Why can any orthogonal matrix be written as $O=e^A$?

Question

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?

Answer 1

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.

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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.