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Assume the formula $\det(e^A)=e^{\operatorname{tr}(A)}$ for all matrices $A \in \mathbb{C}_{n\times n}$.

Show why this implies that the exponential always yields a regular matrix.

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And what is a regular matrix? – copper.hat Mar 4 '13 at 19:28
Since $\exp(\mathbb{C})=\mathbb{C}^*$, we have $\mbox{det}(e^A)=e^{\mbox{Tr}A}\neq 0$. So $e^A$ is invertible. – 1015 Mar 4 '13 at 19:29
A regular nxn matrix. I think. – John Mar 4 '13 at 19:29
The word 'regular' is regularly overused. – copper.hat Mar 4 '13 at 19:33
Do you know that a square matrix is invertible (regular) if and only if its determinant is nonzero? – 1015 Mar 4 '13 at 19:34

Basically, you just combine the following two facts:

1) $e^t\ne 0$ for all $t\in \mathbb C$

2) An $n\times n$ matrix $A$ is regular (meaning invertible) if, and only if, $\det(A)\ne 0$.

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thank you ittay weiss – John Mar 4 '13 at 19:48

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