I read in the book Quantum Field Theory in a Nutshell that any orthogonal real matrix $O$, can be written as $O=e^A$.

I should clarify that $O \in \textrm{SO}(N)$, so $O^TO=1$ and $\det (O)=1$

I wonder why this is true and if there's a simple proof?

  • $\begingroup$ Relevant wikipedia article. $\endgroup$ – Arthur Oct 11 '17 at 13:04
  • $\begingroup$ It was relevant but I don't think it was any proof. $\endgroup$ – Turbotanten Oct 11 '17 at 13:07
  • $\begingroup$ No, but now you know that what you're loking for is called "matrix logarithm". It could give you something to search for if nothing else. $\endgroup$ – Arthur Oct 11 '17 at 13:09

Edit Since this answer was first written, OP has amended the question to include the conditions that $O \in \operatorname{SO}(n, \Bbb R)$ (that is, $O$ is not only orthogonal but real and special orthgonal). The answer as written does not assume that condition holds but does treat it as a special case.

If the matrices are assumed real, this is not true. For any real $n \times n$ matrix $A$, $$\det e^A = e^{\operatorname{tr} A} > 0 ,$$ whereas there are orthogonal matrices with negative determinant, e.g., $I_{n - 1} \oplus \pmatrix{-1}$.

On the other hand, the statement is true if we replace "orthogonal" with "special orthogonal", that is, if we also ask that our orthogonal matrix $O$ satisfy $\det O > 0$ (and hence $\det O = 1$). One can check that the group $\operatorname{SO}(n, \Bbb R)$ of special orthogonal matrices is connected and compact, so any special orthogonal matrix is the exponential of some matrix $A$. In fact, any such $A$ is skew-symmetric, and for any skew-symmetric $A$ the matrix $e^A$ is (special) orthogonal.

On the other hand, if the matrices are assumed complex, even more is true: For any invertible complex matrix $B$ (and hence for any comple orthogonal matrix) there is a complex matrix $A$ such that $B = e^A$.

Again, if $A$ is a skew-symmetric complex matrix then $\exp A$ is orthogonal, but not all complex orthgonal matrices arise this way: $\exp$ is continuous, and so the image of the (connected) space of complex skew-symmetric matrices under that map is connected in the standard topology, but the Lie group of complex orthogonal matrices has two connected components. On the other hand, for any complex special orthogonal matrix $O$ there is a complex skew-symmetric matrix $A$ such that $O = e^A$, and for any complex skew-symmetric matrix $A$ the matrix $e^A$ is special orthogonal.

  • $\begingroup$ If the matrix has negative determinant, you nevertheless have a representation $O = e^A$ with a complex matrix $A$ such that $A^* = -A$. $\endgroup$ – amsmath Oct 11 '17 at 13:18
  • $\begingroup$ I seem to remember a lot of fuss being made about whether or not there is an $A$ which generates a semigroup of linear operators $\{U_a\}_a$ via $U_a = e^{aA}$. I suspect this is the correct place to look for claims of exponential representation. Nonetheless I would have (naively) assumed such a representation existed for any orthogonal group. Does it fail in this case because $A\mapsto e^A$ defines a connected group and $O(n)$ is not connected? $\endgroup$ – SZN Oct 11 '17 at 13:19
  • $\begingroup$ The context of this claim is that it was found in a QFT book. The matrices are almost certainly complex . $\endgroup$ – Spencer Oct 11 '17 at 13:33
  • $\begingroup$ @SZN That's essentially correct, but NB the image of the space of complex matrices under $\exp$ is the full group $\operatorname{GL}(n, \Bbb C)$, so the story is different in the complex case. $\endgroup$ – Travis Oct 11 '17 at 14:06
  • $\begingroup$ @Spencer Yes, you're probably right. I've expanded the answer to explain the complex case, too. $\endgroup$ – Travis Oct 11 '17 at 14:07

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