, for example $\left( \begin{smallmatrix} 0 & \epsilon \\ 0 & 0 \end{smallmatrix} \right)$. But these logarithms have Jordan blocks of size $2$, and thus can't get near $\left( \begin{smallmatrix} 2 \pi i & 0 \\ 0 & -2 \pi i \end{smallmatrix} \right)$.

Similar examples can be made with $\left( \begin{smallmatrix} 0 & 2\pi \\ - 2\pi & 0 \end{smallmatrix} \right)$, if you'd like real entries; and with $\exp \left( \begin{smallmatrix}0 & \pi \\ - \pi & 0 \end{smallmatrix} \right)$ if you'd lke smaller entries.

However, I don't see the connection between this and Hall's statement, and I don't know what the best possible constant in Hall's result is. I would consider the possibility that Hall simply wasn't interested in finding the best constant.