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).

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