I'm trying to prove that the map $f:U \rightarrow V$, where $U$ consists of anti-symmetric matrices $a$ such that $(I_n+a)$ and $(I_n-a)$ are invertible, and V consists of matrices $b \in O(n)$ such that $(I_n + b)$ is invertible, and $f(a) = (I_n + a)(I_n - a)^{-1}$.

To show that the map is surjective, I let $b \in V$ and consider an $a \in U$ such that $a = (b - I_n)(b + I_n)^{-1}$. I can show that $(I+a)$ and $(I-a)$ are invertible, but if I'm not mistaken, I have to show that $a$ is antisymmetric, which I'm having trouble doing.

$U$ and $V$ have the same dimension, but $f$ is not a linear map, so I can't see a way to get out of showing surjectivity directly. Is there a matrix identity that would make it easier to show that $a$ is antisymmetric? Thanks.


There is the exponential map $$\exp\colon \mathfrak{so}(n)\rightarrow O(n)$$ from the Lie algebra consisting of skew-symmetric matrices to the orthogonal group. It is well-known that $\exp$ for this group is surjective, e.g., see here.

To answer your question, the map to consider is the the Cayley transform: Given any rotation matrix, $R \in SO(n)$, if $R$ does not admit $−1$ as an eigenvalue, then there is a unique skew-symmetric matrix $S$ so that $$ R = (S − I)(S + I)^{−1}. $$ This is a classical result of Cayley (1846) and $R$ is called the Cayley transform of $S$. The inverse is given by $$ S = (I+R)(I-R)^{-1}. $$

  • $\begingroup$ Thank you Dietrich. Alternatively, if you're like me and haven't studied Lie Algebras yet, you can just easily show that $a + a^T = 0$ as defined above. $\endgroup$ – user1447447 Sep 21 '16 at 12:38
  • $\begingroup$ You are right, Cayley's result does not need Lie algebras. $\endgroup$ – Dietrich Burde Sep 21 '16 at 12:41
  • $\begingroup$ Does $ R = (S - I)(I + S)^{-1}$ ? $\endgroup$ – user1447447 Sep 21 '16 at 12:41
  • $\begingroup$ Hmm, maybe I'm making an embarrassing mistake: I have $S = (I_n + a)(I_n - a)^{-1}$ so $S(I_n - a) = (I_n + a)$, $S - Sa = I_n + a$, $S - I_n = a + Sa$, $S - I_n = a(I_n + S)$, $a = (S - I_n)(S + I_n)^{-1}$ $\endgroup$ – user1447447 Sep 21 '16 at 12:47
  • $\begingroup$ You are right. I mixed up the two versions with the sign. $\endgroup$ – Dietrich Burde Sep 21 '16 at 12:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.