Equality between matrix exponentials

Question

Let $A,B$ be $n$-th dimensional complex Hermitian matrices, and suppose that $e^{-iA} = e^{-iB}$. Does this necessarily imply that $A = B$?

If we expand the expression, then then $$ I + iA - \frac{A^2}{2} - i\frac{A^3}{3!}+\ldots = I + iB - \frac{B^2}{2} -i\frac{B^3}{3!}+\ldots $$ but how can this imply $A = B$?

score 1 · Accepted Answer · edited Apr 13 '17 at 12:21

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No, because $-iA$ and $-iB$ are skew-Hermitian, and the exponential map $\exp: \mathfrak{u}(n)\rightarrow U(n)$ is not injective in this case - see here.

edited Apr 13 '17 at 12:21

Community

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answered Mar 31 '16 at 13:56

Dietrich Burde

140,055

What are $u(n)$ and $U(n)$? – Joshhh Mar 31 '16 at 14:28
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$\mathfrak{u}(n)$ is the space of skew-Hermitian matrices of size $n$, and $U(n)$ the unitary group. – Dietrich Burde Mar 31 '16 at 14:29

Equality between matrix exponentials

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