## The exponential of a matrix is invertible

Show that over $\mathbb{C}$, $\mathsf{exp}(0) = I$. Deduce that $\mathsf{exp}(A)$ is nonsingular for every matrix $A$ over $\mathbb{C}$.

We have $\mathsf{exp}(0) = \mathsf{exp}( \bigoplus_n [0])$ $= \bigoplus_n \mathsf{exp}([0])$ $= I$ using this previous exercise.

Now using this previous exercise, since $A$ and $-A$ commute, we have $\mathsf{exp}(A) \mathsf{exp}(-A) = \mathsf{exp}(A-A)$ $= \mathsf{exp}(0)$ $= I$. In particular, $\mathsf{exp}(A)$ is invertible with inverse $\mathsf{exp}(-A)$.